DC Circuits and Measurements

8 V

Reason — The five cells are arranged in two rows connected between the same two points A and B.

In the upper row the three cells of emfs 1 V, 2 V and 5 V are joined in series, so the net emf of this row is

1 V + 2 V + 5 V = 8 V

In the lower row the two cells of emf 4 V each are joined in series, so the net emf of this row is

4 V + 4 V = 8 V

The two rows are connected in parallel between A and B. When cells of equal emf are joined in parallel, the emf of the combination is equal to the emf of one row. Hence the emf of the battery AB is 8 V.

0.06 Ω

Reason — For a cell of emf E and internal resistance r sending a current I through an external resistance R,

$\text E = \text I (\text R + \text r)$

Here E = 1.5 V, I = 15 A and R = 0.04 Ω. Substituting,

$$\begin{aligned}1.5 = 15 (0.04 + \text r) \\[1em] 0.04 + \text r = \frac{1.5}{15} \\[1em] 0.04 + \text r = 0.1 \\[1em] \text r = 0.1 - 0.04 \\[1em] \text r = 0.06\ \Omega\end{aligned}$$

1.20

Reason — Inside a cell which is being discharged, the current flows from the negative to the positive electrode. Hence the cell is delivering current and its terminal potential difference is

$\text V = \text E - \text{Ir}$

Substituting E = 1.50 V, I = 2.0 A and r = 0.15 Ω,

$$\begin{aligned}\text V = 1.50 - (2.0 \times 0.15) \\[1em] \text V = 1.50 - 0.30 \\[1em] \text V = 1.20\ \text V\end{aligned}$$

Reason — For a cell of emf E and internal resistance r delivering a current I, the terminal potential difference is

$\text V = \text E - \text{Ir}$

Comparing with the straight-line form y = mx + c, the graph of V against I is a straight line of negative slope (− r) with intercept E on the V-axis. Thus V is maximum and equal to E when I = 0, and falls uniformly as the current drawn increases. Only graph (b) shows a straight line starting at a finite value on the V-axis and decreasing with I.

E/2

Reason — The external resistance and the internal resistance are both equal to r, so the current in the circuit is

$$\begin{aligned}\text I = \frac{\text E}{\text R + \text r} \\[1em] \text I = \frac{\text E}{\text r + \text r} \\[1em] \text I = \frac{\text E}{2\text r}\end{aligned}$$

The terminal potential difference is the potential drop across the external resistance,

$$\begin{aligned}\text V = \text{IR} \\[1em] \text V = \frac{\text E}{2\text r} \times \text r \\[1em] \text V = \frac{\text E}{2}\end{aligned}$$

Potential drop is same in both.

Reason — In a parallel combination both resistances are connected between the same two points of the circuit, so the potential difference across each of them is the same and equal to the terminal potential difference of the cell.

zero

Reason — The four 50 Ω resistors form the four arms of a Wheatstone bridge across the 10 V cell, and the ammeter A is connected in the arm PQ joining the two junctions of the bridge.

For the bridge the ratio of the resistances of the two arms on one side is

$\frac{50}{50} = 1$

and that of the two arms on the other side is also

$\frac{50}{50} = 1$

The condition of balance of the Wheatstone bridge is therefore satisfied. At balance the two junctions P and Q are at the same potential, so no current flows in the arm joining them. Hence the ammeter reads zero.

1 A

Reason — When one cell is reversed, its emf opposes the emfs of the remaining three cells. Hence the net emf of the battery is

$$\begin{aligned}\text E = (3 \times 1.5) - (1 \times 1.5) \\[1em] \text E = 4.5 - 1.5 \\[1em] \text E = 3\ \text V\end{aligned}$$

The internal resistances add up in series irrespective of the direction in which a cell is connected, so the total internal resistance is

$$\begin{aligned}\text r = 4 \times 0.25 \\[1em] \text r = 1\ \Omega\end{aligned}$$

The current sent in the external resistance R = 2 Ω is

$$\begin{aligned}\text I = \frac{\text E}{\text R + \text r} \\[1em] \text I = \frac{3}{2 + 1} \\[1em] \text I = 1\ \text A\end{aligned}$$

4 Ω

Reason — The five resistors form a Wheatstone bridge in which 2 Ω and 3 Ω are the arms of the upper branch, 8 Ω and 12 Ω are the arms of the lower branch, and 6 Ω is the arm joining the two junctions.

Testing the condition of balance,

$\frac{2}{3} = \frac{8}{12} = \frac{2}{3}$

The bridge is balanced, so no current flows through the 6 Ω resistor and it may be omitted.

The upper branch then gives 2 + 3 = 5 Ω and the lower branch gives 8 + 12 = 20 Ω, and these two are in parallel between A and B.

$$\begin{aligned}\frac{1}{\text R} = \frac{1}{5} + \frac{1}{20} \\[1em] \frac{1}{\text R} = \frac{4 + 1}{20} \\[1em] \frac{1}{\text R} = \frac{5}{20} \\[1em] \frac{1}{\text R} = \frac{1}{4} \\[1em] \text R = 4\ \Omega\end{aligned}$$

current in potentiometer wire remains constant

Reason — If the potentiometer wire of length L has resistance R and carries a current I from the driving cell, the potential difference across the whole wire is IR and the potential gradient is

$\text K = \frac{\text{IR}}{\text L}$

For a given wire R and L are fixed, so K depends only on the current I in the wire. Hence the potential gradient stays constant so long as the current in the potentiometer wire remains constant.

result is not affected due to wrong calibration of the meter (galvanometer) or due to imperfection in taking readings

Reason — In a null method the galvanometer is used only as a null detector, that is, merely to indicate whether the current through it is zero or not. The measured quantity is obtained from the balancing condition and not from the magnitude of the deflection.

Hence the wrong calibration of the galvanometer, or an error in reading its scale, cannot affect the result. In a deflection method, on the other hand, the value of the quantity is read from the size of the deflection, so it depends directly on the calibration of the instrument.

+9V

Reason — A current of 2 A flows from A to B. Going from A to B, the potential falls across each resistor in the direction of the current and also falls on passing through the cell from its positive to its negative terminal.

Drop across the 2 Ω resistor = 2 × 2 = 4 V

Drop across the cell = 3 V

Drop across the 1 Ω resistor = 2 × 1 = 2 V

Hence,

$$\begin{aligned}\text V_\text A - \text V_\text B = 4 + 3 + 2 \\[1em] \text V_\text A - \text V_\text B = 9\ \text V\end{aligned}$$

zero

Reason — When the cell is short-circuited the external resistance R becomes zero, so the maximum current is drawn from the cell,

$\text I_\text{max} = \frac{\text E}{\text r}$

The terminal potential difference is then

$$\begin{aligned}\text V = \text E - \text I_\text{max}\ \text r \\[1em] \text V = \text E - \frac{\text E}{\text r} \times \text r \\[1em] \text V = \text E - \text E \\[1em] \text V = 0\end{aligned}$$

Thus the whole of the emf is used up in driving the current through the internal resistance of the cell itself.

The equivalent emf E_eq of the two cells is between E₁ and E₂, i.e., E₁ < E_eq < E₂.

Reason — For two cells joined in parallel, the emf of the equivalent cell is

$\text E_\text{eq} = \frac{\text E_1 \text r_2 + \text E_2 \text r_1}{\text r_1 + \text r_2}$

This expression is a weighted mean of E₁ and E₂, E₁ being weighted by r₂ and E₂ by r₁. A weighted mean of two quantities always lies between them, so E₁ < E_eq < E₂.

zero

Reason — The n cells form a closed circuit among themselves, so there is no external resistance and the only resistance in the circuit is the total internal resistance nr. The current in the circuit is

$$\begin{aligned}\text I = \frac{\text{nE}}{\text{nr}} \\[1em] \text I = \frac{\text E}{\text r}\end{aligned}$$

The potential difference across any one cell is

$$\begin{aligned}\text V = \text E - \text{Ir} \\[1em] \text V = \text E - \frac{\text E}{\text r} \times \text r \\[1em] \text V = 0\end{aligned}$$

charge

Reason — Kirchhoff's current law (junction rule) states that the algebraic sum of the currents meeting at a junction in a circuit is zero, that is, the total current entering a junction equals the total current leaving it.

Since current is the rate of flow of charge, this means that no charge is accumulated at or lost from the junction. Hence the junction rule is a statement of the conservation of charge.

3.7 A

Reason — By Kirchhoff's junction rule the total current entering the network equals the total current leaving it.

Current entering = 4 A + 2 A = 6 A

Current leaving through the upper branch (1 A) and the 1.3 A branch = 1 A + 1.3 A = 2.3 A

Hence the current in the remaining branch is

$$\begin{aligned}\text I = 6 - 2.3 \\[1em] \text I = 3.7\ \text A\end{aligned}$$

Note: In the printed figure the downward branch at the lower junction, whose current is asked for, is labelled 1 A instead of I. Taking that branch as I gives the key value 3.7 A.

0.01

Reason — For a wire of length L, resistivity ρ and area of cross-section A, the resistance is R = ρL/A. If it carries a current I, the potential gradient along it is

$$\begin{aligned}\text K = \frac{\text{IR}}{\text L} \\[1em] \text K = \frac{\text I \rho}{\text A}\end{aligned}$$

Substituting I = 0.2 A, ρ = 40 × 10^-8 Ω m and A = 8 × 10^-6 m²,

$$\begin{aligned}\text K = \frac{0.2 \times 40 \times 10^{-8}}{8 \times 10^{-6}} \\[1em] \text K = \frac{8 \times 10^{-8}}{8 \times 10^{-6}} \\[1em] \text K = 10^{-2} \\[1em] \text K = 0.01\ \text{V m}^{-1}\end{aligned}$$

1 Ω

Reason — In the potentiometer method for the internal resistance of a cell, if l₁ is the balancing length for the emf of the cell and l₂ the balancing length when the cell is shunted by a resistance R, then

$\text r = \text R \left(\frac{\text l_1}{\text l_2} - 1\right)$

Substituting R = 2 Ω, l₁ = 150 cm and l₂ = 100 cm,

$$\begin{aligned}\text r = 2 \left(\frac{150}{100} - 1\right) \\[1em] \text r = 2 (1.5 - 1) \\[1em] \text r = 2 \times 0.5 \\[1em] \text r = 1\ \Omega\end{aligned}$$

3 : 1

Reason — When the two cells support each other the balancing length corresponds to (E₁ + E₂), and when E₂ is reversed it corresponds to (E₁ − E₂). Since the balancing length is proportional to the emf,

$$\begin{aligned}\frac{\text E_1 + \text E_2}{\text E_1 - \text E_2} = \frac{\text l_1}{\text l_2} \\[1em] \frac{\text E_1 + \text E_2}{\text E_1 - \text E_2} = \frac{58}{29} \\[1em] \frac{\text E_1 + \text E_2}{\text E_1 - \text E_2} = 2 \\[1em] \text E_1 + \text E_2 = 2\text E_1 - 2\text E_2 \\[1em] 3\text E_2 = \text E_1 \quad \Rightarrow \quad \frac{\text E_1}{\text E_2} = \frac{3}{1}\end{aligned}$$

The battery of potentiometer can have a voltage of 15 V and R adjusted so that the potential drop across the wire slightly exceeds 10 V.

Reason — A balance point can be obtained on a potentiometer only if the potential difference across the whole wire is greater than the emf being measured. Here the larger emf is 10 V, so the potential drop across the wire must slightly exceed 10 V.

R < r

Reason — In Arrangement A the six cells are in series, so

$\text I_1 = \frac{6\text E}{\text R + 6\text r}$

In Arrangement B the six cells are in parallel, so

$$\begin{aligned}\text I_2 = \frac{\text E}{\text R + \frac{\text r}{6}} \\[1em] \text I_2 = \frac{6\text E}{6\text R + \text r}\end{aligned}$$

For I₂ > I₁,

$$\begin{aligned}\frac{6\text E}{6\text R + \text r} \gt \frac{6\text E}{\text R + 6\text r} \\[1em] 6\text R + \text r \lt \text R + 6\text r \\[1em] 5\text R \lt 5\text r \quad \Rightarrow \quad \text R \lt \text r\end{aligned}$$

B

Reason — For a negative temperature coefficient of resistance, the resistance decreases as the temperature rises. When current is passed for some time, X is heated by the current and its resistance falls.

If l is the balancing length measured from A, the balance condition of the meter bridge is

$\frac{\text R}{\text X} = \frac{\text l}{100 - \text l}$

R remains unchanged while X decreases, so the ratio R/X increases. Hence l must increase, that is, the balance point shifts away from A and towards B.

402.5 Ω

Reason — In the first case the condition of balance of the Wheatstone bridge gives

$$\begin{aligned}\frac{\text P}{\text Q} = \frac{\text R}{\text X} \\[1em] \frac{\text P}{\text Q} = \frac{400}{\text X} \qquad ...(i)\end{aligned}$$

On interchanging P and Q the balance is obtained with R' = 405 Ω, so

$\frac{\text Q}{\text P} = \frac{405}{\text X} \qquad ...(ii)$

Multiplying (i) and (ii),

$$\begin{aligned}1 = \frac{400 \times 405}{\text X^2} \\[1em] \text X^2 = 400 \times 405 \\[1em] \text X^2 = 162000 \\[1em] \text X = \sqrt{162000} \\[1em] \text X = 402.49 \approx 402.5\ \Omega\end{aligned}$$

1 A

Reason — The network is symmetrical about the horizontal line AC. The upper branch A–B–C consists of 2 Ω and 2 Ω in series, and the lower branch consists of 2 Ω and 2 Ω in series, so B and D are both at the mid-potential between A and C. The junction of the two 4 Ω resistors is also at this same potential.

Hence there is no potential difference across either 5 Ω resistor and no current flows through them. The three paths from A to C are then

Upper branch : 2 + 2 = 4 Ω

Middle branch : 4 + 4 = 8 Ω

Lower branch : 2 + 2 = 4 Ω

The current I₁ is the current in the middle branch, across which the potential difference is the full 8 V of the cell.

$$\begin{aligned}\text I_1 = \frac{8}{8} \\[1em] \text I_1 = 1\ \text A\end{aligned}$$

The electrical energy supplied by the constant-voltage supply is being dissipated as heat in the resistor.

Yes, it is a drawback. No, it does not remain constant.

Explanation:
Owing to the internal resistance, some of the energy supplied by the cell is dissipated as heat within the cell, so the terminal potential difference is less than the emf. Further, the internal resistance gradually increases with use of the cell.

Conservation of charge and conservation of energy.

Explanation:
Kirchhoff's first law (junction rule) follows from the conservation of charge, and the second law (loop rule) follows from the conservation of energy.

The graph A represents the parallel combination.

The slope of a V–I graph gives the resistance of the combination. The graph A has the smaller slope, so it corresponds to the smaller resistance. Since the equivalent resistance of a parallel combination is smaller than that of a series combination of the same two resistors, A represents the parallel combination.

For wires of the same metal and the same area of cross-section, R ∝ l, so the resistances are in the ratio 1 : 2 : 3.

In parallel the potential difference across each wire is the same, and I = V/R, so the currents are in the inverse ratio of the resistances,

$$\begin{aligned}\text I_1 : \text I_2 : \text I_3 = 1 : \frac{1}{2} : \frac{1}{3} \\[1em] \text I_1 : \text I_2 : \text I_3 = 6 : 3 : 2\end{aligned}$$

In parallel, the currents are in the ratio 6 : 3 : 2.

In series, the current in all the three wires will be the same.

In parallel, the ratio of the potential differences is 1 : 1 : 1, because all the three wires are connected between the same two points.

In series, the ratio is 1 : 2 : 3, because the current is the same in each wire and V = IR, so V ∝ R.

Volt/metre (or volt/cm); the balance point will be obtained at a greater length.

Explanation:
Since E = Kl, for a given emf the balancing length l is inversely proportional to the potential gradient K, so a smaller K gives a longer balancing length.

A voltmeter draws some current from the cell, so it measures the terminal potential difference and not the emf. In a potentiometer, at the balance point no current is drawn from the cell in the secondary circuit, so it measures the true emf of the cell.

Given,

• Emf of the standard cell, E = 1.018 V
• Balancing length, l = 254.5 cm

At the balance point,

$$\begin{aligned}\text E = \text{Kl} \\[1em] \text K = \frac{\text E}{\text l} \\[1em] \text K = \frac{1.018}{254.5} \\[1em] \text K = 4 \times 10^{-3}\ \text{V/cm}\end{aligned}$$

Hence, the potential gradient across the potentiometer wire is 4 × 10^-3 V/cm = 0.4 V/m.

Potentiometer.

By Kirchhoff's junction rule, the total current entering a junction is equal to the total current leaving it.

(a) At the junction, the currents of 5 A, 7 A and 2 A flow in and the current of 3 A flows out, along with I.

$$\begin{aligned}5 + 7 + 2 = 3 + \text I \\[1em] \text I = 14 - 3 \\[1em] \text I = 11\ \text A\end{aligned}$$

(b) At the first junction, 2 A and 3 A flow in, so the current in the joining branch is 2 + 3 = 5 A.

At the third junction, 1.5 A and 2.5 A flow out, so the current entering it is 1.5 + 2.5 = 4 A.

Applying the junction rule at the second junction, where 5 A enters and I together with 4 A leaves,

$$\begin{aligned}5 = \text I + 4 \\[1em] \text I = 1\ \text A\end{aligned}$$

Hence, I = 11 A in (a) and I = 1 A in (b).

Given,

• Resistance of the whole wire = 16 Ω
• It is cut into 4 equal parts, and each part is stretched to the original length l

The resistance of each part before stretching is

$\frac{16}{4} = 4\ \Omega$

Each part of length l/4 is stretched to length l, that is, its length becomes 4 times. On stretching, the volume remains constant, so R ∝ l². Hence the resistance of each stretched part is

$4 \times (4)^2 = 64\ \Omega$

The four stretched parts are connected in parallel, so

$$\begin{aligned}\frac{1}{\text R} = \frac{4}{64} \\[1em] \frac{1}{\text R} = \frac{1}{16}\end{aligned}$$

Hence, the total resistance of the combination is 16 Ω.

Given,

• Resistance of the whole wire = 4 Ω

The ends of a diameter divide the circle into two equal semicircular halves, each of resistance

$\frac{4}{2} = 2\ \Omega$

These two halves are connected in parallel between the ends of the diameter.

$$\begin{aligned}\frac{1}{\text R} = \frac{1}{2} + \frac{1}{2} \\[1em] \frac{1}{\text R} = 1\end{aligned}$$

Hence, the effective resistance between the ends of a diameter is 1 Ω.

The potentiometer Y is preferred.

The slope of the V–l graph gives the potential gradient ΔV/Δl. The graph Y has the smaller slope, so its potential gradient is smaller. A smaller potential gradient gives a longer balancing length for the same emf, which makes the measurement more sensitive and hence more accurate.

The emf of a cell represents the work done per unit charge. It is equal to the terminal potential difference when the cell is in open circuit (no current drawn).

Explanation:
The emf is the work done by the cell in forcing unit positive-charge to flow round the whole circuit, including the cell itself,

$\text E = \frac{\text{dW}}{\text{dq}}$

In spite of its name, emf is not a force at all; its unit is joule/coulomb, that is, volt.

When no current is drawn from the cell, there is no potential drop Ir inside the cell, so V = E.

Internal resistance : The resistance which the electrolyte between the plates offers to the flow of ions inside a cell is called the internal resistance of the cell. Some of the energy supplied by the cell is dissipated as heat in this resistance.

The internal resistance of a cell depends upon :

(i) the separation between the two plates — it is directly proportional to this separation;

(ii) the area of the plates dipped into the electrolyte — it is inversely proportional to this area;

(iii) the nature, concentration and temperature of the electrolyte — it increases with increase in concentration and decreases with increase in temperature.

The internal resistance of a cell decreases with increase in the temperature of the electrolyte. Hence on a warm day the internal resistance of the car battery is less than on a cold day.

Since the current drawn is I = E/(R + r), a smaller internal resistance allows a larger current to be drawn from the battery, and this larger current starts the engine more easily.

This happens because of the internal resistance r of the cell. The current drawn from the cell is

$\text I = \frac{\text E}{\text R + \text r}$

When R is halved the current becomes

$\text I' = \frac{\text E}{\frac{\text R}{2} + \text r}$

The denominator is reduced from (R + r) to (R/2 + r), which is more than half of (R + r) because r is unchanged. Hence I' is slightly less than 2I.

This happens because of the internal resistance r of the cell. The terminal potential difference is

$\text V = \text E - \text{Ir}$

The emf E and the internal resistance r of the cell are fixed, so as the current I drawn from the cell is increased, the potential drop Ir inside the cell increases and V is lowered further.

Yes. The terminal potential difference of the cell becomes zero when the cell is short-circuited, that is, when the external resistance R = 0.

Explanation:
In this condition the maximum current is drawn from the cell,

$\text I_\text{max} = \frac{\text E}{\text r}$

Therefore,

$$\begin{aligned}\text V = \text E - \text I_\text{max}\ \text r \\[1em] \text V = \text E - \text E \\[1em] \text V = 0\end{aligned}$$

Starting a car needs a very heavy current, and the current that a battery can supply is limited by its internal resistance.

The internal resistance of 8 simple cells joined in series is of the order of 10 Ω, whereas that of a car-battery is only of the order of 0.1 Ω. Hence the simple cells cannot supply the heavy current required to start the car, even though their net emf is also 12 V.

When the motor car is started, the starter draws a heavy current from the car-battery. Because of this large current the potential drop Ir inside the battery increases appreciably, so the potential difference between the plates of the battery falls.

The head-light is connected across the same terminals, so the potential difference across it also falls and it becomes dim.

The heater draws a heavy current from the mains. Because of this large current there is an appreciable potential-drop in the line wires, which have a small but finite resistance.

As a result the potential difference available across the bulbs falls and they become dim.

Kirchhoff's first law (junction rule) : The algebraic sum of the currents meeting at a junction in an electric circuit is zero, that is, the total current flowing towards a junction is equal to the total current flowing away from it.

$\sum \text I = 0$

Proof : Charge cannot accumulate at a junction, because a junction is merely a point of the circuit and has no capacity to store charge. Hence in any time interval dt the charge arriving at the junction must equal the charge leaving it.

If the currents I₁ and I₂ flow towards a junction and I₃ and I₄ flow away from it, then in time dt,

$$\begin{aligned}(\text I_1 + \text I_2)\ \text{dt} = (\text I_3 + \text I_4)\ \text{dt} \\[1em] \text I_1 + \text I_2 = \text I_3 + \text I_4\end{aligned}$$

Thus the junction rule is simply a statement of the conservation of charge.

Kirchhoff's second law (loop rule) : In any closed loop of an electric circuit, the algebraic sum of the emfs is equal to the algebraic sum of the products of the currents and the resistances in that loop.

$\sum \text E = \sum \text{IR}$

Here,

E is the emf of a cell in the loop, taken positive when the loop is traversed from the negative to the positive electrode of the cell and negative otherwise,

I is the current in a resistance of the loop, and

R is that resistance, the product IR being taken positive when the loop is traversed in the direction of the current and negative otherwise.

This law is a statement of the conservation of energy.

$\frac{\text P}{\text Q} = \frac{\text R}{\text S}$

Explanation:
In a Wheatstone bridge ABCD with P = AB, Q = BC, R = AD and S = DC, a cell across AC and a galvanometer across BD, the bridge is balanced when the above relation holds. In this condition no current flows through the galvanometer.

Sensitivity : A potentiometer is said to be sensitive if, on displacing the jockey slightly from the null point, the galvanometer shows a large deflection. The smaller the potential gradient, the greater the sensitivity.

The sensitivity of a potentiometer is increased by decreasing its potential gradient, since

$\text K = \frac{\text{IR}}{\text L}$

where I is the current in the wire, R its resistance and L its length.

This can be done in two ways :

(i) by increasing the length of the potentiometer wire, and

(ii) by decreasing the current in the potentiometer wire, using a rheostat in the primary circuit.

At the balance point E = Kl, so the balancing length is l = E/K.

(i) If the emf to be measured is smaller, the balance point will be obtained at a smaller length of the wire.

(ii) If the current in the potentiometer wire is decreased, the potential gradient K decreases, so the balance point will be obtained at a larger length of the wire.

(iii) If the length of the wire is doubled, the potential difference across the whole wire remaining the same, the potential gradient is halved, so the balance point will be obtained at twice the previous length.

The specific resistance of copper is small and its temperature-coefficient of resistance is large.

Because of its small specific resistance, a copper wire has a very small resistance, so the potential difference across the whole wire is very small (it may be less than the emf to be measured, and then no null point is obtained) and a large current flows, heating the wire. Further, a large temperature-coefficient makes the resistance of the wire change as the current heats it, so the potential gradient does not remain constant. For these reasons copper is not used, and wires of manganin or constantan are used instead.

The voltmeter will read the emf E₂ of the lower cell.

Explanation:
Since the ammeter reads zero, no current flows in the branch containing E₂ and R₂. Hence there is no potential drop across R₂ or across the internal resistance of the cell, and the potential difference between C and D is equal to the emf E₂.

Let E be the emf and r the internal resistance of the equivalent source.

Since the two batteries are connected in parallel, their internal resistances are in parallel,

$\frac{1}{\text r} = \frac{1}{\text r_1} + \frac{1}{\text r_2} \qquad \text{or} \qquad \text r = \frac{\text r_1 \text r_2}{\text r_1 + \text r_2}$

The maximum currents that can be drawn from the two batteries are E₁/r₁ and E₂/r₂, so the total current is

$$\begin{aligned}\text I = \frac{\text E_1}{\text r_1} + \frac{\text E_2}{\text r_2} \\[1em] \text I = \frac{\text E_1 \text r_2 + \text E_2 \text r_1}{\text r_1 \text r_2}\end{aligned}$$

If the same current I is drawn from the equivalent source, its emf is

$$\begin{aligned}\text E = \text{Ir} \\[1em] \text E = \frac{\text E_1 \text r_2 + \text E_2 \text r_1}{\text r_1 \text r_2} \times \frac{\text r_1 \text r_2}{\text r_1 + \text r_2} \\[1em] \text E = \frac{\text E_1 \text r_2 + \text E_2 \text r_1}{\text r_1 + \text r_2}\end{aligned}$$

The resistances R and R' are in parallel across the cell, so the current drawn from the cell is

$$\begin{aligned}\text I = \frac{\text E}{\frac{\text{RR}'}{\text R + \text R'} + \text r} \\[1em] \text I = \frac{\text E (\text R + \text R')}{\text{RR}' + \text r (\text R + \text R')}\end{aligned}$$

The current through the ammeter, that is, through R', is

$$\begin{aligned}\text I' = \frac{\text R}{\text R + \text R'}\ \text I \\[1em] \text I' = \frac{\text E}{\text R' + \text r\left(1 + \frac{\text R'}{\text R}\right)} \qquad ...(i)\end{aligned}$$

(a) If r is negligible, Eq. (i) gives I' = E/R', which does not contain R at all. Hence there will be no effect on the reading of the ammeter on increasing R.

(b) If r is not negligible, then on increasing R the term r(1 + R'/R) in the denominator of Eq. (i) decreases, so the current I' will increase.

After the capacitor C is fully charged, the current in the branch containing C and R₃ stops, and a steady current flows only through R₁ and R₂.

Since there is now no current in R₃, there is no potential difference across it. Hence the potential difference between the plates of C is the same as that between A and B, that is, the potential difference across R₂.

The current in R₁ and R₂ is

$\text I = \frac{\text E}{\text R_1 + \text R_2}$

Therefore the potential difference across R₂, and hence across C, is

$\text V = \text{IR}_2 = \frac{\text{ER}_2}{\text R_1 + \text R_2}$

Given,

• Emf of the cell = 12 V, negative terminal earthed, positive terminal at A
• Current from A through R and the 4.0 Ω resistor = 1.0 A
• Lower end of the 4.0 Ω resistor earthed (zero potential)

The 12 V cell has its lower end earthed, and the earth is at zero potential. Hence the potential at the point A is

$\text V_\text A = 12\ \text V$

A current of 1.0 A flows through the 4.0 Ω resistor, so the potential drop across it is

$1\ \text A \times 4\ \Omega = 4\ \text V$

The lower end of the 4.0 Ω resistor is earthed and is therefore at zero potential. Hence the potential at its upper end B is

$\text V_\text B = 4\ \text V$

Hence, the potential at A is 12 V and that at B is 4 V.

Given,

• E₁ = 10 V, r₁ = 1 Ω; E₂ = 16 V, r₂ = 0.5 Ω (cells opposing)
• Resistances 4 Ω (AB) and 0.5 Ω (DC)
• The wire BC is earthed (zero potential)

The two cells oppose each other, so the net emf driving the current is 16 V − 10 V = 6 V, and the total resistance of the loop is

$$\begin{aligned}0.5 + 1 + 4 + 0.5 = 6\ \Omega \\[1em] \text I = \frac{16 - 10}{6} \\[1em] \text I = 1\ \text A \\ \text{(anticlockwise)}\end{aligned}$$

The point between B and C is earthed, so it is at zero potential.

Potential at A : The 4 Ω resistor lies between A and the earthed point, and the potential drop across it is 1 × 4 = 4 V. Hence

$\text V_\text A = 4\ \text V$

Potential at D : The 0.5 Ω resistor lies between D and the earthed point, and the potential drop across it is 1 × 0.5 = 0.5 V. Since the current flows from the earthed point towards D through this resistor, D is at a lower potential,

$\text V_\text D = -\ 0.5\ \text V$

Terminal voltage across the 10 V cell : The current enters this cell at its positive electrode, so the cell is being charged. Its terminal voltage is

$$\begin{aligned}\text V = \text E + \text{Ir} \\[1em] \text V = 10 + (1 \times 1) \\[1em] \text V = 11\ \text V\end{aligned}$$

Hence, V_A = 4 V, V_D = − 0.5 V and the terminal voltage across the 10 V cell is 11 V.

When the n identical resistors are connected in series,

$$\begin{aligned}\text X = \text R + \text R + \text R + \\ ...\ \text{n terms} \\[1em] \text X = \text{nR} \qquad ...(i)\end{aligned}$$

When they are connected in parallel,

$$\begin{aligned}\frac{1}{\text Y} = \frac{1}{\text R} + \frac{1}{\text R} + \frac{1}{\text R} + \\ ...\ \text{n terms} \\[1em] \frac{1}{\text Y} = \frac{\text n}{\text R} \\[1em] \text Y = \frac{\text R}{\text n} \qquad ...(ii)\end{aligned}$$

Multiplying Eqs. (i) and (ii),

$$\begin{aligned}\text{XY} = \text{nR} \times \frac{\text R}{\text n} \\[1em] \text{XY} = \text R^2 \\[1em] \text R = \sqrt{\text{XY}}\end{aligned}$$

(a) The terminal potential difference is V = E − Ir, so the graph of V against I is a straight line of negative slope.

The Y-intercept of the graph gives the emf E of the cell, since V = E when I = 0.

(b) Comparing V = E − Ir with y = mx + c, the magnitude of the slope of the graph gives the internal resistance,

$\text r = \frac{\Delta \text V}{\Delta \text I}$

The resistance of the fourth arm is the equivalent resistance of S₁ and S₂ in parallel,

$\text S = \frac{\text S_1 \text S_2}{\text S_1 + \text S_2}$

Substituting this in the condition of balance of the Wheatstone bridge, P/Q = R/S,

$$\begin{aligned}\frac{\text P}{\text Q} = \frac{\text R}{\frac{\text S_1 \text S_2}{\text S_1 + \text S_2}} \\[1em] \frac{\text P}{\text Q} = \frac{\text R(\text S_1 + \text S_2)}{\text S_1 \text S_2}\end{aligned}$$

The errors in the circuit are :

(i) The emf of the driver battery is only 10 V. It must be greater than 10 V, so that the potential difference across the wire is more than the emfs of the cells to be compared. Otherwise the balance length for the 10 V cell cannot be determined at all.

(ii) There is no rheostat in the primary circuit. A series variable resistance must be added to the primary circuit, so that the current drawn from the driver cell can be reduced and the potential drop across the potentiometer wire adjusted.

(iii) Both the keys K₁ and K₂ are shown closed. At a time only one key should be closed and the other must be kept open, so that only one cell is included in the secondary circuit while its balance point is being found.

Electric cell : An electric cell is a source of electrical energy which maintains a continuous flow of charge in a circuit. It consists of two rods of different metals, called 'electrodes' or 'plates', immersed in a liquid called 'electrolyte'.

A chemical reaction takes place in the electrolyte of the cell as a result of which positive and negative ions are generated. These move to the electrodes and maintain the supply of charge to them. When the two plates are connected by a wire, the charge begins to flow in the wire. Thus, the cell changes chemical energy into electrical energy.

Emf of a cell : The work done by the cell in forcing unit positive-charge to flow in the whole circuit (including the cell) is called the 'electromotive force' (emf) of the cell.

$\text E = \frac{\text{dW}}{\text{dq}}$

Its unit is joule/coulomb, which is called 'volt'. If in the flow of 1 C of charge in a circuit the work done by the cell be 1 J, then the emf of the cell is 1 V.

Factors on which the emf depends : The emf is a characteristic of the cell which depends upon the nature of the plates and the nature of the electrolyte used in the cell. It does not depend upon the dimensions of the electrodes, their separation or the quantity of electrolyte.

Emf : If dW is the work done by the cell in forcing a charge dq from its negative to its positive electrode, then the emf of the cell is

$\text E = \frac{\text{dW}}{\text{dq}}$

Derivation : Let a cell of emf E and internal resistance r be connected to an external resistance R, and let a steady current I be established in the circuit. In time dt the charge that flows is dq = I dt, so the work done by the cell is

$$\begin{aligned}\text{dW} = \text E\ \text{dq} \\[1em] \text{dW} = \text{EI}\ \text{dt}\end{aligned}$$

This energy is consumed as heat in the external resistance R and in the internal resistance r. By Joule's law, the heat produced in time dt is

$\text{dW}_\text{ext} = \text I^2 \text R\ \text{dt}, \qquad \text{dW}_\text{int} = \text I^2 \text r\ \text{dt}$

By the conservation of energy,

$$\begin{aligned}\text{dW} = \text{dW}_\text{ext} + \text{dW}_\text{int} \\[1em] \text{EI}\ \text{dt} = \text I^2 \text R\ \text{dt} + \text I^2 \text r\ \text{dt}\end{aligned}$$

Dividing throughout by I dt,

$$\begin{aligned}\text E = \text{IR} + \text{Ir} \\[1em] \text E = \text I (\text R + \text r)\end{aligned}$$

The current drawn from the cell is

$\text I = \frac{\text E}{\text R + \text r}$

The voltmeter reads the terminal potential difference, which is the potential drop across R,

$\text V = \text{IR} = \frac{\text{ER}}{\text R + \text r}$

or,

$\text V = \frac{\text E}{1 + \frac{\text r}{\text R}}$

The relation may also be written as

$\text r = \text R\left(\frac{\text E}{\text V} - 1\right)$

Variation of V : As R is reduced, the ratio r/R increases, so the denominator in the expression for V increases. Hence V will decrease when R is reduced.

First law (junction rule) : The algebraic sum of the currents meeting at a junction in an electric circuit is zero,

$\sum \text I = 0$

That is, the total current arriving at a junction is equal to the total current leaving it. Since charge cannot accumulate at a junction, this law is a statement of the conservation of charge.

Second law (loop rule) : In any closed loop of an electric circuit, the algebraic sum of the emfs is equal to the algebraic sum of the products of the currents and the resistances in that loop,

$\sum \text E = \sum \text{IR}$

As one goes round a closed loop and returns to the starting point, the total change of potential must be zero. Hence this law is a statement of the conservation of energy.

Sign convention :

(i) A current flowing towards a junction is taken as positive and one flowing away from it as negative.

(ii) While traversing a loop, the product IR is taken as positive if the loop is traversed in the direction of the current, and negative if traversed against it.

(iii) The emf E of a cell is taken as positive if the loop is traversed from its negative electrode to its positive electrode inside the cell, and negative otherwise.

Application to the given circuit : In the circuit shown, R₁ and the cell E₁ form one branch, R₂ forms a second branch, and R₃ with the cell E₂ forms a third branch, all joined between the same two junctions.

Applying the junction rule at one junction gives the relation among the three branch currents. Applying the loop rule to the closed loop containing R₁, E₁ and R₂, and again to the loop containing R₂, R₃ and E₂, gives two more equations. These three equations are then solved for the unknown currents.

When n identical cells are joined in series, the net emf is nE and the total internal resistance is nr. Hence the current in the circuit is

$$\begin{aligned}\text I = \frac{\text{nE}}{\text R + \text{nr}} \\[1em] \text I (\text R + \text{nr}) = \text{nE} \\[1em] \text{IR} + \text{Inr} = \text{nE} \\[1em] \frac{\text R}{\text n} + \text r = \frac{\text E}{\text I} \\[1em] \text r = \frac{\text E}{\text I} - \frac{\text R}{\text n}\end{aligned}$$

Four resistances R₁, R₂, R₃ and R₄ are connected in the four arms of the bridge (R₁ = AB, R₂ = BC, R₃ = AD, R₄ = DC), with a cell E across AC and a galvanometer G across BD.

The bridge is said to be balanced when the galvanometer shows no deflection. In this condition no current flows through the galvanometer, so the points B and D are at the same potential.

Let the current I₁ flow through R₁ and R₂, and the current I₂ through R₃ and R₄. Since the potentials at the two ends of the galvanometer are equal, the potential drops across R₁ and R₃ must be equal, and so must those across R₂ and R₄.

$$\begin{aligned}\text I_1 \text R_1 = \text I_2 \text R_3 \qquad ...(i) \\[1em] \text I_1 \text R_2 = \text I_2 \text R_4 \qquad ...(ii)\end{aligned}$$

Dividing Eq. (i) by Eq. (ii),

$\frac{\text R_1}{\text R_2} = \frac{\text R_3}{\text R_4}$

This is the required condition of balance of the Wheatstone bridge.

Principle : When four resistances are connected in the four arms of a Wheatstone bridge, with a cell across one diagonal and a galvanometer across the other, the bridge is balanced and the galvanometer shows no deflection when

$\frac{\text P}{\text Q} = \frac{\text R}{\text S}$

Meter bridge : The meter bridge is the practical form of the Wheatstone bridge.

A uniform wire AB of length 100 cm is stretched along a metre scale and its ends are joined to two L-shaped copper strips. A resistance box R is connected in the left gap and the wire S of unknown resistance in the right gap. A cell with a key is connected across AB, and a galvanometer is joined between the central strip and a jockey which can slide along the wire.

A suitable resistance R is taken out from the box and the jockey is moved along the wire until the galvanometer shows no deflection. Let the balance point be at a distance l cm from A. Then the resistances of the two parts of the wire are proportional to their lengths, so

$$\begin{aligned}\frac{\text R}{\text S} = \frac{\text l}{100 - \text l} \\[1em] \text S = \text R\ \frac{(100 - \text l)}{\text l}\end{aligned}$$

Specific resistance : The length L of the given wire is measured with a metre scale and its diameter d is measured with a screw gauge, so that its area of cross-section is

$\text a = \pi \left(\frac{\text d}{2}\right)^2$

Since S = ρL/a, the specific resistance of the material of the wire is

$$\begin{aligned}\rho = \frac{\text{Sa}}{\text L} \\[1em] \rho = \frac{\pi \text d^2 \text S}{4\text L}\end{aligned}$$

Principle : When a constant current is passed through a wire of uniform cross-section, the potential difference across any portion of the wire is directly proportional to the length of that portion (V = Kl).

A potentiometer consists of a long uniform wire AB of high specific resistance and low temperature-coefficient of resistance, stretched on a wooden board.

A driving cell of emf E_D is connected across the ends of the wire through a rheostat, so that a steady current flows through it and a constant potential gradient is set up along its length.

If K is the potential gradient and the balance point for a cell of emf E in the secondary circuit is obtained at a length l from A, then the potential difference across this length is Kl. At balance no current flows through the galvanometer, so

$\text E = \text{Kl}$

Thus the emf of a cell is directly proportional to the balancing length of the potentiometer wire.

Sensitivity : The sensitivity is increased by decreasing the potential gradient K = IR/L (where I is the current in the wire, R its resistance and L its length), that is, by increasing the length of the wire or by decreasing the current in it with the help of the rheostat.

Superiority over a voltmeter : A voltmeter draws some current from the cell whose emf is to be measured, so it always reads the terminal potential difference, which is less than the emf. In a potentiometer no current is drawn from the cell at the balance point, so it measures the true emf. A potentiometer therefore behaves like an ideal voltmeter of infinite resistance.

The potentiometer wire AB is connected to the driving cell E_D through a key and a rheostat, forming the primary circuit. The positive terminals of the two cells E₁ and E₂ are joined to the end A, and their negative terminals are taken through the keys K₁ and K₂ to a two-way key, and then through the galvanometer to the jockey J.

With the two-way key connecting E₁ (K₁ closed, K₂ open), the balance point is obtained at a length l₁ from A. Then, with the two-way key connecting E₂ (K₂ closed, K₁ open), the balance point is obtained at a length l₂.

Working formula :

$\frac{\text E_1}{\text E_2} = \frac{\text l_1}{\text l_2}$

The potentiometer wire AB is connected to the driving cell E_D through a key and a rheostat. The cell of emf E whose internal resistance is to be found is connected between A and the jockey through a galvanometer, and a resistance box R with a key K₁ is connected across this cell.

With the key K₁ open, the balance point is obtained at a length l₁ from A, which corresponds to the emf E of the cell. A resistance R is then introduced and K₁ closed, and the new balance point is obtained at a length l₂, which corresponds to the terminal potential difference V.

Working formula :

$\text r = \text R \left(\frac{\text l_1}{\text l_2} - 1\right)$

(i) The response of the student is correct. The meter bridge is the practical form of the Wheatstone bridge, and the unknown resistance is found from the condition of balance of the Wheatstone bridge.

(ii) The response of the student is correct. The resistance of a wire is

$\text R = \rho\ \frac{\text l}{\text A}$

For the same length and the same material, R is inversely proportional to the area of cross-section. Hence a thicker wire has a smaller resistance.

(iii) The response of the student is incorrect. For the meter bridge,

$\frac{\text R}{\text S} = \frac{\text l}{100 - \text l}$

To shift the null-point towards the left, l must decrease, and for this the ratio R/S must decrease. Since S is fixed, R must be decreased and not increased.

Emf : The work done by the cell in forcing unit positive-charge to flow in the whole circuit, including the cell itself, is called the electromotive force (emf) of the cell.

Terminal potential difference : The work done in carrying unit positive-charge through the external circuit, from one terminal of the cell to the other, is called the terminal potential difference of the cell.

When current is withdrawn from the cell : Let a cell of emf E and internal resistance r send a current I through an external resistance R. Then

$$\begin{aligned}\text E = \text I (\text R + \text r) \\[1em] \text E = \text{IR} + \text{Ir}\end{aligned}$$

The terminal potential difference is V = IR, so

$\text V = \text E - \text{Ir}$

Since Ir is a positive quantity, V is less than E. Thus the emf of a cell is greater than its terminal potential difference when current is drawn from it.

Physically, a part of the energy given by the cell is consumed in driving the current through the internal resistance of the cell itself.

When the cell is being charged : During charging, the current is sent into the cell from an external source and flows inside the cell from the positive to the negative electrode, that is, opposite to the direction in which the cell itself would drive it. Now the potential drop Ir is added to the emf, and

$\text V = \text E + \text{Ir}$

Hence V is greater than E. Thus the emf of a cell is less than its terminal potential difference when the cell is being charged.

(i) Emf and internal resistance of the battery :

In each row the n cells are joined in series, so the emf of one row is nE and its internal resistance is nr.

The m rows are joined in parallel, and all of them have the same emf. Hence the emf of the whole battery is

$\text E_\text{eq} = \text{nE}$

The internal resistances of the m rows, each equal to nr, are in parallel, so the internal resistance of the battery is

$\text r_\text{eq} = \frac{\text{nr}}{\text m}$

(ii) Current through R :

$$\begin{aligned}\text I = \frac{\text E_\text{eq}}{\text R + \text r_\text{eq}} \\[1em] \text I = \frac{\text{nE}}{\text R + \frac{\text{nr}}{\text m}} \\[1em] \text I = \frac{\text{mnE}}{\text{mR} + \text{nr}}\end{aligned}$$

Since N = mn,

$\text I = \frac{\text{NE}}{\text{mR} + \text{nr}}$

Principle : When a steady current flows through a uniform wire of a potentiometer, a constant potential gradient K is set up along its length. If the balance point for a cell of emf E is obtained at a length l, then

$\text E = \text{Kl}$

that is, the emf of the cell is directly proportional to the balancing length.

In this figure B₁ is the driving battery, which sends a steady current through the potentiometer wire AB.

Sensitivity : The sensitivity is increased by decreasing the potential gradient K = IR/L (where I is the current in the wire, R its resistance and L its length), either by increasing the length of the potentiometer wire or by decreasing the current in it by means of a rheostat in the primary circuit.

Why it is called an ideal voltmeter : At the balance point the galvanometer shows no deflection, so no current is drawn from the cell in the secondary circuit. A voltmeter which draws no current is an ideal voltmeter of infinite resistance. Hence a potentiometer behaves like an ideal voltmeter and measures the true emf of the cell.

Determination of internal resistance :

The cell of emf E whose internal resistance r is to be determined is connected in the secondary circuit through a galvanometer, and a resistance box R with a key K₂ is connected across this cell, the primary circuit containing B₁, Rh and K₁.

Open circuit : With K₂ open, no current is drawn from the cell, so the balance point corresponds to its emf E. Let this balancing length be l₁. Then

$\text E = \text{Kl}_1$

Cell sending current through R : A resistance R is now introduced and K₂ is closed. The cell sends a current through R, so the balance point now corresponds to its terminal potential difference V. Let this balancing length be l₂. Then

$\text V = \text{Kl}_2$

Dividing,

$\frac{\text E}{\text V} = \frac{\text l_1}{\text l_2}$

Since V = IR and E = I(R + r),

$\frac{\text E}{\text V} = \frac{\text R + \text r}{\text R}$

Therefore,

$$\begin{aligned}\frac{\text R + \text r}{\text R} = \frac{\text l_1}{\text l_2} \\[1em] \text r = \text R \left(\frac{\text l_1}{\text l_2} - 1\right)\end{aligned}$$

(i) Working principle : When a steady current flows through a wire of uniform area of cross-section, a constant potential gradient K is set up along its length. If a cell of emf E connected in the secondary circuit is balanced against a length l of the wire, then

$\text E = \text{Kl}$

Thus the emf of the cell is directly proportional to the balancing length. At the balance point no current is drawn from the cell.

(ii) The sensitivity of a potentiometer is increased by decreasing its potential gradient, since K = IR/L, where I is the current in the wire, R its resistance and L its length. This is done either by increasing the length of the potentiometer wire, or by decreasing the current in the wire with the help of a rheostat in the primary circuit.

(iii) The potentiometer Y is preferred. The slope of the V–l graph gives the potential gradient. The graph Y has the smaller slope, so its potential gradient is smaller. A smaller potential gradient gives a longer balancing length for the same emf, which makes the measurement more sensitive and hence more accurate.

(iv)

Given,

• Current in the wire, I = 1.0 mA = 1.0 × 10^-3 A
• Length of the wire, L = 4 m
• Resistance per unit length = 4 Ω/m

The resistance of the whole wire is

$$\begin{aligned}\text R = 4 \times 4 \\[1em] \text R = 16\ \Omega\end{aligned}$$

The potential difference across the whole wire is

$$\begin{aligned}\text V = \text{IR} \\[1em] \text V = (1.0 \times 10^{-3}) \times 16 \\[1em] \text V = 1.6 \times 10^{-2}\ \text V\end{aligned}$$

The potential gradient is

$$\begin{aligned}\text K = \frac{\text V}{\text L} \\[1em] \text K = \frac{1.6 \times 10^{-2}}{4} \\[1em] \text K = 4 \times 10^{-3}\ \text{V m}^{-1}\end{aligned}$$

Hence, the potential gradient of the potentiometer wire is 4 × 10^-3 V m^-1.

(v)

Given,

• Balancing length for the first cell, l₁ = 70 cm
• Balancing length for the second cell, l₂ = 60 cm
• E₁ − E₂ = 0.2 V

Since the emf is proportional to the balancing length,

$\frac{\text E_1}{\text E_2} = \frac{\text l_1}{\text l_2}$

Substituting E₂ = E₁ − 0.2,

$$\begin{aligned}\frac{\text E_1}{\text E_1 - 0.2} = \frac{70}{60} \\[1em] 60\ \text E_1 = 70\ \text E_1 - 14 \\[1em] 10\ \text E_1 = 14 \quad \Rightarrow \quad \text E_1 = 1.4\ \text V \\[1em] \text E_2 = 1.4 - 0.2 \\[1em] \text E_2 = 1.2\ \text V\end{aligned}$$

Hence, the emfs of the two cells are 1.4 V and 1.2 V.

Given,

• Potential difference in open circuit = 6 V
• Terminal potential difference, V = 4 V
• Current drawn, I = 2 A

In an open circuit no current is drawn from the cell, so the terminal potential difference is equal to the emf of the cell.

$\text E = 6\ \text V$

When a current I is drawn, the terminal potential difference is

$$\begin{aligned}\text V = \text E - \text{Ir} \\[1em] 4 = 6 - (2 \times \text r) \\[1em] 2\text r = 2 \\[1em] \text r = 1\ \Omega\end{aligned}$$

Hence, the emf of the cell is 6 V and its internal resistance is 1 Ω.

Given,

• Emf of the cell, E = 2.0 V
• External resistance, R = 25 Ω
• Terminal voltage, V = 1.8 V

The terminal voltage is the potential drop across the external resistance,

$$\begin{aligned}\text I = \frac{\text V}{\text R} \\[1em] \text I = \frac{1.8}{25} \\[1em] \text I = 0.072\ \text A\end{aligned}$$

The internal resistance is obtained from V = E − Ir,

$$\begin{aligned}\text r = \frac{\text E - \text V}{\text I} \\[1em] \text r = \frac{2.0 - 1.8}{0.072} \\[1em] \text r = \frac{0.2}{0.072} \\[1em] \text r = 2.78\ \Omega\end{aligned}$$

Hence, the internal resistance of the cell is 2.78 Ω and the current in the 25 Ω resistor is 0.072 A.

Given,

• V₁ = 1.8 V when I₁ = 0.5 A
• V₂ = 1.6 V when I₂ = 1.0 A

Using V = E − Ir for both cases,

$$\begin{aligned}1.8 = \text E - 0.5\ \text r \qquad ...(i) \\[1em] 1.6 = \text E - 1.0\ \text r \qquad ...(ii)\end{aligned}$$

Subtracting Eq. (ii) from Eq. (i),

$$\begin{aligned}0.2 = 0.5\ \text r \\[1em] \text r = 0.4\ \Omega\end{aligned}$$

Substituting r = 0.4 Ω in Eq. (i),

$$\begin{aligned}\text E = 1.8 + (0.5 \times 0.4) \\[1em] \text E = 1.8 + 0.2 \\[1em] \text E = 2.0\ \text V\end{aligned}$$

Hence, the emf of the cell is 2.0 V and its internal resistance is 0.4 Ω.

Given,

• R = 200 Ω
• E = 2.0 V, internal resistance negligible
• Reading of the voltmeter = 1.0 V

The voltmeter is in series with the resistor, so the sum of the potential drops across the two equals the emf of the cell. Hence the potential drop across the 200 Ω resistor is

$2.0 - 1.0 = 1.0\ \text V$

The current in the circuit is

$$\begin{aligned}\text I = \frac{1.0}{200} \\[1em] \text I = 5 \times 10^{-3}\ \text A\end{aligned}$$

The same current flows through the voltmeter, so its resistance is

$$\begin{aligned}\text R_\text V = \frac{1.0}{5 \times 10^{-3}} \\[1em] \text R_\text V = 200\ \Omega\end{aligned}$$

Hence, the resistance of the voltmeter is 200 Ω and the current in the circuit is 5 × 10^-3 A.

Given,

• Reading of the voltmeter alone = 15 V
• Reading of the voltmeter with the ammeter in the circuit, V = 9.0 V
• Reading of the ammeter, I = 1.5 A

A high-resistance voltmeter draws practically no current, so in the first case the reading gives the emf of the battery,

$\text E = 15\ \text V$

(i) Internal resistance :

$$\begin{aligned}\text r = \frac{\text E - \text V}{\text I} \\[1em] \text r = \frac{15 - 9.0}{1.5} \\[1em] \text r = \frac{6.0}{1.5} \\[1em] \text r = 4.0\ \Omega\end{aligned}$$

(ii) Resistance of the ammeter and the connecting wires : The terminal potential difference of 9.0 V appears across them, so

$$\begin{aligned}\text R = \frac{\text V}{\text I} \\[1em] \text R = \frac{9.0}{1.5} \\[1em] \text R = 6.0\ \Omega\end{aligned}$$

Hence, the internal resistance of the battery is 4.0 Ω and the resistance of the ammeter plus the connecting wires is 6.0 Ω.

Given,

• Reading of the voltmeter with K open = 1.53 V
• Reading of the ammeter with K closed, I = 1.00 A
• Reading of the voltmeter with K closed, V = 1.03 V

(i) Emf of the cell : With the key open no current is drawn from the cell, because the voltmeter has a very high resistance. Hence the voltmeter reads the emf itself,

$\text E = 1.53\ \text V$

(ii) Internal resistance : With the key closed,

$$\begin{aligned}\text V = \text E - \text{Ir} \\[1em] \text r = \frac{\text E - \text V}{\text I} \\[1em] \text r = \frac{1.53 - 1.03}{1.00} \\[1em] \text r = 0.50\ \Omega\end{aligned}$$

(iii) Value of R : The terminal potential difference appears across R,

$$\begin{aligned}\text R = \frac{\text V}{\text I} \\[1em] \text R = \frac{1.03}{1.00} \\[1em] \text R = 1.03\ \Omega\end{aligned}$$

Hence, E = 1.53 V, r = 0.50 Ω and R = 1.03 Ω.

Given,

• E₁ = 6.20 V, r₁ = 0.01 Ω
• E₂ = 12.45 V, r₂ = 0.03 Ω
• Charging current, I = 20 A

During charging the current is sent into the battery from outside, so the potential drop Ir is added to the emf and the terminal potential difference is

$\text V = \text E + \text{Ir}$

For the first battery :

$$\begin{aligned}\text V_1 = 6.20 + (20 \times 0.01) \\[1em] \text V_1 = 6.20 + 0.20 \\[1em] \text V_1 = 6.40\ \text V\end{aligned}$$

For the second battery :

$$\begin{aligned}\text V_2 = 12.45 + (20 \times 0.03) \\[1em] \text V_2 = 12.45 + 0.60 \\[1em] \text V_2 = 13.05\ \text V\end{aligned}$$

Hence, the terminal potential differences are 6.40 V and 13.05 V.

Given,

• Emf of each cell, E = 1.5 V
• Current in series combination, I_s = 1.0 A
• Current in parallel combination, I_p = 0.6 A

In series, the net emf is 2E and the total internal resistance is 2r,

$$\begin{aligned}\text I_\text s = \frac{2\text E}{\text R + 2\text r} \\[1em] 1.0 = \frac{3.0}{\text R + 2\text r} \\[1em] \text R + 2\text r = 3.0 \qquad ...(i)\end{aligned}$$

In parallel, the net emf is E and the total internal resistance is r/2,

$$\begin{aligned}\text I_\text p = \frac{\text E}{\text R + \frac{\text r}{2}} \\[1em] 0.6 = \frac{1.5}{\text R + \frac{\text r}{2}} \\[1em] \text R + \frac{\text r}{2} = 2.5 \qquad ...(ii)\end{aligned}$$

Subtracting Eq. (ii) from Eq. (i),

$$\begin{aligned}2\text r - \frac{\text r}{2} = 0.5 \\[1em] \frac{3\text r}{2} = 0.5 \\[1em] \text r = \frac{1}{3}\ \Omega\end{aligned}$$

Hence, the internal resistance of each cell is $\frac{1}{3}$ Ω.

Given,

• External resistance, R = 1 Ω
• Current in series = current in parallel

For two identical cells of emf E and internal resistance r,

$\text I_\text s = \frac{2\text E}{\text R + 2\text r}, \qquad \text I_\text p = \frac{\text E}{\text R + \frac{\text r}{2}}$

Equating the two currents,

$$\begin{aligned}\frac{2\text E}{\text R + 2\text r} = \frac{\text E}{\text R + \frac{\text r}{2}} \\[1em] 2\left(\text R + \frac{\text r}{2}\right) = \text R + 2\text r \\[1em] 2\text R + \text r = \text R + 2\text r \\[1em] \text R = \text r\end{aligned}$$

Hence, the internal resistance of each cell is 1 Ω.

Given,

• Number of bulbs = 50, each 100 V taking 0.6 A
• Number of accumulators = 54, each of emf 2 V and internal resistance 0.005 Ω

Requirement of the bulbs : The bulbs are in parallel, so the total current required, at a potential difference of 100 V, is

$50 \times 0.6 = 30\ \text A$

The resistance of one bulb is 100/0.6 Ω, so the resistance of the parallel combination is

$$\begin{aligned}\text R = \frac{1}{50} \times \frac{100}{0.6} \\[1em] \text R = \frac{100}{30} \\[1em] \text R = \frac{10}{3}\ \Omega\end{aligned}$$

Supply from the battery : The 54 accumulators are in series, so

$$\begin{aligned}\text E = 54 \times 2 \\[1em] \text E = 108\ \text V \\[1em] \text r = 54 \times 0.005 \\[1em] \text r = 0.27\ \Omega \\[1em] \text I = \frac{\text E}{\text R + \text r} \\[1em] \text I = \frac{108}{\frac{10}{3} + 0.27} \\[1em] \text I = \frac{108}{3.6033} \\[1em] \text I = 29.97\ \text A \approx 30\ \text A\end{aligned}$$

The potential difference across the bulbs is

$$\begin{aligned}\text V = \text{IR} \\[1em] \text V = 30 \times \frac{10}{3} \\[1em] \text V = 100\ \text V\end{aligned}$$

Hence, the bulbs will be properly lighted, because the battery gives 30 A at 100 V and the bulbs also require exactly 30 A at 100 V.

Given,

• E₁ = 2.0 V, r₁ = 0.05 Ω
• E₂ = 1.8 V, r₂ = 0.7 Ω
• E₃ = 1.5 V, r₃ = 1.0 Ω
• External resistance, R = 4.0 Ω

(i) In a series combination the emfs add up and so do the internal resistances,

$$\begin{aligned}\text E = 2.0 + 1.8 + 1.5 \\[1em] \text E = 5.3\ \text V \\[1em] \text r = 0.05 + 0.7 + 1.0 \\[1em] \text r = 1.75\ \Omega\end{aligned}$$

The ammeter is of low resistance, so

$$\begin{aligned}\text I = \frac{\text E}{\text R + \text r} \\[1em] \text I = \frac{5.3}{4.0 + 1.75} \\[1em] \text I = \frac{5.3}{5.75} \\[1em] \text I = 0.92\ \text A\end{aligned}$$

Hence, the ammeter would read 0.92 A.

(ii) No. For non-similar cells joined in parallel, Kirchhoff's rules have to be employed to obtain the currents in the different branches.

Given,

• Emf of each cell, E = 2 V
• Number of cells in parallel, n = 3
• External resistance, R = 6 Ω
• Terminal potential difference, V = 1.5 V

For three identical cells in parallel the emf of the combination is E = 2 V and the internal resistance is r/3.

The current in the circuit is

$$\begin{aligned}\text I = \frac{\text V}{\text R} \\[1em] \text I = \frac{1.5}{6} \\[1em] \text I = 0.25\ \text A\end{aligned}$$

Using V = E − I r_eq,

$$\begin{aligned}1.5 = 2 - 0.25 \times \frac{\text r}{3} \\[1em] 0.25 \times \frac{\text r}{3} = 0.5 \\[1em] \text r = \frac{0.5 \times 3}{0.25} \\[1em] \text r = 6\ \Omega\end{aligned}$$

Hence, the internal resistance of each cell is 6 Ω.

Given,

• Number of cells in series, n = 4
• From the graph, the intercept on the V-axis = 5.6 V and V = 0 at I = 2.0 A

(i) Emf of each cell : The intercept of the V–I graph on the V-axis gives the net emf of the combination, so 4E = 5.6 V.

$$\begin{aligned}\text E = \frac{5.6}{4} \\[1em] \text E = 1.4\ \text V\end{aligned}$$

(ii) Current for maximum power dissipation : When V = 0 the combination is short-circuited, so its total internal resistance is

$$\begin{aligned}4\text r = \frac{5.6}{2.0} \\[1em] 4\text r = 2.8\ \Omega\end{aligned}$$

The power dissipated in the external resistance is maximum when the external resistance equals the total internal resistance, that is, when R = 4r = 2.8 Ω. The current is then

$$\begin{aligned}\text I = \frac{4\text E}{\text R + 4\text r} \\[1em] \text I = \frac{5.6}{2.8 + 2.8} \\[1em] \text I = \frac{5.6}{5.6} \\[1em] \text I = 1\ \text A\end{aligned}$$

(iii) Internal resistance of each cell :

$$\begin{aligned}\text r = \frac{2.8}{4} \\[1em] \text r = 0.7\ \Omega\end{aligned}$$

Hence, E = 1.4 V, the current for maximum power dissipation is 1 A, and r = 0.7 Ω.

Given,

• The 2 Ω resistor, the 2 V cell and the branch containing the 3 Ω resistor with the 5 V cell are all connected between the same two points A and B

The middle branch contains only the 2 V cell, whose internal resistance is not given and is therefore taken as negligible. Its positive terminal is towards A, so

$\text V_\text A - \text V_\text B = 2\ \text V$

Let I be the current in the 3 Ω resistor, flowing from right to left. Applying Kirchhoff's loop law ($\sum \text{IR} = \sum \text E$) to the loop formed by the lower branch and the 2 V cell,

$$\begin{aligned}3\text I = 5 - 2 \\[1em] 3\text I = 3 \\[1em] \text I = 1\ \text A\end{aligned}$$

Hence, the current in the 3 Ω resistance is 1 A, flowing from right to left, and the potential difference between A and B is 2 V.

Given,

• E₁ = 2 V, r₁ = 1 Ω
• E₂ = 8 V, r₂ = 2 Ω
• External resistance, R = 4 Ω

The two cells are connected in parallel between the same two points, so they may be replaced by a single equivalent cell of emf

$$\begin{aligned}\text E = \frac{\text E_1 \text r_2 + \text E_2 \text r_1}{\text r_1 + \text r_2} \\[1em] \text E = \frac{(2 \times 2) + (8 \times 1)}{1 + 2} \\[1em] \text E = \frac{4 + 8}{3} \\[1em] \text E = 4\ \text V\end{aligned}$$

and internal resistance

$$\begin{aligned}\text r = \frac{\text r_1 \text r_2}{\text r_1 + \text r_2} \\[1em] \text r = \frac{1 \times 2}{3} \\[1em] \text r = \frac{2}{3}\ \Omega\end{aligned}$$

The current through the 4 Ω resistor is

$$\begin{aligned}\text I = \frac{\text E}{\text R + \text r} \\[1em] \text I = \frac{4}{4 + \frac{2}{3}} \\[1em] \text I = \frac{4}{\frac{14}{3}} \\[1em] \text I = \frac{12}{14} \\[1em] \text I = \frac{6}{7}\ \text A\end{aligned}$$

The potential difference between A and C is the potential drop across the 4 Ω resistor,

$$\begin{aligned}\text V_\text{AC} = \text{IR} \\[1em] \text V_\text{AC} = \frac{6}{7} \times 4 \\[1em] \text V_\text{AC} = \frac{24}{7}\ \text V\end{aligned}$$

Hence, the potential difference between A and C is $\frac{24}{7}$ V.

Given,

• Cell of 3 V with internal resistance 2 Ω in the left branch, carrying I₁
• Cell of 2 V with internal resistance 0.5 Ω in the right branch, carrying I₂
• Resistances 2.5 Ω and 4 Ω in the left loop, 5 Ω and 3.5 Ω in the right loop, and 2 Ω in the middle branch carrying I₃

Junction rule : At the upper junction the currents I₁ and I₂ arrive and the current I₃ leaves through the middle branch,

$\text I_1 + \text I_2 = \text I_3 \qquad ...(i)$

Loop rule for the left loop : Traversing the loop containing the 3 V cell, the 2.5 Ω, the 4 Ω and the middle 2 Ω resistors,

$$\begin{aligned}3 = \text I_1 (2 + 2.5 + 4) + 2\ \text I_3 \\[1em] 3 = 8.5\ \text I_1 + 2\ \text I_3 \qquad ...(ii)\end{aligned}$$

Loop rule for the right loop : Traversing the loop containing the 2 V cell, the 5 Ω, the 3.5 Ω and the middle 2 Ω resistors,

$$\begin{aligned}-2 = \text I_2 (5 + 3.5 + 0.5) + 2\ \text I_3 \\[1em] -2 = 9\ \text I_2 + 2\ \text I_3 \qquad ...(iii)\end{aligned}$$

Substituting I₃ = I₁ + I₂ from Eq. (i) into Eqs. (ii) and (iii),

$$\begin{aligned}3 = 10.5\ \text I_1 + 2\ \text I_2 \qquad ...(iv) \\[1em] -2 = 2\ \text I_1 + 11\ \text I_2 \qquad ...(v)\end{aligned}$$

Multiplying Eq. (iv) by 11 and Eq. (v) by 2 and subtracting,

$$\begin{aligned}33 + 4 = (115.5 - 4)\ \text I_1 \\[1em] \text I_1 = \frac{37}{111.5} \\[1em] \text I_1 = 0.33\ \text A\end{aligned}$$

From Eq. (v),

$$\begin{aligned}\text I_2 = \frac{-2 - (2 \times 0.33)}{11} \\[1em] \text I_2 = \frac{-2.66}{11} \\[1em] \text I_2 = -\ 0.24\ \text A\end{aligned}$$

From Eq. (i),

$$\begin{aligned}\text I_3 = 0.33 - 0.24 \\[1em] \text I_3 = 0.09\ \text A\end{aligned}$$

Hence, I₁ = 0.33 A, I₂ = − 0.24 A and I₃ = 0.09 A. The negative sign for I₂ shows that this current actually flows opposite to the direction assumed.

Given,

• E₁ = 6 V, r₁ = 1 Ω

• E₂ = 12 V, r₂ = 2 Ω

• External resistance, R = 15 Ω

Let I₁ be the current supplied by the 6 V cell, I₂ be the current supplied by the 12 V cell, and I be the current through the 15 Ω resistance.

By Kirchhoff's junction law,

$\text I = \text I_1 + \text I_2 \qquad ...(i)$

Applying Kirchhoff's loop law to the loop containing the 6 V cell,

$$\begin{aligned}\text E_1 = \text{IR} + \text I_1 \text r_1 \\[1em] 6 = 15\text I + \text I_1 \\[1em] \text I_1 = 6 - 15\text I \qquad ...(ii)\end{aligned}$$

Applying Kirchhoff's loop law to the loop containing the 12 V cell,

$$\begin{aligned}\text E_2 = \text{IR} + \text I_2 \text r_2 \\[1em] 12 = 15\text I + 2\text I_2 \\[1em] \text I_2 = \frac{12 - 15\text I}{2} \qquad ...(iii)\end{aligned}$$

Substituting Eqs. (ii) and (iii) in Eq. (i),

$$\begin{aligned}\text I = 6 - 15\text I + \frac{12 - 15\text I}{2} \\[1em] 2\text I = 12 - 30\text I + 12 - 15\text I \\[1em] 47\text I = 24 \\[1em] \text I = \frac{24}{47}\ \text A\end{aligned}$$

Therefore, the potential difference across the 15 Ω resistance is

$$\begin{aligned}\text V = \text{IR} \\[1em] \text V = \frac{24}{47} \times 15 \\[1em] \text V = \frac{360}{47} \\[1em] \text V = 7.66\ \text V\end{aligned}$$

Current in the 6 V branch :

From Eq. (ii),

$$\begin{aligned}\text I_1 = 6 - 15\left(\frac{24}{47}\right) \\[1em] \text I_1 = \frac{282 - 360}{47} \\[1em] \text I_1 = -\frac{78}{47}\ \text A\end{aligned}$$

The negative sign shows that the actual current flows opposite to the assumed direction. Therefore, the current through the 6 V cell is

$\frac{78}{47}\ \text A$

from its positive terminal to its negative terminal, i.e., the 6 V cell is being charged.

Current in the 12 V branch :

From Eq. (iii),

$$\begin{aligned}\text I_2 = \frac{12 - 15\left(\frac{24}{47}\right)}{2} \\[1em] \text I_2 = \frac{564 - 360}{94} \\[1em] \text I_2 = \frac{102}{47}\ \text A\end{aligned}$$

Hence, $\frac{78}{47}$ A flows through the 6 V cell in the charging direction, $\frac{102}{47}$ A flows through the 12 V cell in the discharging direction, $\frac{24}{47}$ A flows through the 15 Ω resistor, and the PD across it is 7.66 V.

Given,

• E₁ = E₂ = 1.5 V
• r₁ = 1.5 Ω, r₂ = 0.5 Ω
• External resistance, R = 1.0 Ω

Since the two cells have equal emfs, the emf of the parallel combination is

$\text E = 1.5\ \text V$

and its internal resistance is

$$\begin{aligned}\text r = \frac{\text r_1 \text r_2}{\text r_1 + \text r_2} \\[1em] \text r = \frac{1.5 \times 0.5}{2.0} \\[1em] \text r = 0.375\ \Omega\end{aligned}$$

The total current in the external resistance is

$$\begin{aligned}\text I = \frac{\text E}{\text R + \text r} \\[1em] \text I = \frac{1.5}{1.0 + 0.375} \\[1em] \text I = \frac{1.5}{1.375} \\[1em] \text I = \frac{12}{11}\ \text A\end{aligned}$$

The terminal potential difference of each cell is the potential drop across the 1.0 Ω resistor,

$$\begin{aligned}\text V = \text{IR} \\[1em] \text V = \frac{12}{11} \times 1.0 \\[1em] \text V = \frac{12}{11}\ \text V\end{aligned}$$

Current from cell A :

$$\begin{aligned}\text I_1 = \frac{\text E - \text V}{\text r_1} \\[1em] \text I_1 = \frac{1.5 - \frac{12}{11}}{1.5} \\[1em] \text I_1 = \frac{\frac{4.5}{11}}{1.5} \\[1em] \text I_1 = \frac{3}{11}\ \text A\end{aligned}$$

Current from cell B :

$$\begin{aligned}\text I_2 = \frac{\text E - \text V}{\text r_2} \\[1em] \text I_2 = \frac{\frac{4.5}{11}}{0.5} \\[1em] \text I_2 = \frac{9}{11}\ \text A\end{aligned}$$

Hence, the currents drawn are $\frac{3}{11}$ A from A and $\frac{9}{11}$ A from B, and the terminal PD of each cell is $\frac{12}{11}$ V.

Given,

• Cells of 2 V and 1 V, both of negligible internal resistance
• Resistances 1 Ω (BC), 2 Ω (AC) and 1 Ω (DC)

Junction rule : At the junction A the current I₃ arrives through the 2 Ω resistor and the currents I₁ and I₂ leave it,

$\text I_3 = \text I_1 + \text I_2 \qquad ...(i)$

Loop ABCA : The 2 V cell drives the current I₁ through the 1 Ω resistor BC, and the loop is closed through the 2 Ω resistor carrying I₃,

$2 = 1 \times \text I_1 + 2\ \text I_3 \qquad ...(ii)$

Loop ADCA : The 1 V cell drives the current I₂ through the 1 Ω resistor DC,

$1 = 1 \times \text I_2 + 2\ \text I_3 \qquad ...(iii)$

Substituting Eq. (i) in Eqs. (ii) and (iii),

$$\begin{aligned}2 = 3\ \text I_1 + 2\ \text I_2 \qquad ...(iv) \\[1em] 1 = 2\ \text I_1 + 3\ \text I_2 \qquad ...(v)\end{aligned}$$

Multiplying Eq. (iv) by 3, Eq. (v) by 2 and subtracting,

$$\begin{aligned}6 - 2 = (9 - 4)\ \text I_1 \\[1em] \text I_1 = \frac{4}{5} \\[1em] \text I_1 = 0.8\ \text A\end{aligned}$$

From Eq. (v),

$$\begin{aligned}\text I_2 = \frac{1 - (2 \times 0.8)}{3} \\[1em] \text I_2 = \frac{-0.6}{3} \\[1em] \text I_2 = -\ 0.2\ \text A\end{aligned}$$

From Eq. (i),

$$\begin{aligned}\text I_3 = 0.8 - 0.2 \\[1em] \text I_3 = 0.6\ \text A\end{aligned}$$

Hence, I₁ = 0.8 A, I₂ = − 0.2 A and I₃ = 0.6 A.

Given,

• V_A = 0
• Currents of 1 A and 2 A as marked in the figure, a 2 Ω resistor in the branch DC, and cells of 1 V and 2 V

Junction rule at D : A current of 2 A arrives at D and a current of 1 A leaves it along the upper branch, so the current flowing from D to C through the 2 Ω resistor is

$2 - 1 = 1\ \text A$

Potential at C : Going from A to C through the 1 V cell,

$$\begin{aligned}\text V_\text C = \text V_\text A + 1 \\[1em] \text V_\text C = 0 + 1 \\[1em] \text V_\text C = 1\ \text V\end{aligned}$$

Potential at D : The current of 1 A flows from D to C through the 2 Ω resistor, so D is at the higher potential,

$$\begin{aligned}\text V_\text D = \text V_\text C + (1 \times 2) \\[1em] \text V_\text D = 1 + 2 \\[1em] \text V_\text D = 3\ \text V\end{aligned}$$

Potential at B : Going from B to D through the 2 V cell, the potential rises by 2 V,

$$\begin{aligned}\text V_\text D = \text V_\text B + 2 \\[1em] \text V_\text B = 3 - 2 \\[1em] \text V_\text B = 1\ \text V\end{aligned}$$

Hence, the potential at point B is 1 V.

Given,

• Resistances 1 Ω (AF), 1 Ω (FE) and R (DC)
• Middle branch containing the 4 Ω resistor in series with a 6 V cell
• Cells of 9 V and 3 V in the lower branch

Since no current flows through the 4 Ω resistor, there is no potential drop across it, and the same current I flows round the outer loop AFEDCBA. The arm EB carries no current, so this loop may be traversed in two parts, AFEBA and EDCBE.

Value of R : Applying Kirchhoff's loop law ($\sum \text{IR} = \sum \text E$) to the loop AFEBA,

$$\begin{aligned}\text I \times 1 + \text I \times 1 = 9 - 6 \\[1em] 2\text I = 3 \\[1em] \text I = 1.5\ \text A\end{aligned}$$

Applying Kirchhoff's loop law to the loop EDCBE, in which the current I flows from D to C through R,

$$\begin{aligned}\text{IR} = 6 - 3 \\[1em] 1.5\ \text R = 3 \\[1em] \text R = 2\ \Omega\end{aligned}$$

Potential difference between A and D : E and D are joined by a wire of no resistance, so V_D = V_E.

$$\begin{aligned}\text V_\text A - \text V_\text D = \text I \times (1 + 1) \\[1em] \text V_\text A - \text V_\text D = 1.5 \times 2 \\[1em] \text V_\text A - \text V_\text D = 3\ \text V\end{aligned}$$

Hence, R = 2 Ω and V_A − V_D = 3 V.

Given,

• Initial balancing length, l = 40 cm
• After connecting 10 Ω in series with R₁, the resistance in the left gap increases, so the null point shifts 10 cm away from A, giving l' = 40 + 10 = 50 cm

For a metre bridge the condition of balance is

$\frac{\text R_1}{\text R_2} = \frac{\text l}{100 - \text l}$

Initially,

$$\begin{aligned}\frac{\text R_1}{\text R_2} = \frac{40}{60} \\[1em] \frac{\text R_1}{\text R_2} = \frac{2}{3} \qquad ...(i)\end{aligned}$$

After adding 10 Ω in series,

$$\begin{aligned}\frac{\text R_1 + 10}{\text R_2} = \frac{50}{50} \\[1em] \frac{\text R_1 + 10}{\text R_2} = 1 \\[1em] \text R_1 + 10 = \text R_2 \qquad ...(ii)\end{aligned}$$

Substituting R₁ = (2/3)R₂ from Eq. (i) in Eq. (ii),

$$\begin{aligned}\frac{2}{3}\text R_2 + 10 = \text R_2 \\[1em] \frac{\text R_2}{3} = 10 \\[1em] \text R_2 = 30\ \Omega \\[1em] \text R_1 = \frac{2}{3} \times 30 \\[1em] \text R_1 = 20\ \Omega\end{aligned}$$

To bring the null point back to 40 cm, the resistance of the left gap must again become 20 Ω. Let X be connected in parallel with (R₁ + 10) = 30 Ω,

$$\begin{aligned}\frac{30\ \text X}{30 + \text X} = 20 \\[1em] 30\ \text X = 600 + 20\ \text X \\[1em] 10\ \text X = 600 \\[1em] \text X = 60\ \Omega\end{aligned}$$

Hence, a resistance of 60 Ω should be connected in parallel.

Given,

• Arms of the bridge : 6 Ω and the parallel combination of 15 Ω and R in the upper branch, 3 Ω and 1.5 Ω in the lower branch

Let S be the equivalent resistance of 15 Ω and R in parallel,

$\text S = \frac{15\ \text R}{15 + \text R}$

For the balanced Wheatstone bridge,

$$\begin{aligned}\frac{6}{\text S} = \frac{3}{1.5} \\[1em] \frac{6}{\text S} = 2 \\[1em] \text S = 3\ \Omega\end{aligned}$$

Therefore,

$$\begin{aligned}\frac{15\ \text R}{15 + \text R} = 3 \\[1em] 15\ \text R = 45 + 3\ \text R \\[1em] 12\ \text R = 45 \\[1em] \text R = 3.75\ \Omega\end{aligned}$$

Hence, R = 3.75 Ω.

Given,

• Arms of the bridge : 2 Ω and 4 Ω in the upper branch, 3 Ω and 6 Ω in the lower branch
• Galvanometer of 40 Ω across the bridge arm
• Emf of the battery, E = 6 V
• Internal resistance of the battery, r = 0.4 Ω

Test of balance :

$\frac{2}{4} = \frac{1}{2} \qquad \text{and} \qquad \frac{3}{6} = \frac{1}{2}$

Since,

$\frac{2}{4} = \frac{3}{6}$

The Wheatstone bridge is balanced. Therefore, no current flows through the galvanometer, and the galvanometer branch may be omitted.

The resistance of the upper branch is

$$\begin{aligned}\text R_1 = 2 + 4 \\[1em] \text R_1 = 6\ \Omega\end{aligned}$$

The resistance of the lower branch is

$$\begin{aligned}\text R_2 = 3 + 6 \\[1em] \text R_2 = 9\ \Omega\end{aligned}$$

These two branches are connected in parallel. Therefore, their equivalent resistance is

$$\begin{aligned}\text R = \frac{6 \times 9}{6 + 9} \\[1em] \text R = \frac{54}{15} \\[1em] \text R = 3.6\ \Omega\end{aligned}$$

Including the internal resistance of the battery, the total resistance of the circuit is

$$\begin{aligned}\text R_{\text{total}} = 3.6 + 0.4 \\[1em] \text R_{\text{total}} = 4\ \Omega\end{aligned}$$

Therefore, the total current supplied by the battery is

$$\begin{aligned}\text I = \frac{\text E}{\text R_{\text{total}}} \\[1em] \text I = \frac{6}{4} \\[1em] \text I = 1.5\ \text A\end{aligned}$$

The potential difference across the parallel combination of the two branches is

$$\begin{aligned}\text V = \text{IR} \\[1em] \text V = 1.5 \times 3.6 \\[1em] \text V = 5.4\ \text V\end{aligned}$$

Since both branches are in parallel, the potential difference across each branch is 5.4 V.

Current in the upper branch :

Using Ohm's law,

$$\begin{aligned}\text I_1 = \frac{\text V}{\text R_1} \\[1em] \text I_1 = \frac{5.4}{6} \\[1em] \text I_1 = 0.9\ \text A\end{aligned}$$

Current in the lower branch :

Using Ohm's law,

$$\begin{aligned}\text I_2 = \frac{\text V}{\text R_2} \\[1em] \text I_2 = \frac{5.4}{9} \\[1em] \text I_2 = 0.6\ \text A\end{aligned}$$

Also,

$$\begin{aligned}\text I_1 + \text I_2 = 0.9 + 0.6 \\[1em] \text I_1 + \text I_2 = 1.5\ \text A \\[1em] \text I_1 + \text I_2 = \text I\end{aligned}$$

Hence, I = 1.5 A, I₁ = 0.9 A and I₂ = 0.6 A.

Given,

• AB = 10 Ω, BC = 20 Ω, AD = 3 Ω, DC = 7 Ω
• Emf of the cell = 60 V, and V_C = 0

Currents in the two branches : The branch ABC has resistance 10 + 20 = 30 Ω and the branch ADC has resistance 3 + 7 = 10 Ω, and both are connected across the 60 V cell.

$$\begin{aligned}\text I_\text{ABC} = \frac{60}{30} \\[1em] \text I_\text{ABC} = 2\ \text A \\[1em] \text I_\text{ADC} = \frac{60}{10} \\[1em] \text I_\text{ADC} = 6\ \text A\end{aligned}$$

(a) Potentials at B and D : Since C is earthed, V_C = 0.

$$\begin{aligned}\text V_\text B - \text V_\text C = \text I_\text{ABC} \times 20 \\[1em] \text V_\text B - \text V_\text C = 2 \times 20 \\[1em] \text V_\text B - \text V_\text C = 40\ \text V \\[1em] \text V_\text B = +\ 40\ \text V \\[1em] \text V_\text D - \text V_\text C = \text I_\text{ADC} \times 7 \\[1em] \text V_\text D - \text V_\text C = 6 \times 7 \\[1em] \text V_\text D - \text V_\text C = 42\ \text V \\[1em] \text V_\text D = +\ 42\ \text V\end{aligned}$$

(b) Since V_D > V_B, the current in the galvanometer would flow from D to B.

(c) Value of DC for balance : For the balanced bridge,

$$\begin{aligned}\frac{\text{AB}}{\text{BC}} = \frac{\text{AD}}{\text{DC}} \\[1em] \frac{10}{20} = \frac{3}{\text{DC}} \\[1em] \text{DC} = \frac{3 \times 20}{10} \\[1em] \text{DC} = 6\ \Omega\end{aligned}$$

Hence, V_B = + 40 V, V_D = + 42 V, the current would flow from D to B, and DC should be 6 Ω.

Given,

• R = 2 Ω, balancing length l = 40 cm
• Required balancing length in the second case, l' = 60 cm

Taking the resistance box (R) in the left gap and the wire (S) in the right gap, with the balancing length l measured from the left end, the condition of balance of the metre bridge is

$\frac{\text R}{\text S} = \frac{\text l}{100 - \text l}$

Resistance of the wire :

$$\begin{aligned}\frac{2}{\text S} = \frac{40}{60} \\[1em] \text S = \frac{2 \times 60}{40} \\[1em] \text S = 3\ \Omega\end{aligned}$$

Resistance for a null-point at 60 cm :

$$\begin{aligned}\frac{\text R'}{3} = \frac{60}{40} \\[1em] \frac{\text R'}{3} = 1.5 \\[1em] \text R' = 4.5\ \Omega\end{aligned}$$

Hence, the resistance of the wire is 3 Ω, and 4.5 Ω will have to be taken out from the resistance box.

Given,

• Resistance in the left gap = 5 Ω shunted by a wire of resistance X
• Resistance in the right gap = 2 Ω
• Balancing length, l = 62.5 cm

The resistance of the left gap is the parallel combination of 5 Ω and X,

$\text R = \frac{5\ \text X}{5 + \text X}$

Applying the balance condition of the metre bridge,

$$\begin{aligned}\frac{\text R}{2} = \frac{62.5}{100 - 62.5} \\[1em] \frac{\text R}{2} = \frac{62.5}{37.5} \\[1em] \frac{\text R}{2} = \frac{5}{3} \\[1em] \text R = \frac{10}{3}\ \Omega\end{aligned}$$

Therefore,

$$\begin{aligned}\frac{5\ \text X}{5 + \text X} = \frac{10}{3} \\[1em] 15\ \text X = 50 + 10\ \text X \\[1em] 5\ \text X = 50 \\[1em] \text X = 10\ \Omega\end{aligned}$$

Hence, the resistance of the shunt is 10 Ω.

Given,

• Initial balancing length, l = 60.0 cm
• Balancing length after adding 5 Ω in series with S, l' = 50 cm

Initially,

$$\begin{aligned}\frac{\text R}{\text S} = \frac{60}{40} \\[1em] \frac{\text R}{\text S} = \frac{3}{2} \\[1em] \text R = \frac{3}{2}\ \text S \qquad ...(i)\end{aligned}$$

After adding 5 Ω in series with S,

$$\begin{aligned}\frac{\text R}{\text S + 5} = \frac{50}{50} \\[1em] \frac{\text R}{\text S + 5} = 1 \\[1em] \text R = \text S + 5 \qquad ...(ii)\end{aligned}$$

From Eqs. (i) and (ii),

$$\begin{aligned}\frac{3}{2}\ \text S = \text S + 5 \\[1em] \frac{\text S}{2} = 5 \\[1em] \text S = 10\ \Omega \\[1em] \text R = 10 + 5 \\[1em] \text R = 15\ \Omega\end{aligned}$$

Hence, R = 15 Ω and S = 10 Ω.

Given,

• Resistance in the left gap, R = 6 Ω
• Balancing length, l = 40 cm
• Required balancing length, l' = 50 cm

Value of S :

$$\begin{aligned}\frac{6}{\text S} = \frac{40}{60} \\[1em] \text S = \frac{6 \times 60}{40} \\[1em] \text S = 9\ \Omega\end{aligned}$$

For the null point at the middle of the wire, l' = 50 cm, so the resistance S' of the right gap must be

$$\begin{aligned}\frac{6}{\text S'} = \frac{50}{50} \\[1em] \frac{6}{\text S'} = 1 \\[1em] \text S' = 6\ \Omega\end{aligned}$$

Since the required resistance is less than S, a resistance X must be joined in parallel with S,

$$\begin{aligned}\frac{9\ \text X}{9 + \text X} = 6 \\[1em] 9\ \text X = 54 + 6\ \text X \\[1em] 3\ \text X = 54 \\[1em] \text X = 18\ \Omega\end{aligned}$$

Hence, a resistance of 18 Ω should be connected in parallel with S.

Given,

• E = 3 V, r = 1 Ω
• Length of the wire = 100 cm, resistance of the wire = 5 Ω
• Voltmeter connected across 60 cm of the wire

The resistance of 60 cm of the wire is

$5 \times \frac{60}{100} = 3\ \Omega$

and that of the remaining 40 cm is 2 Ω.

(a) Voltmeter taking negligible current : The whole wire carries the same current,

$$\begin{aligned}\text I = \frac{\text E}{5 + \text r} \\[1em] \text I = \frac{3}{5 + 1} \\[1em] \text I = 0.5\ \text A \\[1em] \text V = 0.5 \times 3 \\[1em] \text V = 1.5\ \text V\end{aligned}$$

(b) Voltmeter of resistance 150 Ω : The voltmeter is in parallel with the 3 Ω portion,

$$\begin{aligned}\text R_\text p = \frac{3 \times 150}{3 + 150} \\[1em] \text R_\text p = \frac{450}{153} \\[1em] \text R_\text p = 2.941\ \Omega\end{aligned}$$

The total resistance of the circuit is

$$\begin{aligned}2.941 + 2 + 1 = 5.941\ \Omega \\[1em] \text I = \frac{3}{5.941} \\[1em] \text I = 0.505\ \text A \\[1em] \text V = 0.505 \times 2.941 \\[1em] \text V = 1.485\ \text V\end{aligned}$$

Hence, the readings are (a) 1.5 V and (b) 1.485 V.

Given,

• Length of the wire, L = 10 m = 1000 cm
• Potential gradient, K = 0.0025 V/cm
• Emf of the standard cell, E₁ = 1.025 V
• Balancing length for the second cell, l₂ = 860 cm

Balancing length for the standard cell :

$$\begin{aligned}\text l_1 = \frac{\text E_1}{\text K} \\[1em] \text l_1 = \frac{1.025}{0.0025} \\[1em] \text l_1 = 410\ \text{cm}\end{aligned}$$

Emf of the second cell :

$$\begin{aligned}\text E_2 = \text{Kl}_2 \\[1em] \text E_2 = 0.0025 \times 860 \\[1em] \text E_2 = 2.15\ \text V\end{aligned}$$

Maximum PD that can be measured : This corresponds to the whole length of the wire,

$$\begin{aligned}\text V_\text{max} = \text{KL} \\[1em] \text V_\text{max} = 0.0025 \times 1000 \\[1em] \text V_\text{max} = 2.5\ \text V\end{aligned}$$

Hence, the null-point is at 410 cm, the emf of the other cell is 2.15 V and the maximum PD that can be measured is 2.5 V.

Given,

• R = 10.0 Ω, l₁ = 58.3 cm
• l₂ = 68.5 cm

The same current flows through both the resistors, so the potential difference across each is proportional to its resistance, and the balancing length is proportional to the potential difference. Hence

$$\begin{aligned}\frac{\text X}{\text R} = \frac{\text l_2}{\text l_1} \\[1em] \text X = 10.0 \times \frac{68.5}{58.3} \\[1em] \text X = 11.75\ \Omega\end{aligned}$$

If no null-point is obtained, it means that the potential drop across R or X is greater than the potential difference across the whole wire AB. The current must then be reduced, and hence the potential drop, by putting a suitable resistor in series with R and X.

Hence, X = 11.75 Ω.

Given,

• l₁ = 85 cm, l₂ = 75 cm
• Shunt resistance, R = 7.5 Ω

$$\begin{aligned}\text r = \text R \left(\frac{\text l_1}{\text l_2} - 1\right) \\[1em] \text r = 7.5 \left(\frac{85}{75} - 1\right) \\[1em] \text r = 7.5 \times \frac{10}{75} \\[1em] \text r = 1.0\ \Omega\end{aligned}$$

Hence, the internal resistance of the cell is 1.0 Ω.

Given,

• E₁ = 1.25 V, l₁ = 35.0 cm
• l₂ = 63.0 cm

Since the emf is proportional to the balancing length,

$$\begin{aligned}\frac{\text E_1}{\text E_2} = \frac{\text l_1}{\text l_2} \\[1em] \text E_2 = \text E_1 \times \frac{\text l_2}{\text l_1} \\[1em] \text E_2 = 1.25 \times \frac{63.0}{35.0} \\[1em] \text E_2 = 2.25\ \text V\end{aligned}$$

Hence, the emf of the second cell is 2.25 V.

Given,

• l₁ = 120 cm, when the two cells oppose each other
• l₂ = 360 cm, when the two cells support each other

Since the balancing length is proportional to the net emf in the secondary circuit,

$$\begin{aligned}\text E_1 - \text E_2 = \text K \times 120 \qquad ...(i) \\[1em] \text E_1 + \text E_2 = \text K \times 360 \qquad ...(ii)\end{aligned}$$

(i) Ratio of the emfs : Adding Eqs. (i) and (ii),

$$\begin{aligned}2\text E_1 = \text K \times 480 \\[1em] \text E_1 = 240\ \text K\end{aligned}$$

Subtracting Eq. (i) from Eq. (ii),

$$\begin{aligned}2\text E_2 = \text K \times 240 \\[1em] \text E_2 = 120\ \text K \\[1em] \frac{\text E_1}{\text E_2} = \frac{240}{120} \\[1em] \frac{\text E_1}{\text E_2} = 2\end{aligned}$$

(ii) Null point for E₁ alone :

$$\begin{aligned}\text l = \frac{\text E_1}{\text K} \\[1em] \text l = \frac{240\ \text K}{\text K} \\[1em] \text l = 240\ \text{cm}\end{aligned}$$

Hence, E₁/E₂ = 2 and the null point for E₁ alone is at 240 cm from A.

Given,

• E₁ = 1.5 V, l₁ = 60 cm
• l₂ = 80 cm

(i) Unknown emf :

$$\begin{aligned}\frac{\text E_1}{\text E_2} = \frac{\text l_1}{\text l_2} \\[1em] \text E_2 = 1.5 \times \frac{80}{60} \\[1em] \text E_2 = 2.0\ \text V\end{aligned}$$

(ii) The circuit will not work if the driver cell is of emf 1 V, because the total potential difference across the wire AB would then be only 1 V, which cannot balance the 1.5 V of the cell in the secondary circuit. No balance point could be obtained anywhere on the wire.

(iii) No, the high resistance R does not affect the balance point. At the balance point no current flows through the galvanometer, so the cell in the secondary circuit remains in an open circuit and there is no potential drop across R.

Hence, the unknown emf is 2.0 V.

Given,

• Current entering at A, I = 2 A
• The 9 Ω resistor joins A directly to B, while the 4 Ω, 6 Ω and 2 Ω resistors form the other path from A to B

(i) Equivalent resistance : The upper path contains 4 Ω, 6 Ω and 2 Ω in series,

$4 + 6 + 2 = 12\ \Omega$

This is in parallel with the 9 Ω resistor,

$$\begin{aligned}\text R = \frac{9 \times 12}{9 + 12} \\[1em] \text R = \frac{108}{21} \\[1em] \text R = \frac{36}{7}\ \Omega\end{aligned}$$

(ii) PD between A and B :

$$\begin{aligned}\text V = \text{IR} \\[1em] \text V = 2 \times \frac{36}{7} \\[1em] \text V = \frac{72}{7}\ \text V\end{aligned}$$

(iii) Currents in the resistors :

Current in the 9 Ω resistor,

$$\begin{aligned}\text I_1 = \frac{\text V}{9} \\[1em] \text I_1 = \frac{72}{7 \times 9} \\[1em] \text I_1 = \frac{8}{7}\ \text A\end{aligned}$$

Current in the 12 Ω path, that is, in the 4 Ω, 6 Ω and 2 Ω resistors,

$$\begin{aligned}\text I_2 = \frac{\text V}{12} \\[1em] \text I_2 = \frac{72}{7 \times 12} \\[1em] \text I_2 = \frac{6}{7}\ \text A\end{aligned}$$

Hence, R = $\frac{36}{7}$ Ω, V = $\frac{72}{7}$ V, the current in the 9 Ω resistor is $\frac{8}{7}$ A and that in the 4 Ω, 6 Ω and 2 Ω resistors is $\frac{6}{7}$ A.

Given,

• Emf of the cell = 9 V with a 2 Ω resistor in series
• 4 Ω resistor in the branch containing the ammeter, and 8 Ω resistor in the branch containing the key K

(a) When the key K is off : The 8 Ω branch is broken, so the current flows only through the 2 Ω and the 4 Ω resistors in series,

$$\begin{aligned}\text I = \frac{9}{2 + 4} \\[1em] \text I = \frac{9}{6} \\[1em] \text I = 1.5\ \text A\end{aligned}$$

(b) When the key K is on : The 8 Ω resistor comes in parallel with the 4 Ω resistor,

$$\begin{aligned}\text R_\text p = \frac{4 \times 8}{4 + 8} \\[1em] \text R_\text p = \frac{32}{12} \\[1em] \text R_\text p = \frac{8}{3}\ \Omega\end{aligned}$$

The total resistance of the circuit is

$$\begin{aligned}2 + \frac{8}{3} = \frac{14}{3}\ \Omega \\[1em] \text I = \frac{9}{\frac{14}{3}} \\[1em] \text I = \frac{27}{14}\ \text A\end{aligned}$$

The ammeter carries the current in the 4 Ω branch,

$$\begin{aligned}\text I_\text A = \frac{27}{14} \times \frac{8}{4 + 8} \\[1em] \text I_\text A = \frac{27}{14} \times \frac{2}{3} \\[1em] \text I_\text A = \frac{9}{7}\ \text A\end{aligned}$$

Hence, the ammeter reads 1.5 A when K is off and $\frac{9}{7}$ A when K is on.

Given,

• Capacitance, C = 2 µF = 2 × 10^-6 F
• Current in the circuit, I = 2 A
• Resistances 2 Ω, 3 Ω and 5 Ω in series in the loop

In the steady state no current flows through the branch containing the capacitor, so the whole current of 2 A flows through the 2 Ω, 3 Ω and 5 Ω resistors in series. The potential difference across the plates of the capacitor is therefore equal to the potential drop across these three resistors,

$$\begin{aligned}\text V = \text I (2 + 3 + 5) \\[1em] \text V = 2 \times 10 \\[1em] \text V = 20\ \text V\end{aligned}$$

The energy stored in the capacitor is

$$\begin{aligned}\text U = \frac{1}{2}\ \text{CV}^2 \\[1em] \text U = \frac{1}{2} \times (2 \times 10^{-6}) \times (20)^2 \\[1em] \text U = \frac{1}{2} \times 2 \times 10^{-6} \times 400 \\[1em] \text U = 4 \times 10^{-4}\ \text J\end{aligned}$$

Hence, the energy stored in the capacitor is 4 × 10^-4 J.

Given,

• Emf of the cell = 2 V, in series with a 2 Ω resistor and the ammeter in one branch
• Three other branches, each of 2 Ω, connected between the same two points

The three 2 Ω resistors are in parallel,

$$\begin{aligned}\frac{1}{\text R_\text p} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} \\[1em] \frac{1}{\text R_\text p} = \frac{3}{2} \\[1em] \text R_\text p = \frac{2}{3}\ \Omega\end{aligned}$$

This combination is in series with the 2 Ω resistor of the branch containing the cell,

$$\begin{aligned}\text R = 2 + \frac{2}{3} \\[1em] \text R = \frac{8}{3}\ \Omega\end{aligned}$$

The ammeter reads the current drawn from the cell,

$$\begin{aligned}\text I = \frac{\text E}{\text R} \\[1em] \text I = \frac{2}{\frac{8}{3}} \\[1em] \text I = \frac{6}{8} \\[1em] \text I = 0.75\ \text A\end{aligned}$$

Hence, the ammeter reads 0.75 A.

Given,

• Ammeter reading, I = 2 A
• Voltmeter reading, V = 120 V
• R = 75 Ω

The voltmeter is connected across R, so both have the same potential difference of 120 V.

Current through R,

$$\begin{aligned}\text I_\text R = \frac{\text V}{\text R} \\[1em] \text I_\text R = \frac{120}{75} \\[1em] \text I_\text R = 1.6\ \text A\end{aligned}$$

The ammeter reads the total current, so the current through the voltmeter is

$$\begin{aligned}\text I_\text V = \text I - \text I_\text R \\[1em] \text I_\text V = 2 - 1.6 \\[1em] \text I_\text V = 0.4\ \text A\end{aligned}$$

The resistance of the voltmeter is

$$\begin{aligned}\text R_\text V = \frac{\text V}{\text I_\text V} \\[1em] \text R_\text V = \frac{120}{0.4} \\[1em] \text R_\text V = 300\ \Omega\end{aligned}$$

Hence, the resistance of the voltmeter is 300 Ω.

Given,

• R₁ = 10 Ω, R₂ = R₃ = 5 Ω
• E = 5 V, r = 0

(i) f, h, j

The lower wire h–g–f contains no resistance, so all the points on it are at the same potential. The point j lies at the lower end of R₁ and is joined to g by a plain wire, so j is also at this same potential.

(ii) $\frac{1}{2}$ A

Since the internal resistance of the cell is zero, the whole emf appears across R₁, which is connected between the upper and the lower wires,

$$\begin{aligned}\text I = \frac{\text E}{\text R_1} \\[1em] \text I = \frac{5}{10} \\[1em] \text I = \frac{1}{2}\ \text A\end{aligned}$$

(iii) 2.5 W

$$\begin{aligned}\text P = \frac{\text E^2}{\text R_1} \\[1em] \text P = \frac{(5)^2}{10} \\[1em] \text P = 2.5\ \text W\end{aligned}$$

(iv) 2.5 V

R₂ and R₃ are joined in series across the same 5 V, and they are equal, so the potential difference is shared equally between them,

$$\begin{aligned}\text V_3 = \frac{5}{2} \\[1em] \text V_3 = 2.5\ \text V\end{aligned}$$

Given,

• Emf of the standard cell, E = 1.08 V
• Balancing length, l = 91 cm = 0.91 m
• Emf of the driving cell = 2 V, internal resistance = 0
• Series resistance = 2 Ω
• Length of the potentiometer wire, L = 1 m

Let R be the resistance of the whole potentiometer wire. The current in the primary circuit is

$\text I = \frac{2}{2 + \text R}$

The resistance of 91 cm of the wire is 0.91 R, so the potential difference across this length is

$$\begin{aligned}\text V = \text I \times 0.91\ \text R \\[1em] \text V = \frac{2 \times 0.91\ \text R}{2 + \text R}\end{aligned}$$

At the balance point this equals the emf of the standard cell,

$$\begin{aligned}\frac{1.82\ \text R}{2 + \text R} = 1.08 \\[1em] 1.82\ \text R = 2.16 + 1.08\ \text R \\[1em] 0.74\ \text R = 2.16 \\[1em] \text R = 2.92\ \Omega\end{aligned}$$

Since the wire is 1 m long, the resistance per unit length is

$$\begin{aligned}\frac{\text R}{\text L} = \frac{2.92}{1} \\[1em] \frac{\text R}{\text L} = 2.92\ \Omega\text{/m}\end{aligned}$$

Hence, the resistance per unit length of the potentiometer wire is 2.92 Ω/m.

Given,

• E = 2 V, r = 0.1 Ω
• Charging current, I = 5 A

During charging the current is sent into the battery from outside, so

$$\begin{aligned}\text V = \text E + \text{Ir} \\[1em] \text V = 2 + (5 \times 0.1) \\[1em] \text V = 2 + 0.5 \\[1em] \text V = 2.5\ \text V\end{aligned}$$

Hence, the potential difference between the terminals of the battery is 2.5 V.

Given,

• Emf of each cell = E
• Internal resistances r₁ and r₂
• Voltmeter connected across the first cell reads zero

The two cells are joined in series, so the current in the circuit is

$\text I = \frac{2\text E}{\text R + \text r_1 + \text r_2}$

Expression for R : The voltmeter reads the terminal potential difference of the first cell, and this is zero,

$$\begin{aligned}\text E - \text{Ir}_1 = 0 \\[1em] \text I = \frac{\text E}{\text r_1}\end{aligned}$$

Equating the two expressions for I,

$$\begin{aligned}\frac{2\text E}{\text R + \text r_1 + \text r_2} = \frac{\text E}{\text r_1} \\[1em] 2\text r_1 = \text R + \text r_1 + \text r_2 \\[1em] \text R = \text r_1 - \text r_2\end{aligned}$$

Voltage across the cell of internal resistance r₂ :

$$\begin{aligned}\text V' = \text E - \text{Ir}_2 \\[1em] \text V' = \text E - \frac{\text E}{\text r_1}\ \text r_2 \\[1em] \text V' = \text E\left(\frac{\text r_1 - \text r_2}{\text r_1}\right)\end{aligned}$$

Given,

• Resistances in the two upper arms = 8 Ω and 12 Ω
• Length of the wire AB = 400 cm

The wire AB forms the lower two arms of a Wheatstone bridge, and the resistance of a portion of a uniform wire is proportional to its length. Hence at zero deflection,

$$\begin{aligned}\frac{8}{12} = \frac{\text{AJ}}{\text{JB}} \\[1em] \frac{8}{12} = \frac{\text{AJ}}{400 - \text{AJ}} \\[1em] 3\ \text{AJ} = 2 (400 - \text{AJ}) \\[1em] 3\ \text{AJ} + 2\ \text{AJ} = 800 \\[1em] \text{AJ} = 160\ \text{cm}\end{aligned}$$

Hence, the jockey should be placed at 160 cm from the end A.

A voltmeter draws some current from the cell, so it measures the terminal potential difference and not the emf. In a potentiometer, at the balance point no current is drawn from the cell in the secondary circuit, so it measures the true emf of the cell.

Emf : If dW is the work done by the cell in forcing a charge dq from its negative to its positive electrode, then the emf of the cell is

$\text E = \frac{\text{dW}}{\text{dq}}$

Derivation : Let a cell of emf E and internal resistance r send a steady current I through an external resistance R. In time dt the charge that flows is dq = I dt, so the work done by the cell is

$$\begin{aligned}\text{dW} = \text E\ \text{dq} \\[1em] \text{dW} = \text{EI}\ \text{dt}\end{aligned}$$

By Joule's law, the heat produced in time dt in the external resistance and in the internal resistance is

$\text{dW}_\text{ext} = \text I^2 \text R\ \text{dt}, \qquad \text{dW}_\text{int} = \text I^2 \text r\ \text{dt}$

By the conservation of energy,

$\text{EI}\ \text{dt} = \text I^2 \text R\ \text{dt} + \text I^2 \text r\ \text{dt}$

Dividing throughout by I dt,

$\text E = \text I (\text R + \text r)$

Given,

• E₁ = 4 V, E₂ = 6 V, internal resistances negligible
• R₁ = 2 Ω, R₂ = 12 Ω, R₃ = 4 Ω

Junction rule : The currents I₁ and I₂ meet at the junction B and the current (I₁ + I₂) flows through R₃.

Loop containing E₁ and R₃ :

$$\begin{aligned}4 = 2\ \text I_1 + 4 (\text I_1 + \text I_2) \\[1em] 4 = 6\ \text I_1 + 4\ \text I_2 \qquad ...(i)\end{aligned}$$

Loop containing E₂ and R₃ :

$$\begin{aligned}6 = 12\ \text I_2 + 4 (\text I_1 + \text I_2) \\[1em] 6 = 4\ \text I_1 + 16\ \text I_2 \qquad ...(ii)\end{aligned}$$

Dividing Eq. (i) by 2 and Eq. (ii) by 2,

$$\begin{aligned}2 = 3\ \text I_1 + 2\ \text I_2 \qquad ...(iii) \\[1em] 3 = 2\ \text I_1 + 8\ \text I_2 \qquad ...(iv)\end{aligned}$$

From Eq. (iii), I₂ = 1 − 1.5 I₁. Substituting in Eq. (iv),

$$\begin{aligned}3 = 2\ \text I_1 + 8 (1 - 1.5\ \text I_1) \\[1em] 3 = 8 - 10\ \text I_1 \\[1em] 10\ \text I_1 = 5 \\[1em] \text I_1 = 0.5\ \text A \\[1em] \text I_2 = 1 - (1.5 \times 0.5) \\[1em] \text I_2 = 0.25\ \text A\end{aligned}$$

Hence, I₁ = 0.5 A and I₂ = 0.25 A.

Given,

• R₁ = 2 Ω, R₂ = 5 Ω, Y = 15 Ω

When M and N are at the same potential, no current flows between them and the network is a balanced Wheatstone bridge,

$$\begin{aligned}\frac{\text R_1}{\text R_2} = \frac{\text X}{\text Y} \\[1em] \frac{2}{5} = \frac{\text X}{15} \\[1em] \text X = \frac{2 \times 15}{5} \\[1em] \text X = 6\ \Omega\end{aligned}$$

Hence, X = 6 Ω.

Given,

• Resistance of the wire XY = 9 Ω, length = 100 cm
• Emf of the accumulator = 4 V, internal resistance = 1 Ω
• Emf of the cell E = 1.8 V, balancing length XJ = 80 cm

The resistance of 80 cm of the wire is

$9 \times \frac{80}{100} = 7.2\ \Omega$

At the balance point the potential difference across XJ equals the emf of the cell E,

$$\begin{aligned}\text I \times 7.2 = 1.8 \\[1em] \text I = 0.25\ \text A\end{aligned}$$

The current in the primary circuit is

$$\begin{aligned}\text I = \frac{4}{9 + 1 + \text R} \\[1em] 0.25 = \frac{4}{10 + \text R} \\[1em] 10 + \text R = 16 \\[1em] \text R = 6\ \Omega\end{aligned}$$

Hence, the value of R is 6 Ω.

Given,

• Current of 3 A through the 2 Ω resistor from A to B
• Current of 2 A through the 6 Ω resistor from A to C
• Current I through the 4 Ω resistor from B to C

Applying Kirchhoff's second law to the closed loop ABCA,

$$\begin{aligned}(3 \times 2) + (\text I \times 4) - (2 \times 6) = 0 \\[1em] 6 + 4\ \text I - 12 = 0 \\[1em] 4\ \text I = 6 \\[1em] \text I = 1.5\ \text A\end{aligned}$$

Hence, the current flowing through the 4 Ω resistor is 1.5 A.

Given,

• Length of PQ = 4 m, resistance of PQ = 20 Ω
• X = 1 Ω, Y = 4 Ω
• Emf of the battery = 10 V, internal resistance = 1 Ω

(i) When the jockey does not touch the wire, no current flows through the galvanometer branch. The path P–X–Y–Q of resistance 1 + 4 = 5 Ω is then in parallel with the wire PQ of resistance 20 Ω,

$$\begin{aligned}\text R = \frac{5 \times 20}{5 + 20} \\[1em] \text R = \frac{100}{25} \\[1em] \text R = 4\ \Omega \\[1em] \text I = \frac{10}{4 + 1} \\[1em] \text I = 2\ \text A\end{aligned}$$

(ii) For no deflection in the galvanometer the bridge must be balanced,

$$\begin{aligned}\frac{\text X}{\text Y} = \frac{\text{PJ}}{\text{JQ}} \\[1em] \frac{1}{4} = \frac{\text{PJ}}{4 - \text{PJ}} \\[1em] 4 - \text{PJ} = 4\ \text{PJ} \\[1em] \text{PJ} = 0.8\ \text m\end{aligned}$$

Hence, the ammeter reads 2 A, and the jockey should be pressed at 0.8 m from P.

Given,

• E = 1.018 V
• Balancing length, l = 254.5 cm

At the balance point,

$$\begin{aligned}\text E = \text{Kl} \\[1em] \text K = \frac{\text E}{\text l} \\[1em] \text K = \frac{1.018}{254.5} \\[1em] \text K = 0.004\ \text{V/cm}\end{aligned}$$

Hence, the potential gradient across the potentiometer wire is 0.004 V/cm.

Given,

• E₁ = 4.0 V, r₁ = 1 Ω, with a 4 Ω resistor in its branch
• E₂ = 3.5 V, r₂ = 2 Ω, with a 3 Ω resistor in its branch
• Middle branch of 10 Ω carrying I₃

Junction rule :

$\text I_3 = \text I_1 + \text I_2 \qquad ...(i)$

Loop containing E₁ :

$$\begin{aligned}4.0 = \text I_1 (1 + 4) + 10\ \text I_3 \\[1em] 4.0 = 5\ \text I_1 + 10\ \text I_3 \qquad ...(ii)\end{aligned}$$

Loop containing E₂ :

$$\begin{aligned}3.5 = \text I_2 (2 + 3) + 10\ \text I_3 \\[1em] 3.5 = 5\ \text I_2 + 10\ \text I_3 \qquad ...(iii)\end{aligned}$$

Substituting Eq. (i) in Eqs. (ii) and (iii),

$$\begin{aligned}4.0 = 15\ \text I_1 + 10\ \text I_2 \qquad ...(iv) \\[1em] 3.5 = 10\ \text I_1 + 15\ \text I_2 \qquad ...(v)\end{aligned}$$

Multiplying Eq. (iv) by 3 and Eq. (v) by 2 and subtracting,

$$\begin{aligned}12 - 7 = (45 - 20)\ \text I_1 \\[1em] \text I_1 = \frac{5}{25} \\[1em] \text I_1 = 0.2\ \text A\end{aligned}$$

From Eq. (iv),

$$\begin{aligned}10\ \text I_2 = 4.0 - (15 \times 0.2) \\[1em] 10\ \text I_2 = 1.0 \\[1em] \text I_2 = 0.1\ \text A \\[1em] \text I_3 = 0.2 + 0.1 \\[1em] \text I_3 = 0.3\ \text A\end{aligned}$$

Hence, I₁ = 0.2 A, I₂ = 0.1 A and I₃ = 0.3 A.

Given,

• E = 6 V, r = 1 Ω
• R₁ = 3 Ω, R₂ = 7 Ω, R₃ = 2 Ω, R₄ = 12 Ω, R₅ = 8 Ω, R₆ = 5 Ω

The path from the junction of R₁ and R₂ to the junction of R₃ and R₄, taken through R₂, R₆ and R₄, has resistance

$7 + 5 + 12 = 24\ \Omega$

This is in parallel with R₅ = 8 Ω joining the same two junctions,

$$\begin{aligned}\text R_\text p = \frac{8 \times 24}{8 + 24} \\[1em] \text R_\text p = \frac{192}{32} \\[1em] \text R_\text p = 6\ \Omega\end{aligned}$$

This combination is in series with R₁ and R₃,

$$\begin{aligned}\text R = 3 + 6 + 2 \\[1em] \text R = 11\ \Omega\end{aligned}$$

The reading of the ammeter is the current drawn from the battery,

$$\begin{aligned}\text I = \frac{\text E}{\text R + \text r} \\[1em] \text I = \frac{6}{11 + 1} \\[1em] \text I = \frac{6}{12} \\[1em] \text I = 0.5\ \text A\end{aligned}$$

Hence, the ammeter reads 0.5 A.

4 Ω

The five resistors form a Wheatstone bridge with 2 Ω and 3 Ω in the upper branch, 8 Ω and 12 Ω in the lower branch and 6 Ω as the bridge arm. Testing the condition of balance,

$$\begin{aligned}\frac{2}{3} = \frac{8}{12} \\[1em] = \frac{2}{3}\end{aligned}$$

The bridge is balanced, so no current flows through the 6 Ω resistor and it may be omitted. The upper branch gives 2 + 3 = 5 Ω and the lower branch 8 + 12 = 20 Ω, and these are in parallel,

$$\begin{aligned}\text R = \frac{5 \times 20}{5 + 20} \\[1em] \text R = \frac{100}{25} \\[1em] \text R = 4\ \Omega\end{aligned}$$

(i) In each row the n cells are joined in series, so the emf of one row is ne and its internal resistance is nr.

The m rows are joined in parallel and all have the same emf, so the emf of the whole battery is

$\text e_\text{eq} = \text{ne}$

The internal resistances of the m rows, each equal to nr, are in parallel, so

$\text r_\text{eq} = \frac{\text{nr}}{\text m}$

(ii) The current through the external resistance is

$$\begin{aligned}\text I = \frac{\text e_\text{eq}}{\text R + \text r_\text{eq}} \\[1em] \text I = \frac{\text{ne}}{\text R + \frac{\text{nr}}{\text m}}\end{aligned}$$

Multiplying the numerator and the denominator by m,

$\text I = \frac{\text{mne}}{\text{mR} + \text{nr}}$

Since N = mn,

$\text I = \frac{\text{Ne}}{\text{mR} + \text{nr}}$

Given,

• Arms of the network : 3 Ω and 15 Ω in the upper branch, 30 Ω and 60 Ω in the lower branch

For M and N to be at the same potential the bridge must be balanced. If X is the required resistance of the second upper arm, then

$$\begin{aligned}\frac{3}{\text X} = \frac{30}{60} \\[1em] \frac{3}{\text X} = \frac{1}{2} \\[1em] \text X = 6\ \Omega\end{aligned}$$

Since 6 Ω is less than 15 Ω, a resistance Y must be connected in parallel with the 15 Ω resistor,

$$\begin{aligned}\frac{15\ \text Y}{15 + \text Y} = 6 \\[1em] 15\ \text Y = 90 + 6\ \text Y \\[1em] 9\ \text Y = 90 \\[1em] \text Y = 10\ \Omega\end{aligned}$$

Hence, a resistance of 10 Ω should be connected in parallel with the 15 Ω resistor.

Given,

• E₁ = 17 V, E₂ = 21 V
• R₁ = 2 Ω, R₂ = 3 Ω, R₃ = 5 Ω

Junction rule :

$\text I_3 = \text I_1 + \text I_2 \qquad ...(i)$

Loop containing E₁, R₁ and R₃ :

$17 = 2\ \text I_1 + 5\ \text I_3 \qquad ...(ii)$

Loop containing E₂, R₂ and R₃ :

$21 = 3\ \text I_2 + 5\ \text I_3 \qquad ...(iii)$

Substituting Eq. (i) in Eqs. (ii) and (iii),

$$\begin{aligned}17 = 7\ \text I_1 + 5\ \text I_2 \qquad ...(iv) \\[1em] 21 = 5\ \text I_1 + 8\ \text I_2 \qquad ...(v)\end{aligned}$$

Multiplying Eq. (iv) by 8 and Eq. (v) by 5 and subtracting,

$$\begin{aligned}136 - 105 = (56 - 25)\ \text I_1 \\[1em] 31 = 31\ \text I_1 \\[1em] \text I_1 = 1\ \text A\end{aligned}$$

From Eq. (iv),

$$\begin{aligned}5\ \text I_2 = 17 - 7 \\[1em] 5\ \text I_2 = 10 \\[1em] \text I_2 = 2\ \text A \\[1em] \text I_3 = 1 + 2 \\[1em] \text I_3 = 3\ \text A\end{aligned}$$

Hence, the currents through R₁, R₂ and R₃ are 1 A, 2 A and 3 A respectively.

The battery consists of two rows connected in parallel between the same two terminals.

In the upper row the cells of 1 V, 2 V and 3 V are joined in series, so the emf of this row is

$1 + 2 + 3 = 6\ \text V$

The lower row contains a single cell of 6 V.

Since the two rows have the same emf and are connected in parallel, the emf of the combination is equal to the emf of one row.

Hence, the emf of the battery is 6 V.

The potentiometer wire AB is connected to the driving cell B₁ through a key and a rheostat, forming the primary circuit. The positive terminals of the two cells E₁ and E₂ are joined to the end A, and their negative terminals are taken through the keys K₁ and K₂ and the galvanometer to the jockey J.

With K₁ closed and K₂ open the balance point is obtained at a length l₁, and with K₂ closed and K₁ open it is obtained at a length l₂.

Working formula :

$\frac{\text E_1}{\text E_2} = \frac{\text l_1}{\text l_2}$

The current in the circuit is I = V/R, and the terminal potential difference is V = e − Ir. Therefore,

$$\begin{aligned}\text V = \text e - \frac{\text V}{\text R}\ \text r \\[1em] \text e = \text V\left(1 + \frac{\text r}{\text R}\right) \\[1em] \text e = \frac{\text V (\text R + \text r)}{\text R}\end{aligned}$$

The relation may also be written as

$\text r = \text R\left(\frac{\text e}{\text V} - 1\right)$

Given,

• Emf of each cell = 2 V, internal resistance of each cell = 1 Ω, three cells in series
• Two resistors of 6 Ω each in parallel

Emf and internal resistance of the battery :

$$\begin{aligned}\text E = 3 \times 2 \\[1em] \text E = 6\ \text V \\[1em] \text r = 3 \times 1 \\[1em] \text r = 3\ \Omega\end{aligned}$$

External resistance :

$$\begin{aligned}\text R = \frac{6 \times 6}{6 + 6} \\[1em] \text R = 3\ \Omega\end{aligned}$$

Current delivered :

$$\begin{aligned}\text I = \frac{\text E}{\text R + \text r} \\[1em] \text I = \frac{6}{3 + 3} \\[1em] \text I = 1\ \text A\end{aligned}$$

Hence, the current delivered by the battery is 1 A.

Given,

• Length of the wire, L = 10 m, resistance of the wire = 20 Ω
• Series resistance = 480 Ω
• Emf of the battery = 5 V, internal resistance negligible
• Balancing length, l = 6 m

(i) Potential gradient : The current in the primary circuit is

$$\begin{aligned}\text I = \frac{5}{20 + 480} \\[1em] \text I = \frac{5}{500} \\[1em] \text I = 0.01\ \text A\end{aligned}$$

The potential difference across the potentiometer wire is

$$\begin{aligned}\text V = 0.01 \times 20 \\[1em] \text V = 0.2\ \text V \\[1em] \text K = \frac{\text V}{\text L} \\[1em] \text K = \frac{0.2}{10} \\[1em] \text K = 0.02\ \text{V/m}\end{aligned}$$

(ii) Unknown emf :

$$\begin{aligned}\text e = \text{Kl} \\[1em] \text e = 0.02 \times 6 \\[1em] \text e = 0.12\ \text V\end{aligned}$$

Hence, the potential gradient is 0.02 V/m and the unknown emf is 0.12 V.

Given,

• Resistances 3 Ω and 6 Ω in the upper branch, 5 Ω and X in the lower branch

For A and B to be at the same potential no current flows between them, so the network is a balanced Wheatstone bridge,

$$\begin{aligned}\frac{3}{6} = \frac{5}{\text X} \\[1em] \text X = \frac{5 \times 6}{3} \\[1em] \text X = 10\ \Omega\end{aligned}$$

Hence, X = 10 Ω.

Given,

• E₁ = 2 V, E₂ = 3 V, internal resistances negligible
• R₁ = 1 Ω, R₂ = 6 Ω, R₃ = 2 Ω

Junction rule : The current through R₃ is (I₁ + I₂).

Loop containing E₁, R₁ and R₃ :

$$\begin{aligned}2 = 1 \times \text I_1 + 2 (\text I_1 + \text I_2) \\[1em] 2 = 3\ \text I_1 + 2\ \text I_2 \qquad ...(i)\end{aligned}$$

Loop containing E₂, R₂ and R₃ :

$$\begin{aligned}3 = 6\ \text I_2 + 2 (\text I_1 + \text I_2) \\[1em] 3 = 2\ \text I_1 + 8\ \text I_2 \qquad ...(ii)\end{aligned}$$

Multiplying Eq. (i) by 4 and subtracting Eq. (ii),

$$\begin{aligned}8 - 3 = (12 - 2)\ \text I_1 \\[1em] 5 = 10\ \text I_1 \\[1em] \text I_1 = 0.5\ \text A\end{aligned}$$

From Eq. (i),

$$\begin{aligned}2\ \text I_2 = 2 - (3 \times 0.5) \\[1em] 2\ \text I_2 = 0.5 \\[1em] \text I_2 = 0.25\ \text A\end{aligned}$$

Hence, I₁ = 0.5 A and I₂ = 0.25 A.

The junction rule follows from the conservation of charge, and the loop rule follows from the conservation of energy.

Given,

• Resistance in the left gap, R = 2 Ω
• Balancing length, l = 40 cm
• Required balancing length, l' = 50 cm

Value of X :

$$\begin{aligned}\frac{2}{\text X} = \frac{40}{60} \\[1em] \text X = 3\ \Omega\end{aligned}$$

For the null point at 50 cm, the resistance X' of the right gap must be

$$\begin{aligned}\frac{2}{\text X'} = \frac{50}{50} \\[1em] \frac{2}{\text X'} = 1 \\[1em] \text X' = 2\ \Omega\end{aligned}$$

Since 2 Ω is less than 3 Ω, a resistance Y must be connected in parallel with X,

$$\begin{aligned}\frac{3\ \text Y}{3 + \text Y} = 2 \\[1em] 3\ \text Y = 6 + 2\ \text Y \\[1em] \text Y = 6\ \Omega\end{aligned}$$

Hence, a resistance of 6 Ω should be connected in parallel with X.

Given,

• R₁ = 200 Ω, R₂ = 880 Ω
• Emf of the battery = 40 V, internal resistance negligible
• Resistance of the voltmeter = 300 Ω

The voltmeter is in parallel with R₁,

$$\begin{aligned}\text R_\text p = \frac{200 \times 300}{200 + 300} \\[1em] \text R_\text p = \frac{60000}{500} \\[1em] \text R_\text p = 120\ \Omega\end{aligned}$$

The total resistance of the circuit is

$$\begin{aligned}120 + 880 = 1000\ \Omega \\[1em] \text I = \frac{40}{1000} \\[1em] \text I = 0.04\ \text A\end{aligned}$$

The reading of the voltmeter is the potential difference across the parallel combination,

$$\begin{aligned}\text V = 0.04 \times 120 \\[1em] \text V = 4.8\ \text V\end{aligned}$$

Hence, the voltmeter reads 4.8 V.

Given,

• E₁ = 2 V, l₁ = 120 cm
• E₂ = 1.5 V

Since the balancing length is proportional to the emf,

$$\begin{aligned}\frac{\text E_1}{\text E_2} = \frac{\text l_1}{\text l_2} \\[1em] \text l_2 = \text l_1 \times \frac{\text E_2}{\text E_1} \\[1em] \text l_2 = 120 \times \frac{1.5}{2} \\[1em] \text l_2 = 90\ \text{cm}\end{aligned}$$

Hence, the balancing length for the other cell is 90 cm.

Given,

• ε₁ = 34 V, r₁ = 1 Ω ; ε₂ = 10 V, r₂ = 2 Ω
• Resistances 4 Ω and 4 Ω in the left branch, 7 Ω and 7 Ω in the right branch, and 5 Ω in the middle branch carrying (I₁ + I₂)

Loop containing ε₁ :

$$\begin{aligned}34 = \text I_1 (1 + 4 + 7) + 5 (\text I_1 + \text I_2) \\[1em] 34 = 17\ \text I_1 + 5\ \text I_2 \qquad ...(i)\end{aligned}$$

Loop containing ε₂ :

$$\begin{aligned}10 = \text I_2 (2 + 4 + 7) + 5 (\text I_1 + \text I_2) \\[1em] 10 = 5\ \text I_1 + 18\ \text I_2 \qquad ...(ii)\end{aligned}$$

Multiplying Eq. (i) by 18 and Eq. (ii) by 5 and subtracting,

$$\begin{aligned}612 - 50 = (306 - 25)\ \text I_1 \\[1em] 562 = 281\ \text I_1 \\[1em] \text I_1 = 2\ \text A\end{aligned}$$

From Eq. (ii),

$$\begin{aligned}18\ \text I_2 = 10 - (5 \times 2) \\[1em] 18\ \text I_2 = 0 \\[1em] \text I_2 = 0\end{aligned}$$

Hence, I₁ = 2 A and I₂ = 0.

Given,

• R = 2 Ω, l₁ = 100 cm
• l₂ = 500 cm

The same current flows through both the resistors, so the potential difference across each is proportional to its resistance, and hence to the balancing length,

$$\begin{aligned}\frac{\text X}{\text R} = \frac{\text l_2}{\text l_1} \\[1em] \text X = 2 \times \frac{500}{100} \\[1em] \text X = 10\ \Omega\end{aligned}$$

Hence, the unknown resistance is 10 Ω.

The arrangement consists of two rows connected in parallel between A and B.

In the upper row the two cells of 4 V each are joined in series, so the emf of this row is

$4 + 4 = 8\ \text V$

The lower row contains a single cell of 8 V.

Since both the rows have the same emf and are connected in parallel, the net emf of the combination is equal to the emf of one row.

Hence, the net emf across A and B is 8 V.

(i) Balancing condition : Four resistances P, Q, R and S are connected in the four arms of the bridge, a cell E is connected across one diagonal and a galvanometer G across the other.

The bridge is balanced when the galvanometer shows no deflection. In this condition no current flows through the galvanometer, so the two junctions joined by it are at the same potential.

Let the current I₁ flow through P and Q, and the current I₂ through R and S. Then the potential drops across P and R are equal, and so are those across Q and S,

$$\begin{aligned}\text I_1 \text P = \text I_2 \text R \qquad ...(i) \\[1em] \text I_1 \text Q = \text I_2 \text S \qquad ...(ii)\end{aligned}$$

Dividing Eq. (i) by Eq. (ii),

$\frac{\text P}{\text Q} = \frac{\text R}{\text S}$

(ii) Potentiometer for internal resistance :

The cell of emf E whose internal resistance is to be measured is joined in the secondary circuit through a galvanometer, and a resistance box R with a key is connected across it. With the key open the balancing length is l₁, and with a resistance R introduced and the key closed it is l₂.

Working formula :

$\text r = \text R \left(\frac{\text l_1}{\text l_2} - 1\right)$

Given,

• E₁ = 3 V, r₁ = 2 Ω, with R₁ = 4 Ω in its branch
• E₂ = 4 V, r₂ = 1 Ω, with R₂ = 7 Ω in its branch
• R₃ = 8 Ω in the middle branch carrying (I₁ + I₂)

Loop containing E₁ :

$$\begin{aligned}3 = \text I_1 (2 + 4) + 8 (\text I_1 + \text I_2) \\[1em] 3 = 14\ \text I_1 + 8\ \text I_2 \qquad ...(i)\end{aligned}$$

Loop containing E₂ :

$$\begin{aligned}4 = \text I_2 (1 + 7) + 8 (\text I_1 + \text I_2) \\[1em] 4 = 8\ \text I_1 + 16\ \text I_2 \qquad ...(ii)\end{aligned}$$

Multiplying Eq. (i) by 2 and subtracting Eq. (ii),

$$\begin{aligned}6 - 4 = (28 - 8)\ \text I_1 \\[1em] 2 = 20\ \text I_1 \\[1em] \text I_1 = 0.1\ \text A\end{aligned}$$

From Eq. (ii),

$$\begin{aligned}16\ \text I_2 = 4 - (8 \times 0.1) \\[1em] 16\ \text I_2 = 3.2 \\[1em] \text I_2 = 0.2\ \text A\end{aligned}$$

Hence, I₁ = 0.1 A and I₂ = 0.2 A.

Given,

• Length of AB = 2 m, resistance of AB = 8 Ω
• E₁ = 4 V, r₁ = 1 Ω, series resistance R = 7 Ω
• E₂ = 1.5 V, r₂ = 2 Ω

(i) (a) Current through AB : When the jockey does not touch the wire, no current is drawn by the secondary circuit, so the current in the primary circuit is

$$\begin{aligned}\text I = \frac{\text E_1}{\text r_1 + \text R + \text R_\text{AB}} \\[1em] \text I = \frac{4}{1 + 7 + 8} \\[1em] \text I = \frac{4}{16} \\[1em] \text I = 0.25\ \text A\end{aligned}$$

(b) Potential gradient : The potential difference across AB is

$$\begin{aligned}\text V = 0.25 \times 8 \\[1em] \text V = 2\ \text V \\[1em] \text K = \frac{\text V}{\text L} \\[1em] \text K = \frac{2}{2} \\[1em] \text K = 1\ \text{V m}^{-1}\end{aligned}$$

(ii) Length AC : At the balance point the potential difference across AC equals the emf of E₂,

$$\begin{aligned}\text E_2 = \text K \times \text{AC} \\[1em] \text{AC} = \frac{1.5}{1} \\[1em] \text{AC} = 1.5\ \text m\end{aligned}$$

Hence, I = 0.25 A, K = 1 V m^-1 and AC = 1.5 m.

Four resistances P, Q, R and S are connected in the four arms of the bridge. A cell E with a key is connected across one diagonal and a galvanometer G across the other diagonal.

The bridge is said to be balanced when the galvanometer shows no deflection. In this condition no current flows through the galvanometer, so the two junctions joined by it are at the same potential.

Let the current I₁ flow through P and Q, and the current I₂ through R and S. Since the two junctions are at the same potential, the potential drops across P and R are equal, and so are those across Q and S,

$$\begin{aligned}\text I_1 \text P = \text I_2 \text R \qquad ...(i) \\[1em] \text I_1 \text Q = \text I_2 \text S \qquad ...(ii)\end{aligned}$$

Dividing Eq. (i) by Eq. (ii),

$\frac{\text P}{\text Q} = \frac{\text R}{\text S}$

This is the balancing condition of the Wheatstone bridge.

The potentiometer wire AB is joined to the driving cell B₁ through a key K₁ and a rheostat, forming the primary circuit. The cell of emf E whose internal resistance is to be determined is connected between the end A and the jockey J through a galvanometer, and a resistance box R with a key K₂ is connected across this cell.

With the key K₂ open the balancing length l₁ corresponds to the emf E, and with a resistance R introduced and K₂ closed the balancing length l₂ corresponds to the terminal potential difference V. The internal resistance is then

$\text r = \text R \left(\frac{\text l_1}{\text l_2} - 1\right)$

Given,

• Resistance in the left gap = 4 Ω
• Balancing length, l = 40 cm

(i) Value of R :

$$\begin{aligned}\frac{4}{\text R} = \frac{40}{100 - 40} \\[1em] \frac{4}{\text R} = \frac{40}{60} \\[1em] \text R = \frac{4 \times 60}{40} \\[1em] \text R = 6\ \Omega\end{aligned}$$

(ii) For the null point at the mid-point, l' = 50 cm, so the resistance of the right gap must be

$$\begin{aligned}\frac{4}{\text R'} = \frac{50}{50} \\[1em] \frac{4}{\text R'} = 1 \\[1em] \text R' = 4\ \Omega\end{aligned}$$

A resistance Y must therefore be connected in parallel with R,

$$\begin{aligned}\frac{6\ \text Y}{6 + \text Y} = 4 \\[1em] 6\ \text Y = 24 + 4\ \text Y \\[1em] 2\ \text Y = 24 \\[1em] \text Y = 12\ \Omega\end{aligned}$$

Hence, R = 6 Ω, and a 12 Ω resistance should be connected in parallel with R so that their equivalent resistance becomes 4 Ω.

Given,

• Number of cells, n = 6, each of emf 1.5 V and internal resistance 0.5 Ω
• External resistance, R = 9 Ω

(i) Emf of the battery :

$$\begin{aligned}\text E = 6 \times 1.5 \\[1em] \text E = 9\ \text V\end{aligned}$$

(ii) Current in the circuit : The total internal resistance is

$$\begin{aligned}\text r = 6 \times 0.5 \\[1em] \text r = 3\ \Omega \\[1em] \text I = \frac{\text E}{\text R + \text r} \\[1em] \text I = \frac{9}{9 + 3} \\[1em] \text I = \frac{9}{12} \\[1em] \text I = 0.75\ \text A\end{aligned}$$

Hence, the emf of the battery is 9 V and the current in the circuit is 0.75 A.

The current drawn from a cell of constant emf E and internal resistance r is

$\text I = \frac{\text E}{\text R + \text r}$

Thus I decreases as R increases, but not linearly. When R = 0 the current has its maximum value E/r, and as R becomes very large the current tends to zero without ever becoming negative. The graph is therefore a curve falling steeply at first and then flattening out, which is shown only in option (d).

2.25V

Given, I = 1.5 A, R = 1.0 Ω and r = 0.5 Ω.

$$\begin{aligned}\text E = \text I (\text R + \text r) \\[1em] \text E = 1.5 (1.0 + 0.5) \\[1em] \text E = 1.5 \times 1.5 \\[1em] \text E = 2.25\ \text V\end{aligned}$$

MNe/(MR + Nr)

Each row of N cells in series has an emf Ne and an internal resistance Nr. The M rows in parallel give an emf Ne and an internal resistance Nr/M. Hence

$$\begin{aligned}\text I = \frac{\text{Ne}}{\text R + \frac{\text{Nr}}{\text M}} \\[1em] \text I = \frac{\text{MNe}}{\text{MR} + \text{Nr}}\end{aligned}$$

charge

The junction rule states that the total current arriving at a junction equals the total current leaving it. Since current is the rate of flow of charge, no charge is accumulated at the junction, so the rule follows from the conservation of charge.

resistance of a wire

At balance the Wheatstone bridge gives P/Q = R/S, so an unknown resistance can be found from three known resistances. The meter bridge, which is the practical form of the Wheatstone bridge, is used to determine the resistance of a wire and hence the specific resistance of its material.

30.0 cm

Since the balancing length is proportional to the emf,

$$\begin{aligned}\frac{\text E_1}{\text E_2} = \frac{\text l_1}{\text l_2} \\[1em] \text l_2 = 21.6 \times \frac{1.50}{1.08} \\[1em] \text l_2 = 30.0\ \text{cm}\end{aligned}$$

2.50 Vm^-1

$$\begin{aligned}\text K = \frac{\text V}{\text l} \\[1em] \text K = \frac{2}{80\ \text{cm}} \\[1em] \text K = \frac{2}{0.80\ \text m} \\[1em] \text K = 2.50\ \text{V m}^{-1}\end{aligned}$$

0

Testing the condition of balance for the bridge as connected in the figure,

$$\begin{aligned}\frac{\text P}{\text Q} = \frac{10}{20} \\[1em] \frac{\text P}{\text Q} = \frac{1}{2} \\[1em] \frac{\text S}{\text R} = \frac{100}{200} \\[1em] \frac{\text S}{\text R} = \frac{1}{2}\end{aligned}$$

The two ratios are equal, so the bridge is balanced. The two junctions joined by the galvanometer are then at the same potential and no current flows through it.

Given,

• n = 8 cells in each row, each of emf 2 V and internal resistance 3 Ω
• m = 6 rows in parallel
• External resistance, R = 6 Ω

(i) Emf of the battery : All the rows have the same emf, so

$$\begin{aligned}\text E = \text{ne} \\[1em] \text E = 8 \times 2 \\[1em] \text E = 16\ \text V\end{aligned}$$

(ii) Internal resistance : Each row has an internal resistance of 8 × 3 = 24 Ω, and the six rows are in parallel,

$$\begin{aligned}\text r = \frac{\text{nr}}{\text m} \\[1em] \text r = \frac{24}{6} \\[1em] \text r = 4\ \Omega\end{aligned}$$

(iii) Current through R :

$$\begin{aligned}\text I = \frac{\text E}{\text R + \text r} \\[1em] \text I = \frac{16}{6 + 4} \\[1em] \text I = 1.6\ \text A\end{aligned}$$

Hence, E = 16 V, r = 4 Ω and I = 1.6 A.

Given,

• E₁ = 25 V, r₁ = 1 Ω, with a 4 Ω resistor in its branch
• E₂ = 26 V, r₂ = 5 Ω, with a 3 Ω resistor in its branch
• A 2 Ω resistor in the common branch carrying (I₁ + I₂)

Loop containing E₁ :

$$\begin{aligned}25 = \text I_1 (1 + 4) + 2 (\text I_1 + \text I_2) \\[1em] 25 = 7\ \text I_1 + 2\ \text I_2 \qquad ...(i)\end{aligned}$$

Loop containing E₂ :

$$\begin{aligned}26 = \text I_2 (5 + 3) + 2 (\text I_1 + \text I_2) \\[1em] 26 = 2\ \text I_1 + 10\ \text I_2 \qquad ...(ii)\end{aligned}$$

Multiplying Eq. (i) by 5 and subtracting Eq. (ii),

$$\begin{aligned}125 - 26 = (35 - 2)\ \text I_1 \\[1em] 99 = 33\ \text I_1 \\[1em] \text I_1 = 3\ \text A\end{aligned}$$

From Eq. (i),

$$\begin{aligned}2\ \text I_2 = 25 - 21 \\[1em] 2\ \text I_2 = 4 \\[1em] \text I_2 = 2\ \text A\end{aligned}$$

Hence, I₁ = 3 A and I₂ = 2 A.

Given,

• External resistance, R = 10 Ω
• Current in series = current in parallel

For two identical cells of emf E and internal resistance r,

$$\begin{aligned}\frac{2\text E}{\text R + 2\text r} = \frac{\text E}{\text R + \frac{\text r}{2}} \\[1em] 2\left(\text R + \frac{\text r}{2}\right) = \text R + 2\text r \\[1em] 2\text R + \text r = \text R + 2\text r \\[1em] \text R = \text r\end{aligned}$$

Hence, the internal resistance of each cell is 10 Ω.

Given,

• Cell of 29 V with internal resistance 1 Ω, with 3 Ω and 4 Ω resistors in its branch, carrying I₁
• Cell of 47 V with internal resistance 2 Ω, with 5 Ω and 6 Ω resistors in its branch, carrying I₂
• A 7 Ω resistor in the common branch BD carrying (I₁ + I₂)

Loop containing the 29 V cell :

$$\begin{aligned}29 = \text I_1 (1 + 3 + 4) + 7 (\text I_1 + \text I_2) \\[1em] 29 = 15\ \text I_1 + 7\ \text I_2 \qquad ...(i)\end{aligned}$$

Loop containing the 47 V cell :

$$\begin{aligned}47 = \text I_2 (2 + 5 + 6) + 7 (\text I_1 + \text I_2) \\[1em] 47 = 7\ \text I_1 + 20\ \text I_2 \qquad ...(ii)\end{aligned}$$

Multiplying Eq. (i) by 20 and Eq. (ii) by 7 and subtracting,

$$\begin{aligned}580 - 329 = (300 - 49)\ \text I_1 \\[1em] 251 = 251\ \text I_1 \\[1em] \text I_1 = 1\ \text A\end{aligned}$$

From Eq. (i),

$$\begin{aligned}7\ \text I_2 = 29 - 15 \\[1em] 7\ \text I_2 = 14 \\[1em] \text I_2 = 2\ \text A\end{aligned}$$

Hence, I₁ = 1 A and I₂ = 2 A.

remain same

The balancing condition of the bridge, R/S = l/(100 − l), is obtained from the fact that at balance no current flows through the galvanometer and the two junctions across it are at the same potential.

This condition involves only the four resistances of the arms, and is unaltered if the cell and the galvanometer are interchanged. Hence the balance point remains at the same position.

The emf of the battery is e.

When identical cells are connected in parallel, the positive terminals are joined together and so are the negative terminals. The emf of the combination is therefore the same as that of a single cell, that is, e. Only the internal resistance is reduced, becoming r/3.

Given,

• Total number of cells = 40, each of emf 1.45 V and internal resistance 0.5 Ω
• The cells are arranged in 4 rows of 10 cells each, the rows being in parallel

(i) Emf of the battery : In each row the 10 cells are in series,

$$\begin{aligned}\text E = 10 \times 1.45 \\[1em] \text E = 14.5\ \text V\end{aligned}$$

All the four rows have the same emf and are in parallel, so the emf of the battery is 14.5 V.

(ii) Internal resistance : Each row has an internal resistance of 10 × 0.5 = 5 Ω, and the four rows are in parallel,

$$\begin{aligned}\text r = \frac{5}{4} \\[1em] \text r = 1.25\ \Omega\end{aligned}$$

Hence, the emf of the battery is 14.5 V and its internal resistance is 1.25 Ω.

(i)

Given,

• E₁ = 18 V, r₁ = 1 Ω ; E₂ = 10 V, r₂ = 2 Ω
• W₁ (AC) = 8 Ω, W₂ (FD) = 6 Ω, W₃ (BE) = 10 Ω

Since B and E are the mid-points of W₁ and W₂,

$$\begin{aligned}\text{AB} = \text{BC} = \frac{8}{2} \\[1em] \text{AB} = \text{BC} = 4\ \Omega \\[1em] \text{FE} = \text{ED} = \frac{6}{2} \\[1em] \text{FE} = \text{ED} = 3\ \Omega\end{aligned}$$

Let I₁ be the current in the upper loop and I₂ that in the lower loop, so that the current in W₃ is I₃ = I₁ + I₂.

Upper loop :

$$\begin{aligned}18 = \text I_1 (1 + 4 + 3) + 10 (\text I_1 + \text I_2) \\[1em] 18 = 18\ \text I_1 + 10\ \text I_2 \qquad ...(i)\end{aligned}$$

Lower loop :

$$\begin{aligned}10 = \text I_2 (2 + 4 + 3) + 10 (\text I_1 + \text I_2) \\[1em] 10 = 10\ \text I_1 + 19\ \text I_2 \qquad ...(ii)\end{aligned}$$

Multiplying Eq. (i) by 19 and Eq. (ii) by 10 and subtracting,

$$\begin{aligned}342 - 100 = (342 - 100)\ \text I_1 \\[1em] 242 = 242\ \text I_1 \\[1em] \text I_1 = 1\ \text A\end{aligned}$$

From Eq. (i),

$$\begin{aligned}10\ \text I_2 = 18 - 18 \\[1em] 10\ \text I_2 = 0 \\[1em] \text I_2 = 0 \\[1em] \text I_3 = 1 + 0 \\[1em] \text I_3 = 1\ \text A\end{aligned}$$

Hence, the current flowing in the wire W₃ is 1 A.

(ii)

Given,

• Emf of the driver cell, E_D = 6 V, internal resistance = 2 Ω
• Length of AB = 10 m, resistance of AB = 28 Ω
• Series resistance, R_s = 20 Ω
• Balancing length, AC = 4.5 m

(a) Current in the potentiometer wire : When the jockey does not touch the wire, no current is drawn by the secondary circuit,

$$\begin{aligned}\text I_\text p = \frac{6}{2 + 20 + 28} \\[1em] \text I_\text p = \frac{6}{50} \\[1em] \text I_\text p = 0.12\ \text A\end{aligned}$$

(b) Emf of the cell X : The potential difference across AB is

$$\begin{aligned}\text V = 0.12 \times 28 \\[1em] \text V = 3.36\ \text V \\[1em] \text K = \frac{3.36}{10} \\[1em] \text K = 0.336\ \text{V m}^{-1} \\[1em] \text E_\text X = \text K \times \text{AC} \\[1em] \text E_\text X = 0.336 \times 4.5 \\[1em] \text E_\text X = 1.512\ \text V\end{aligned}$$

Hence, I_p = 0.12 A and the emf of the cell X is 1.512 V.

Kirchhoff's junction rule is based on the conservation of charge.

$\frac{\text E_0 \text r}{(\text r + \text r_1)} \cdot \frac{\text l}{\text L}$

Reason — The current in the potentiometer wire is

$\text I = \frac{\text E_0}{\text r + \text r_1}$

The potential difference across the whole wire of resistance r is

$$\begin{aligned}\text V = \text{Ir} \\[1em] \text V = \frac{\text E_0 \text r}{\text r + \text r_1}\end{aligned}$$

Hence the potential gradient along the wire is

$$\begin{aligned}\text K = \frac{\text V}{\text L} \\[1em] \text K = \frac{\text E_0 \text r}{(\text r + \text r_1)\text L}\end{aligned}$$

At the balance point,

$$\begin{aligned}\text E = \text{Kl} \\[1em] \text E = \frac{\text E_0 \text r}{(\text r + \text r_1)} \cdot \frac{\text l}{\text L}\end{aligned}$$

V_A = V_B = V_C

Reason — The voltmeters B and C are connected in parallel, so their equivalent resistance is

$$\begin{aligned}\text R_\text p = \frac{1.5\ \text R \times 3\ \text R}{1.5\ \text R + 3\ \text R} \\[1em] \text R_\text p = \frac{4.5\ \text R^2}{4.5\ \text R} \\[1em] \text R_\text p = \text R\end{aligned}$$

This parallel combination is in series with A, whose resistance is also R. Since the same current passes through A and through the combination, and their resistances are equal, the potential difference across each is the same.

Being in parallel, B and C have the same potential difference across them. Hence V_A = V_B = V_C.

32 Ω

Reason — The required potential gradient is

$$\begin{aligned}\text K = 1\ \text{mV/cm} \\[1em] \text K = 0.1\ \text{V/m}\end{aligned}$$

The potential difference across the whole wire of length 4 m is

$$\begin{aligned}\text V = \text{KL} \\[1em] \text V = 0.1 \times 4 \\[1em] \text V = 0.4\ \text V\end{aligned}$$

The current in the wire is

$$\begin{aligned}\text I = \frac{\text V}{\text R_\text{wire}} \\[1em] \text I = \frac{0.4}{8} \\[1em] \text I = 0.05\ \text A\end{aligned}$$

The total resistance of the circuit is

$\frac{2}{0.05} = 40\ \Omega$

Hence the resistance to be connected in series is

$40 - 8 = 32\ \Omega$

+9 V

Reason — A current of 2 A flows from A to B. Going from A to B, the potential falls across each resistor in the direction of the current, and also falls on passing through the cell from its positive to its negative terminal.

Drop across the 2 Ω resistor = 2 × 2 = 4 V

Drop across the cell = 3 V

Drop across the 1 Ω resistor = 2 × 1 = 2 V

$$\begin{aligned}\text V_\text A - \text V_\text B = 4 + 3 + 2 \\[1em] \text V_\text A - \text V_\text B = 9\ \text V\end{aligned}$$

3 : 2

Reason — When the two cells support each other the balancing length corresponds to (E₁ + E₂), and when they oppose each other it corresponds to (E₁ − E₂). Since the balancing length is proportional to the net emf,

$$\begin{aligned}\frac{\text E_1 + \text E_2}{\text E_1 - \text E_2} = \frac{50}{10} \\[1em] \frac{\text E_1 + \text E_2}{\text E_1 - \text E_2} = 5 \\[1em] \text E_1 + \text E_2 = 5\text E_1 - 5\text E_2 \\[1em] 6\text E_2 = 4\text E_1 \\[1em] \frac{\text E_1}{\text E_2} = \frac{3}{2}\end{aligned}$$

$\text{CE} \frac{\text r_2}{(\text r + \text r_2)}$

Reason — In the steady state no current flows through the branch containing the capacitor, so there is no potential drop across r₁. The potential difference across the plates of the capacitor is therefore the same as that between A and B, that is, the potential drop across r₂.

The current in the circuit is

$$\begin{aligned}\text I = \frac{\text E}{\text r + \text r_2} \\[1em] \text V = \text{Ir}_2 \\[1em] \text V = \frac{\text{Er}_2}{\text r + \text r_2} \\[1em] \text q = \text{CV} \\[1em] \text q = \text{CE}\ \frac{\text r_2}{\text r + \text r_2}\end{aligned}$$

2 A

Reason — The current through an inductor cannot change instantaneously, so just after the switch is closed the current in each inductor is zero. Both the branches containing the inductors therefore carry no current and behave as open branches.

Hence just after the switch is closed the battery drives current only through the single resistor of the middle branch,

$$\begin{aligned}\text I = \frac{\text E}{\text R} \\[1em] \text I = \frac{18}{9.0} \\[1em] \text I = 2\ \text A\end{aligned}$$

In a balanced Wheatstone bridge if the cell and the galvanometer are exchanged, the null-point is disturbed.

Reason — The condition of balance of a Wheatstone bridge, P/Q = R/S, involves only the four resistances of the arms. It is therefore unaltered if the cell and the galvanometer are interchanged, and the null-point remains undisturbed. Hence this statement is false.

The other three statements are correct. The loop rule follows from the conservation of energy, the bridge is most sensitive when the four resistances are of the same order of magnitude, and a rheostat connected with its three terminals can indeed be used as a potential divider.

a condition of no current flow through the galvanometer

Reason — At the balance point of a potentiometer the galvanometer shows no deflection, so no current is drawn from the cell whose emf is being measured. The cell therefore remains in an open circuit and there is no potential drop across its internal resistance, so the potentiometer measures the true emf and not the terminal potential difference. This null condition is what makes the method accurate.

Reason — When n identical cells are joined in series, the net emf is nE and the total internal resistance is nr. On short-circuiting, the external resistance is zero, so

$$\begin{aligned}\text I = \frac{\text{nE}}{\text{nr}} \\[1em] \text I = \frac{\text E}{\text r}\end{aligned}$$

The current is thus independent of n. The graph of I against n is therefore a straight line parallel to the n-axis, which is shown in option (a).

10

Reason — In the series combination the external resistance is nR, so

$\text I = \frac{\text E}{\text{nR} + \text R}$

In the parallel combination the external resistance is R/n, so

$10\ \text I = \frac{\text E}{\frac{\text R}{\text n} + \text R}$

Dividing the second equation by the first,

$$\begin{aligned}10 = \frac{\text{nR} + \text R}{\frac{\text R}{\text n} + \text R} \\[1em] 10 = \frac{(\text n + 1)\text R}{\frac{(1 + \text n)\text R}{\text n}} \\[1em] 10 = \text n\end{aligned}$$

Hence, n = 10.

550 Ω

Reason — Let the initial balancing length from the left end be l. On interchanging the two resistances the balance point goes over to (100 − l).

Since this new balance point is 10 cm to the left of the original one,

$$\begin{aligned}100 - \text l = \text l - 10 \\[1em] 2\ \text l = 110 \\[1em] \text l = 55\ \text{cm}\end{aligned}$$

For the metre bridge,

$\frac{\text R}{\text S} = \frac{55}{45}$

and it is given that R + S = 1000 Ω. Hence

$$\begin{aligned}\text R = 1000 \times \frac{55}{100} \\[1em] \text R = 550\ \Omega\end{aligned}$$

1.5 Ω

Reason — In the potentiometer method for the internal resistance of a cell,

$\text r = \text R \left(\frac{\text l_1}{\text l_2} - 1\right)$

Substituting R = 5 Ω, l₁ = 52 cm and l₂ = 40 cm,

$$\begin{aligned}\text r = 5 \left(\frac{52}{40} - 1\right) \\[1em] \text r = 5 \times \frac{12}{40} \\[1em] \text r = 1.5\ \Omega\end{aligned}$$

300 Ω

Reason — The current through the branch containing R₃ and R₄ in series is

$$\begin{aligned}\text I_1 = \frac{5}{\text R_4} \\[1em] \text I_1 = \frac{5}{500} \\[1em] \text I_1 = 0.01\ \text A\end{aligned}$$

The potential difference across this branch is

$$\begin{aligned}\text V_2 = \text I_1 (\text R_3 + \text R_4) \\[1em] \text V_2 = 0.01 \times 600 \\[1em] \text V_2 = 6\ \text V\end{aligned}$$

Since R₂ is connected in parallel with this branch, the potential difference across R₂ is also 6 V.

The remaining potential difference appears across R₁,

$$\begin{aligned}\text V_1 = 18 - 6 \\[1em] \text V_1 = 12\ \text V \\[1em] \text I = \frac{12}{400} \\[1em] \text I = 0.03\ \text A\end{aligned}$$

The current through R₂ is

$$\begin{aligned}\text I_2 = 0.03 - 0.01 \\[1em] \text I_2 = 0.02\ \text A \\[1em] \text R_2 = \frac{6}{0.02} \\[1em] \text R_2 = 300\ \Omega\end{aligned}$$

2 V

Reason — A and B are the free ends of the network, so no current is drawn from them. Hence no current flows through the 5 Ω and the 10 Ω resistors and there is no potential drop across either of them.

The three branches, each of 1 Ω with cells of 1 V, 2 V and 3 V, are connected in parallel between the points D and C. The equivalent emf of this parallel combination is

$$\begin{aligned}\text E = \frac{\frac{1}{1} + \frac{2}{1} + \frac{3}{1}}{\frac{1}{1} + \frac{1}{1} + \frac{1}{1}} \\[1em] \text E = \frac{6}{3} \\[1em] \text E = 2\ \text V\end{aligned}$$

Since V_A = V_D and V_B = V_C, the potential difference between A and B is 2 V.

The emf of the battery is 1.5 V and its internal resistance is 1.5 Ω

Reason — The voltmeter is connected across the battery, so it reads the terminal potential difference,

$\text V = \text E - \text{Ir}$

The intercept of the graph on the V-axis gives the emf, so E = 1.5 V.

The graph falls to V₀ ≈ 0 when the current is 1000 mA = 1 A. Putting V = 0 and I = 1 A,

$$\begin{aligned}0 = 1.5 - (1 \times \text r) \\[1em] \text r = 1.5\ \Omega\end{aligned}$$

5 A

Reason — Let the potential of the junction C be V, the earthed end being at zero potential. Applying Kirchhoff's junction rule at C,

$\frac{20 - \text V}{2} + \frac{10 - \text V}{4} = \frac{\text V}{2}$

Multiplying throughout by 4,

$$\begin{aligned}2(20 - \text V) + (10 - \text V) = 2\ \text V \\[1em] 40 - 2\text V + 10 - \text V = 2\ \text V \\[1em] 50 = 5\ \text V \\[1em] \text V = 10\ \text V\end{aligned}$$

The current through the 2 Ω resistor in the branch containing the switch is

$$\begin{aligned}\text I = \frac{\text V}{2} \\[1em] \text I = \frac{10}{2} \\[1em] \text I = 5\ \text A\end{aligned}$$

1.1 A, 0.4 A, 0.4 A

Reason — Applying Kirchhoff's junction rule at each junction in turn.

At the junction P : the current I₅ arrives and the current I₆ leaves,

$\text I_6 = \text I_5 = 0.4\ \text A$

At the junction S : the current I₄ arrives while I₅ and I₃ leave,

$$\begin{aligned}\text I_3 = \text I_4 - \text I_5 \\[1em] \text I_3 = 0.8 - 0.4 \\[1em] \text I_3 = 0.4\ \text A\end{aligned}$$

At the junction Q : the currents I₆ and I₃ arrive while I₁ and I₂ leave,

$$\begin{aligned}\text I_6 + \text I_3 = \text I_1 + \text I_2 \\[1em] 0.4 + 0.4 = (-0.3) + \text I_2 \\[1em] \text I_2 = 0.8 + 0.3 \\[1em] \text I_2 = 1.1\ \text A\end{aligned}$$

4

Reason — For the meter bridge the unknown resistance is

$\text X = \text R\ \frac{(100 - \text l)}{\text l}$

Calculating X for each reading,

Reading 1 : $\text X = 1000 \times \frac{40}{60} = 666.7\ \Omega$

Reading 2 : $\text X = 100 \times \frac{87}{13} = 669.2\ \Omega$

Reading 3 : $\text X = 10 \times \frac{98.5}{1.5} = 656.7\ \Omega$

Reading 4 : $\text X = 1 \times \frac{99}{1} = 99\ \Omega$

The first three readings give nearly the same value of X, whereas the fourth gives a value quite different from them. Hence reading 4 is inconsistent.

60 Ω

Reason — Initially,

$$\begin{aligned}\frac{\text R_1}{\text R_2} = \frac{40}{60} \\[1em] \frac{\text R_1}{\text R_2} = \frac{2}{3}\end{aligned}$$

After adding 10 Ω in series the null point shifts to 50 cm, so

$$\begin{aligned}\frac{\text R_1 + 10}{\text R_2} = \frac{50}{50} \\[1em] \frac{\text R_1 + 10}{\text R_2} = 1 \\[1em] \text R_1 + 10 = \text R_2\end{aligned}$$

Substituting R₁ = (2/3)R₂,

$$\begin{aligned}\frac{2}{3}\text R_2 + 10 = \text R_2 \\[1em] \text R_2 = 30\ \Omega \\[1em] \text R_1 = 20\ \Omega\end{aligned}$$

To bring the null point back to 40 cm the left gap must again offer 20 Ω. Let X be joined in parallel with 30 Ω,

$$\begin{aligned}\frac{30\ \text X}{30 + \text X} = 20 \\[1em] 10\ \text X = 600 \\[1em] \text X = 60\ \Omega\end{aligned}$$

0.25 m

Reason — Since

$\frac{\text{dR}}{\text{dl}} \propto \frac{1}{\sqrt{\text l}}$

let dR/dl = k/√l. Integrating from 0 to l,

$$\begin{aligned}\text R = \int_0^{\text l} \frac{\text k}{\sqrt{\text l}}\ \text{dl} \\[1em] \text R = \text k \left[2\sqrt{\text l}\ \right]_0^{\text l} \\[1em] \text R = 2\text k \sqrt{\text l}\end{aligned}$$

Thus the resistance of a portion of the wire is proportional to the square root of its length.

The two resistances in the gaps are equal, so at balance the resistances of the two parts of the wire must also be equal,

$$\begin{aligned}\text R_\text{AP} = \text R_\text{PB} \\[1em] 2\text k \sqrt{\text l} = 2\text k \left(\sqrt{1} - \sqrt{\text l}\ \right) \\[1em] 2\sqrt{\text l} = 1 \\[1em] \sqrt{\text l} = 0.5 \\[1em] \text l = 0.25\ \text m\end{aligned}$$

brown, blue, brown

Reason — For the colour code, orange stands for 3, red for 2 and brown for a multiplier of 10¹. Hence

$$\begin{aligned}\text R_1 = 32 \times 10^1 \\[1em] \text R_1 = 320\ \Omega\end{aligned}$$

For the balanced bridge,

$$\begin{aligned}\frac{\text R_1}{\text R_2} = \frac{\text R_3}{\text R_4} \\[1em] \frac{320}{80} = \frac{\text R_3}{40} \\[1em] \text R_3 = 4 \times 40 \\[1em] \text R_3 = 160\ \Omega\end{aligned}$$

Writing 160 Ω as 16 × 10¹ Ω, the first digit 1 is brown, the second digit 6 is blue and the multiplier 10¹ is brown.

0.3 V

Reason — The primary circuit contains the 6 V cell, the rheostat R_h and the bridge wire AB of resistance 4 Ω.

First case :

$$\begin{aligned}\text I_1 = \frac{6}{2 + 4} \\[1em] \text I_1 = 1\ \text A\end{aligned}$$

If R_AJ is the resistance of the balancing portion,

$$\begin{aligned}\text E = \text I_1\ \text R_\text{AJ} \\[1em] 0.5 = 1 \times \text R_\text{AJ} \\[1em] \text R_\text{AJ} = 0.5\ \Omega\end{aligned}$$

Second case : The null point J is the same, so R_AJ is unchanged,

$$\begin{aligned}\text I_2 = \frac{6}{6 + 4} \\[1em] \text I_2 = 0.6\ \text A \\[1em] \text E_2 = \text I_2\ \text R_\text{AJ} \\[1em] \text E_2 = 0.6 \times 0.5 \\[1em] \text E_2 = 0.3\ \text V\end{aligned}$$

+1 V

Reason — Take V_A = 0.

At the junction D : a current of 2 A arrives and a current of 1 A leaves along the upper branch, so the current flowing from D to C through the 2 Ω resistor is

$2 - 1 = 1\ \text A$

Potential at C : going from A to C through the 1 V cell,

$$\begin{aligned}\text V_\text C = 0 + 1 \\[1em] \text V_\text C = 1\ \text V\end{aligned}$$

Potential at D : the current of 1 A flows from D to C through the 2 Ω resistor, so

$$\begin{aligned}\text V_\text D = \text V_\text C + (1 \times 2) \\[1em] \text V_\text D = 3\ \text V\end{aligned}$$

Potential at B : going from B to D through the 2 V cell the potential rises by 2 V,

$$\begin{aligned}\text V_\text B = \text V_\text D - 2 \\[1em] \text V_\text B = 3 - 2 \\[1em] \text V_\text B = 1\ \text V\end{aligned}$$

Hence the potential at B with respect to A is + 1 V.

I₂R₂ + E₂ − E₃ − I₃R₁ = 0

Reason — The loop BCDEB is traversed in the order B → C → D → E → B.

Going from B to C the loop is traversed in the direction of the current I₂ through R₂, so the term I₂R₂ is taken as positive.

Going from C to D through the cell E₂ the loop is traversed from its positive to its negative electrode, so E₂ is taken as negative.

Going from D to E through the cell E₃ the loop is traversed from its negative to its positive electrode, so E₃ is taken as positive.

Going from E to B the loop is traversed against the direction of the current I₃ through R₁, so the term I₃R₁ is taken as negative.

Applying Kirchhoff's second law ($\sum \text{IR} = \sum \text E$),

$$\begin{aligned}\text I_2 \text R_2 - \text I_3 \text R_1 = -\text E_2 + \text E_3 \\[1em] \text I_2 \text R_2 + \text E_2 - \text E_3 - \text I_3 \text R_1 = 0\end{aligned}$$

1 A

Reason — The two cells of 2 V and 4 V are so connected that their emfs support each other in the loop, so the net emf is

$2 + 4 = 6\ \text V$

The resistances in the loop are 1 Ω, 1 Ω and 4 Ω in series,

$$\begin{aligned}\text R = 1 + 1 + 4 \\[1em] \text R = 6\ \Omega \\[1em] \text I = \frac{6}{6} \\[1em] \text I = 1\ \text A\end{aligned}$$

0.02 V/cm

Reason — The null position is 49 cm from Q, so the balancing length measured from P is

$$\begin{aligned}\text l = 100 - 49 \\[1em] \text l = 51\ \text{cm}\end{aligned}$$

With the switch S open no current is drawn from the cell E₂, so the balancing length corresponds to its emf,

$$\begin{aligned}\text E_2 = \text{Kl} \\[1em] \text K = \frac{\text E_2}{\text l} \\[1em] \text K = \frac{1.02}{51} \\[1em] \text K = 0.02\ \text{V/cm}\end{aligned}$$

60 cm

Reason — Since the balancing length is proportional to the emf,

$$\begin{aligned}\frac{\text E_1}{\text E_2} = \frac{\text l_1}{\text l_2} \\[1em] \text l_2 = \text l_1 \times \frac{\text E_2}{\text E_1} \\[1em] \text l_2 = 36 \times \frac{2.5}{1.5} \\[1em] \text l_2 = 60\ \text{cm}\end{aligned}$$

5 V

Reason — The resistance of the whole potentiometer wire is

$$\begin{aligned}\text R_\text{AB} = 0.1\ \Omega\text{/cm} \times 1000\ \text{cm} \\[1em] \text R_\text{AB} = 100\ \Omega\end{aligned}$$

The primary circuit contains the 6 V battery and the 20 Ω resistance in series with the wire,

$$\begin{aligned}\text I = \frac{6}{20 + 100} \\[1em] \text I = 0.05\ \text A\end{aligned}$$

The maximum emf that can be measured is the potential difference across the whole wire,

$$\begin{aligned}\text V_\text{max} = \text I\ \text R_\text{AB} \\[1em] \text V_\text{max} = 0.05 \times 100 \\[1em] \text V_\text{max} = 5\ \text V\end{aligned}$$

$\frac{40}{9}$ V

Reason — The sliding contact is at the middle of the wire, so the wire is divided into two halves of 5 Ω each. The 2 Ω resistor is connected between the sliding contact and the end B, so it is in parallel with the second half,

$$\begin{aligned}\text R_\text p = \frac{5 \times 2}{5 + 2} \\[1em] \text R_\text p = \frac{10}{7}\ \Omega\end{aligned}$$

The total resistance across the 20 V supply is

$$\begin{aligned}\text R = 5 + \frac{10}{7} \\[1em] \text R = \frac{45}{7}\ \Omega \\[1em] \text I = \frac{20}{\frac{45}{7}} \\[1em] \text I = \frac{28}{9}\ \text A\end{aligned}$$

The potential drop across the parallel combination, and hence across the 2 Ω resistor, is

$$\begin{aligned}\text V = \text I\ \text R_\text p \\[1em] \text V = \frac{28}{9} \times \frac{10}{7} \\[1em] \text V = \frac{40}{9}\ \text V\end{aligned}$$

4.87 mA

Reason — Testing the condition of balance,

$$\begin{aligned}\frac{100}{10} = 10 \\[1em] \frac{60}{5} = 12\end{aligned}$$

The bridge is not balanced, so a current flows through the galvanometer.

Potential difference between B and D with the galvanometer removed :

$$\begin{aligned}\text V_\text B - \text V_\text C = 10 \times \frac{10}{100 + 10} \\[1em] \text V_\text B - \text V_\text C = 0.909\ \text V \\[1em] \text V_\text D - \text V_\text C = 10 \times \frac{5}{60 + 5} \\[1em] \text V_\text D - \text V_\text C = 0.769\ \text V \\[1em] \text V_\text B - \text V_\text D = 0.909 - 0.769 \\[1em] \text V_\text B - \text V_\text D = 0.140\ \text V\end{aligned}$$

Resistance of the network between B and D :

$$\begin{aligned}\frac{100 \times 10}{110} + \frac{60 \times 5}{65} = 9.09 + 4.62 \\[1em] \frac{100 \times 10}{110} + \frac{60 \times 5}{65} = 13.71\ \Omega\end{aligned}$$

Current through the galvanometer :

$$\begin{aligned}\text I_\text g = \frac{0.140}{13.71 + 15} \\[1em] \text I_\text g = \frac{0.140}{28.71} \\[1em] \text I_\text g = 4.87 \times 10^{-3}\ \text A \\[1em] \text I_\text g = 4.87\ \text{mA}\end{aligned}$$

10 A

Reason — Let V be the potential of the junction to which the 6 Ω resistor is connected, the lower rail being at zero potential. Applying Kirchhoff's junction rule,

$\frac{140 - \text V}{20} + \frac{90 - \text V}{5} = \frac{\text V}{6}$

Multiplying throughout by 60,

$$\begin{aligned}3(140 - \text V) + 12(90 - \text V) = 10\ \text V \\[1em] 420 - 3\text V + 1080 - 12\text V = 10\ \text V \\[1em] 1500 = 25\ \text V \\[1em] \text V = 60\ \text V\end{aligned}$$

The current in the 6 Ω resistance is

$$\begin{aligned}\text I = \frac{60}{6} \\[1em] \text I = 10\ \text A\end{aligned}$$

All the four statements are correct.

Reason — Let A be the upper vertex, B the lower vertex, C the left vertex, D the right vertex and O the central junction of the network.

Folding the network : The network is symmetrical about the horizontal line COD, which contains both batteries: R₆ is the mirror image of R₅, R₇ of R₈ and R₂ of R₄. Hence A and B are at the same potential and may be joined into a single point AB. Each mirrored pair of 1 Ω resistances then comes in parallel, giving $\frac{1}{2}\ \Omega$ between C and AB, between AB and D, and between AB and O.

Let x be the current in the branch containing ε₂ and R₃, flowing from O towards C, and y the current in the branch containing ε₁ and R₁, flowing from O towards D. By Kirchhoff's first law, the current in the $\frac{1}{2}\ \Omega$ branch from AB to O is (x + y).

Loop OCABO (through ε₂) : Traversed in the direction of x, the cell ε₂ is crossed from its negative to its positive electrode. Applying Kirchhoff's second law ($\sum \text{IR} = \sum \text E$),

$$\begin{aligned}1 \times x + \frac{1}{2}x + \frac{1}{2}(x + y) = 6 \\[1em] 2x + \frac{y}{2} = 6 \\[1em] 4x + y = 12 \qquad ...(i)\end{aligned}$$

Loop ODABO (through ε₁) : Traversed in the direction of y, the cell ε₁ is crossed from its positive to its negative electrode,

$$\begin{aligned}1 \times y + \frac{1}{2}y + \frac{1}{2}(x + y) = -12 \\[1em] \frac{x}{2} + 2y = -12 \\[1em] x + 4y = -24 \qquad ...(ii)\end{aligned}$$

Multiplying Eq. (i) by 4 and subtracting Eq. (ii),

$$\begin{aligned}15x = 48 + 24 \\[1em] 15x = 72 \\[1em] x = 4.8\ \text A\end{aligned}$$

From Eq. (i),

$$\begin{aligned}y = 12 - (4 \times 4.8) \\[1em] y = -7.2\ \text A\end{aligned}$$

The negative sign shows only that the current in the R₁ branch flows from D to O. The current (x + y) = −2.4 A divides equally between R₂ and R₄, and the current x divides equally between R₆ and R₅,

$$\begin{aligned}|\text I_1| = |y| \\[1em] |\text I_1| = 7.2\ \text A \\[1em] |\text I_3| = x \\[1em] |\text I_3| = 4.8\ \text A \\[1em] |\text I_2| = \frac{|x + y|}{2} \\[1em] |\text I_2| = \frac{2.4}{2} \\[1em] |\text I_2| = 1.2\ \text A \\[1em] |\text I_5| = \frac{x}{2} \\[1em] |\text I_5| = \frac{4.8}{2} \\[1em] |\text I_5| = 2.4\ \text A\end{aligned}$$

Each of the four statements therefore agrees with the calculated value, and all of them are correct.

should be approximately equal and are small

Reason — A Wheatstone bridge is most sensitive when all the four resistances of its arms are of the same order of magnitude. If P and Q are approximately equal, the ratio arms are balanced and a small change in X produces the largest change in the galvanometer deflection.

Keeping P and Q small also allows a larger current to flow through the bridge, which makes the galvanometer deflection more pronounced and the detection of the null point more precise.

0.4 A

Reason — Let V be the potential of the junction where the three resistors meet. Applying Kirchhoff's junction rule,

$\frac{30 - \text V}{10} = \frac{\text V - 12}{20} + \frac{\text V - 2}{30}$

Multiplying throughout by 60,

$$\begin{aligned}6(30 - \text V) = 3(\text V - 12) + 2(\text V - 2) \\[1em] 180 - 6\text V = 3\text V - 36 + 2\text V - 4 \\[1em] 220 = 11\ \text V \\[1em] \text V = 20\ \text V\end{aligned}$$

The current through the 20 Ω resistor is

$$\begin{aligned}\text I = \frac{\text V - 12}{20} \\[1em] \text I = \frac{20 - 12}{20} \\[1em] \text I = 0.4\ \text A\end{aligned}$$

$\frac{4}{5}$

Reason — In the steady state no current flows through the branches containing the capacitors. Hence the whole current flows along the path containing the 6 Ω resistor, the galvanometer of 2 Ω and the 8 Ω resistor in series,

$$\begin{aligned}\text I = \frac{4}{6 + 2 + 8} \\[1em] \text I = \frac{4}{16} \\[1em] \text I = 0.25\ \text A\end{aligned}$$

The potential difference across C₁ is that between A and C,

$$\begin{aligned}\text V_1 = (0.25 \times 6) + (0.25 \times 2) \\[1em] \text V_1 = 1.5 + 0.5 \\[1em] \text V_1 = 2\ \text V\end{aligned}$$

The potential difference across C₂ is that between B and D,

$$\begin{aligned}\text V_2 = (0.25 \times 2) + (0.25 \times 8) \\[1em] \text V_2 = 0.5 + 2 \\[1em] \text V_2 = 2.5\ \text V \\[1em] \frac{\text V_1}{\text V_2} = \frac{2}{2.5} \\[1em] \frac{\text V_1}{\text V_2} = \frac{4}{5}\end{aligned}$$

100

Reason — With the ten resistors in series the external resistance is 10R, so

$\text I_\text s = \frac{\text E}{10\ \text R}$

With them in parallel the external resistance is R/10, so

$$\begin{aligned}\text I_\text p = \frac{\text E}{\frac{\text R}{10}} \\[1em] \text I_\text p = \frac{10\ \text E}{\text R} \\[1em] \text n = \frac{\text I_\text p}{\text I_\text s} \\[1em] \text n = \frac{\frac{10\ \text E}{\text R}}{\frac{\text E}{10\ \text R}} \\[1em] \text n = 100\end{aligned}$$

All the four statements are correct.

Reason — Let L be the left junction and N the right junction joining the four branches. Both cells have their positive electrodes towards N.

Before the key is closed : The capacitor branch carries no current. The current of 1 A in the 1 Ω resistor flows from N to L, so going from L to N along this branch,

$$\begin{aligned}\text V_\text N - \text V_\text L = 5 + (1 \times 1) \\[1em] \text V_\text N - \text V_\text L = 6\ \text V\end{aligned}$$

The same potential difference drives the current I₁ through the 3 Ω branch,

$$\begin{aligned}\text I_1 \times 3 = 6 \\[1em] \text I_1 = 2\ \text A\end{aligned}$$

By Kirchhoff's first law at L, the current in the branch containing R is I = I₁ + 1 = 3 A, flowing from L to N. For this branch,

$$\begin{aligned}15 - \text{IR} = 6 \\[1em] 15 - 3\text R = 6 \\[1em] \text R = 3\ \Omega\end{aligned}$$

After the key is closed : The three branches without the capacitor act as one equivalent cell,

$$\begin{aligned}\frac{1}{\text r_\text{eq}} = \frac{1}{3} + \frac{1}{1} + \frac{1}{3} \\[1em] \text r_\text{eq} = \frac{3}{5}\ \Omega \\[1em] \text E_\text{eq} = \text r_\text{eq}\left(\frac{15}{3} + \frac{5}{1} + \frac{0}{3}\right) \\[1em] \text E_\text{eq} = \frac{3}{5} \times 10 \\[1em] \text E_\text{eq} = 6\ \text V\end{aligned}$$

The capacitor charges through r_eq and the 3 Ω resistor in series, so the time constant is

$$\begin{aligned}\text{CR} = (2 \times 10^{-6}) \times \left(\frac{3}{5} + 3\right) \\[1em] \text{CR} = 7.2 \times 10^{-6}\ \text s\end{aligned}$$

The charging current is $\text I = \frac{\text E_\text{eq}}{\text r_\text{eq} + 3}\ \text e^{-(\text t - \text t_0)/\text{CR}}$. At t = t₀ + 7.2 µs, (t − t₀) = CR,

$$\begin{aligned}\text I = \frac{6}{3.6} \times \text e^{-1} \\[1em] \text I = \frac{5}{3} \times 0.36 \\[1em] \text I = 0.6\ \text A\end{aligned}$$

As t → ∞ no current flows in the capacitor branch, and the potential difference across the capacitor equals E_eq = 6 V,

$$\begin{aligned}\text Q = \text{CV} \\[1em] \text Q = 2\ \mu\text F \times 6\ \text V \\[1em] \text Q = 12\ \mu\text C\end{aligned}$$

Each of the four statements therefore agrees with the calculated value, and all of them are correct.

66

Reason — For the meter bridge,

$$\begin{aligned}\frac{4.5}{\text S} = \frac{60}{40} \\[1em] \frac{4.5}{\text S} = 1.5 \\[1em] \text S = \frac{4.5}{1.5} \\[1em] \text S = 3\ \Omega\end{aligned}$$

The area of cross-section of the wire is

$$\begin{aligned}\text A = \pi \text r^2 \\[1em] \text A = \frac{22}{7} \times (\sqrt{7} \times 10^{-4})^2 \\[1em] \text A = \frac{22}{7} \times 7 \times 10^{-8} \\[1em] \text A = 22 \times 10^{-8}\ \text m^2\end{aligned}$$

The resistivity is

$$\begin{aligned}\rho = \frac{\text{SA}}{\text l} \\[1em] \rho = \frac{3 \times 22 \times 10^{-8}}{0.10} \\[1em] \rho = 66 \times 10^{-7}\ \Omega\ \text m\end{aligned}$$

Comparing with ρ = R × 10^-7 Ω m, we get R = 66.

8V

Reason — The current in the circuit is

$$\begin{aligned}\text I = \frac{\text E}{\text R + \text r} \\[1em] \text I = \frac{10}{4 + 1} \\[1em] \text I = 2\ \text A\end{aligned}$$

The terminal voltage is

$$\begin{aligned}\text V = \text E - \text{Ir} \\[1em] \text V = 10 - (2 \times 1) \\[1em] \text V = 8\ \text V\end{aligned}$$

Given,

• Emf of the ideal cell, E = 10 V
• Each resistance R₁ = R₂ = R₃ = R₄ = R₅ = 2 Ω

When the capacitor is fully charged, no current flows in the branch containing the capacitor, and hence no current flows through R₅ which lies in that branch.

The current therefore flows from the cell along two paths from A to the junction M, namely through R₃ alone and through R₁ and R₂ in series, and then through R₄ back to the cell.

$$\begin{aligned}\text R_1 + \text R_2 = 2 + 2 \\[1em] \text R_1 + \text R_2 = 4\ \Omega \\[1em] \text R_\text{AM} = \frac{2 \times 4}{2 + 4} \\[1em] \text R_\text{AM} = \frac{8}{6} \\[1em] \text R_\text{AM} = \frac{4}{3}\ \Omega \\[1em] \text R = \frac{4}{3} + 2 \\[1em] \text R = \frac{10}{3}\ \Omega \\[1em] \text I = \frac{10}{\frac{10}{3}} \\[1em] \text I = 3\ \text A\end{aligned}$$

The potential difference between A and M is

$$\begin{aligned}\text V_\text{AM} = 3 \times \frac{4}{3} \\[1em] \text V_\text{AM} = 4\ \text V\end{aligned}$$

The current in the branch containing R₁ and R₂ is

$\frac{4}{4} = 1\ \text A$

Taking the negative terminal of the cell as zero potential, V_A = 10 V. The potential of the junction Q between R₁ and the capacitor is

$$\begin{aligned}\text V_\text Q = 10 - (1 \times 2) \\[1em] \text V_\text Q = 8\ \text V\end{aligned}$$

Since no current flows through R₅, the other plate of the capacitor is at the potential of the negative terminal, that is, at 0 V.

$$\begin{aligned}\text V_\text C = 8 - 0 \\[1em] \text V_\text C = 8\ \text V\end{aligned}$$

Hence, the potential difference across the capacitor is 8 V.

Given,

• Each resistor = 10 Ω, number of resistors = n
• Emf of the battery, E = 20 V, internal resistance r = 10 Ω
• Current in parallel = 20 × current in series

In series,

$\text I = \frac{20}{10\ \text n + 10}$

In parallel,

$20\ \text I = \frac{20}{\frac{10}{\text n} + 10}$

Dividing the second equation by the first,

$$\begin{aligned}20 = \frac{10\ \text n + 10}{\frac{10}{\text n} + 10} \\[1em] 20\left(\frac{10}{\text n} + 10\right) = 10\ \text n + 10 \\[1em] \frac{200}{\text n} + 200 = 10\ \text n + 10\end{aligned}$$

Multiplying throughout by n,

$$\begin{aligned}200 + 200\ \text n = 10\ \text n^2 + 10\ \text n \\[1em] 10\ \text n^2 - 190\ \text n - 200 = 0 \\[1em] \text n^2 - 19\ \text n - 20 = 0 \\[1em] (\text n - 20)(\text n + 1) = 0\end{aligned}$$

Since n cannot be negative, n = 20.

Hence, the value of n is 20.

Given,

• C = 50 µF
• Three resistors of 2 kΩ each in series across a 6 V supply
• The capacitor is connected across the lowest 2 kΩ resistor

In the steady state no current flows through the capacitor branch, so the current flows through the three resistors in series,

$$\begin{aligned}\text I = \frac{6}{3 \times 2000} \\[1em] \text I = \frac{6}{6000} \\[1em] \text I = 1 \times 10^{-3}\ \text A\end{aligned}$$

The potential difference across the lowest 2 kΩ resistor, and hence across the capacitor, is

$$\begin{aligned}\text V = (1 \times 10^{-3}) \times 2000 \\[1em] \text V = 2\ \text V \\[1em] \text q = \text{CV} \\[1em] \text q = 50 \times 2 \\[1em] \text q = 100\ \mu\text C\end{aligned}$$

Hence, the charge on the upper plate of the capacitor is 100 µC.

Given,

• Emf of the cell = 12 V with a 1 Ω resistor in series
• Upper path : 4 Ω ∥ 4 Ω, then 2 Ω, then 15 Ω ∥ 10 Ω
• Lower path : 8 Ω ∥ 8 Ω, then 12 Ω ∥ 12 Ω

Upper path :

$$\begin{aligned}\frac{4 \times 4}{8} = 2\ \Omega \\[1em] \frac{15 \times 10}{25} = 6\ \Omega \\[1em] \text R_\text{upper} = 2 + 2 + 6 \\[1em] \text R_\text{upper} = 10\ \Omega\end{aligned}$$

Lower path :

$$\begin{aligned}\frac{8 \times 8}{16} = 4\ \Omega \\[1em] \frac{12 \times 12}{24} = 6\ \Omega \\[1em] \text R_\text{lower} = 4 + 6 \\[1em] \text R_\text{lower} = 10\ \Omega\end{aligned}$$

The two paths are in parallel,

$$\begin{aligned}\text R_\text p = \frac{10 \times 10}{20} \\[1em] \text R_\text p = 5\ \Omega \\[1em] \text I = \frac{12}{5 + 1} \\[1em] \text I = 2\ \text A\end{aligned}$$

Since the two paths have equal resistance, the current divides equally and 1 A flows through the upper path. The voltage drop across the parallel combination of 15 Ω and 10 Ω is

$$\begin{aligned}\text V = 1 \times 6 \\[1em] \text V = 6\ \text V\end{aligned}$$

Hence, the voltage drop across the 15 Ω resistance is 6 V.

Given,

• V₁ = 1.25 V across R₁ = 5 Ω
• V₂ = 1 V across R₂ = 2 Ω

First case :

$$\begin{aligned}\text I_1 = \frac{1.25}{5} \\[1em] \text I_1 = 0.25\ \text A \\[1em] \text E = 1.25 + 0.25\ \text r \qquad ...(i)\end{aligned}$$

Second case :

$$\begin{aligned}\text I_2 = \frac{1}{2} \\[1em] \text I_2 = 0.5\ \text A \\[1em] \text E = 1 + 0.5\ \text r \qquad ...(ii)\end{aligned}$$

Equating Eqs. (i) and (ii),

$$\begin{aligned}1.25 + 0.25\ \text r = 1 + 0.5\ \text r \\[1em] 0.25 = 0.25\ \text r \\[1em] \text r = 1\ \Omega \\[1em] \text E = 1.25 + (0.25 \times 1) \\[1em] \text E = 1.5\ \text V \\[1em] \text E = \frac{15}{10}\ \text V\end{aligned}$$

Hence, the value of x is 15.

Given,

• Resistance of the whole wire = 16 Ω
• Emf of the supply = 9 V, internal resistance = 1 Ω

Each side of the square has a resistance of

$\frac{16}{4} = 4\ \Omega$

The supply is connected across one side AB, so this side of 4 Ω is in parallel with the other three sides in series, that is, with 12 Ω,

$$\begin{aligned}\text R = \frac{4 \times 12}{4 + 12} \\[1em] \text R = 3\ \Omega \\[1em] \text I = \frac{9}{3 + 1} \\[1em] \text I = 2.25\ \text A\end{aligned}$$

The potential difference across AB is

$$\begin{aligned}\text V_\text{AB} = 2.25 \times 3 \\[1em] \text V_\text{AB} = 6.75\ \text V\end{aligned}$$

The current in the path A–D–C–B of 12 Ω is

$\frac{6.75}{12} = 0.5625\ \text A$

Taking V_B = 0 so that V_A = 6.75 V, the potential drop across each of the sides AD and DC is

$$\begin{aligned}0.5625 \times 4 = 2.25\ \text V \\[1em] \text V_\text D = 6.75 - 2.25 \\[1em] \text V_\text D = 4.5\ \text V \\[1em] \text V_\text C = 4.5 - 2.25 \\[1em] \text V_\text C = 2.25\ \text V\end{aligned}$$

The potential drop across the diagonal AC is

$$\begin{aligned}\text V_\text A - \text V_\text C = 6.75 - 2.25 \\[1em] \text V_\text A - \text V_\text C = 4.5\ \text V \\[1em] \text V_\text A - \text V_\text C = 45 \times 10^{-1}\ \text V\end{aligned}$$

Hence, the required value is 45.

Given,

• Balancing length for E₁, l₁ = 20 cm
• Balancing length for E₂, l₂ = 60 cm

Since the emf is proportional to the balancing length,

$$\begin{aligned}\frac{\text E_1}{\text E_2} = \frac{\text l_1}{\text l_2} \\[1em] \frac{\text E_1}{\text E_2} = \frac{20}{60} \\[1em] \frac{\text E_1}{\text E_2} = \frac{1}{3}\end{aligned}$$

Comparing with a/b, we get a = 1 and b = 3.

Hence, the value of a is 1.

Given,

• Three branches, each of 1 kΩ, containing cells of 2 V, 4 V and 6 V, joined in parallel

Since no current is drawn at the output terminal, the three branches form a parallel combination of cells. The emf of the equivalent cell is

$\text V_0 = \frac{\frac{\text E_1}{\text R_1} + \frac{\text E_2}{\text R_2} + \frac{\text E_3}{\text R_3}}{\frac{1}{\text R_1} + \frac{1}{\text R_2} + \frac{1}{\text R_3}}$

All the three resistances are equal, so

$$\begin{aligned}\text V_0 = \frac{2 + 4 + 6}{3} \\[1em] \text V_0 = \frac{12}{3} \\[1em] \text V_0 = 4\ \text V\end{aligned}$$

Hence, the value of V₀ is 4 V.

Given,

• E = 10 V, r = 10 Ω
• R = 100 Ω, R' = 200 Ω
• C = 1.1 µF

In the steady state no current flows in the branch containing the capacitor, so no current flows through R' and there is no potential drop across it.

The current in the circuit is

$$\begin{aligned}\text I = \frac{\text E}{\text r + \text R} \\[1em] \text I = \frac{10}{10 + 100} \\[1em] \text I = \frac{1}{11}\ \text A\end{aligned}$$

The potential difference across R, and hence across the capacitor, is

$$\begin{aligned}\text V = \text{IR} \\[1em] \text V = \frac{1}{11} \times 100 \\[1em] \text V = \frac{100}{11}\ \text V \\[1em] \text q = \text{CV} \\[1em] \text q = (1.1 \times 10^{-6}) \times \frac{100}{11} \\[1em] \text q = 10 \times 10^{-6}\ \text C\end{aligned}$$

Hence, the charge stored in the capacitor is 10 × 10^-6 C.

Given,

• Cell of 15 V with a 1 Ω resistor in the branch AB
• A 2 Ω resistor in the other branch between A and B

The two branches form a closed loop, so the current in the loop is

$$\begin{aligned}\text I = \frac{15}{1 + 2} \\[1em] \text I = 5\ \text A\end{aligned}$$

The potential difference between B and A is the potential drop across the 2 Ω resistor,

$$\begin{aligned}\text V_\text B - \text V_\text A = \text I \times 2 \\[1em] \text V_\text B - \text V_\text A = 5 \times 2 \\[1em] \text V_\text B - \text V_\text A = 10\ \text V\end{aligned}$$

Hence, V_B − V_A = 10 V.

Given,

• P = 40 Ω in the left gap
• Initial balancing length, l₁ = 40 cm
• New balancing length, l₂ = 80 cm

Initially,

$$\begin{aligned}\frac{\text P}{\text Q} = \frac{\text l_1}{100 - \text l_1} \\[1em] \frac{\text P}{\text Q} = \frac{40}{60} \\[1em] \frac{\text P}{\text Q} = \frac{2}{3} \\[1em] \text Q = \frac{3}{2} \times 40 \\[1em] \text Q = 60\ \Omega\end{aligned}$$

After connecting x in series with P,

$$\begin{aligned}\frac{\text P + \text x}{\text Q} = \frac{\text l_2}{100 - \text l_2} \\[1em] \frac{\text P + \text x}{\text Q} = \frac{80}{20} \\[1em] \frac{\text P + \text x}{\text Q} = 4 \\[1em] \text P + \text x = 4 \times 60 \\[1em] \text P + \text x = 240\ \Omega \\[1em] \text x = 240 - 40 \\[1em] \text x = 200\ \Omega\end{aligned}$$

Hence, the value of x is 200 Ω.

Note: The printed answer key gives 20 for this question, which is the same value as that printed for the next question. The balancing conditions for the two positions of the null point lead to x = 200 Ω, so the key appears to be in error here.

Given,

• Length of the wire, L = 300 cm
• Series resistance = 780 Ω, emf of the standard cell = 4 V
• Emf of the cell in the secondary circuit = 20 mV = 20 × 10^-3 V at l = 60 cm

The potential gradient along the wire is

$\text K = \frac{20 \times 10^{-3}}{60}\ \text{V/cm}$

The potential difference across the whole wire is

$$\begin{aligned}\text V = \text{KL} \\[1em] \text V = \frac{20 \times 10^{-3}}{60} \times 300 \\[1em] \text V = 0.1\ \text V\end{aligned}$$

If R is the resistance of the potentiometer wire, the current in the primary circuit is

$$\begin{aligned}\text I = \frac{4}{780 + \text R} \\[1em] \text V = \text{IR} \\[1em] 0.1 = \frac{4\ \text R}{780 + \text R} \\[1em] 78 + 0.1\ \text R = 4\ \text R \\[1em] 3.9\ \text R = 78 \\[1em] \text R = 20\ \Omega\end{aligned}$$

Hence, the resistance of the potentiometer wire is 20 Ω.

Given,

• R = 5.6 kΩ = 5600 Ω
• Balancing length from the left end, l = 30 cm

The unknown resistance S is in the left gap, so

$$\begin{aligned}\frac{\text S}{\text R} = \frac{\text l}{100 - \text l} \\[1em] \frac{\text S}{\text R} = \frac{30}{70} \\[1em] \text S = 5600 \times \frac{30}{70} \\[1em] \text S = 5600 \times \frac{3}{7} \\[1em] \text S = 2400\ \Omega\end{aligned}$$

Hence, the value of the unknown resistance S is 2400 Ω.

Given,

• l₁ = 75 cm
• E₁ : E₂ = 3 : 2

Since the balancing length is proportional to the emf,

$$\begin{aligned}\frac{\text E_1}{\text E_2} = \frac{\text l_1}{\text l_2} \\[1em] \frac{\text E_1}{\text E_2} = \frac{3}{2} \\[1em] \text l_2 = \frac{2}{3} \times 75 \\[1em] \text l_2 = 50\ \text{cm}\end{aligned}$$

The difference in the balancing lengths is

$$\begin{aligned}\text l_1 - \text l_2 = 75 - 50 \\[1em] \text l_1 - \text l_2 = 25\ \text{cm}\end{aligned}$$

Hence, the difference in the balancing length is 25 cm.

Given,

• R₁ = 8 Ω, l₁ = 3 m
• R₂ = 4 Ω, l₂ = 2 m

When the cell of emf E and internal resistance r is shunted by a resistance R, the terminal potential difference is

$\text V = \frac{\text{ER}}{\text R + \text r}$

Since the balancing length is proportional to this potential difference,

$$\begin{aligned}\frac{\text l_1}{\text l_2} = \frac{\frac{\text E \text R_1}{\text R_1 + \text r}}{\frac{\text E \text R_2}{\text R_2 + \text r}} \\[1em] \frac{\text l_1}{\text l_2} = \frac{\text R_1 (\text R_2 + \text r)}{\text R_2 (\text R_1 + \text r)} \\[1em] \frac{3}{2} = \frac{8 (4 + \text r)}{4 (8 + \text r)} \\[1em] \frac{3}{2} = \frac{2 (4 + \text r)}{8 + \text r} \\[1em] 3 (8 + \text r) = 4 (4 + \text r) \\[1em] 24 + 3\ \text r = 16 + 4\ \text r \\[1em] \text r = 8\ \Omega\end{aligned}$$

Hence, the internal resistance of the cell is 8 Ω.

Given,

• Known resistance = 15 Ω in the left gap
• Balancing length from A = 43 cm
• End correction for the end A = 2 cm

The corrected balancing length from A is

$$\begin{aligned}\text l = 43 + 2 \\[1em] \text l = 45\ \text{cm}\end{aligned}$$

and the length of the other portion is 100 − 43 = 57 cm.

Applying the balance condition,

$$\begin{aligned}\frac{15}{\text R} = \frac{45}{57} \\[1em] \text R = 15 \times \frac{57}{45} \\[1em] \text R = 19\ \Omega\end{aligned}$$

Hence, the determined value of R is 19 Ω.

Given,

• Resistance of the bridge wire, R₀ = 50 Ω
• Series resistance = R₀/2
• Emf of the driving cell = E with internal resistance r₁
• Emf of the cell in the secondary circuit = E/2
• Balancing length, l = 72 cm

The current in the main circuit is

$$\begin{aligned}\text I = \frac{\text E}{\text r_1 + \frac{\text R_0}{2} + \text R_0} \\[1em] \text I = \frac{\text E}{\text r_1 + \frac{3\text R_0}{2}}\end{aligned}$$

Let A and B be the left and right ends of the bridge wire, C the null point on it (AC = 72 cm, so the resistance of AC is 0.72 R₀) and M the junction of r₁ and R₀/2, to which the cell E/2 is connected. At the null point no current flows through the galvanometer, so the current I flows through the whole wire from A to B.

Applying Kirchhoff's second law ($\sum \text{IR} = \sum \text E$) to the mesh ACMA, the cell E is crossed from its negative to its positive electrode and the cell E/2 from its positive to its negative electrode,

$$\begin{aligned}0.72\ \text{IR}_0 + \text{Ir}_1 = \text E - \frac{\text E}{2} \\[1em] \frac{\text E}{2} = \text I\ (0.72\ \text R_0 + \text r_1)\end{aligned}$$

Substituting for I,

$$\begin{aligned}\frac{\text E}{2} = \frac{\text E\ (0.72\ \text R_0 + \text r_1)}{\text r_1 + 1.5\ \text R_0} \\[1em] \text r_1 + 1.5\ \text R_0 = 2\ \text r_1 + 1.44\ \text R_0 \\[1em] \text r_1 = 0.06\ \text R_0 \\[1em] \text r_1 = 0.06 \times 50 \\[1em] \text r_1 = 3\ \Omega\end{aligned}$$

Hence, the value of r₁ is 3 Ω.

Given,

• Resistance of each wire = 2 Ω
• Emf of the battery = 6 V, connected across the face diagonal ac

For a cube of twelve equal resistances, the equivalent resistance across a face diagonal is

$$\begin{aligned}\text R_\text{eq} = \frac{3}{4}\ \text R \\[1em] \text R_\text{eq} = \frac{3}{4} \times 2 \\[1em] \text R_\text{eq} = \frac{3}{2}\ \Omega\end{aligned}$$

The current drawn from the battery is

$$\begin{aligned}\text I = \frac{6}{\frac{3}{2}} \\[1em] \text I = 4\ \text A\end{aligned}$$

By the symmetry of the cube about the diagonal ac, the current entering at a divides so that a current I₂ = 1 A flows along the branch ae, and this current divides equally at e.

Hence the current in the wire ef is

$\frac{\text I_2}{2} = 0.5\ \text A$

The potential difference between e and f is

$$\begin{aligned}\Delta \text V = 0.5 \times 2 \\[1em] \Delta \text V = 1\ \text V\end{aligned}$$

Hence, the voltage difference between e and f is 1 V.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: In the absence of an electric field the free electrons in a conductor move about randomly with thermal velocities. Their velocities are distributed uniformly in all directions, so the vector sum of all of them is zero and the average thermal velocity is zero.

Reason (R) is also correct: The direction of motion of the electrons is indeed randomly oriented, so for every electron moving in one direction there is, on the average, another moving in the opposite direction. This random orientation is precisely the cause of the average velocity being zero, so the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If both assertion and reason are true but reason is not the correct explanation of assertion.

Explanation

Assertion (A) is correct: On raising the temperature the free electrons collide more frequently with the ions of the lattice, so the relaxation time decreases. Since the drift velocity is proportional to the relaxation time, it decreases.

Reason (R) is also correct: On increasing the temperature the resistivity of a metallic wire increases, and since conductivity is the reciprocal of resistivity, the conductivity decreases.

However, the fall in conductivity is itself a consequence of the decrease in relaxation time, and is not the cause of the decrease in the drift velocity. Hence the Reason does not explain the Assertion.

Therefore, both assertion and reason are true but reason is not the correct explanation of assertion.

If both assertion and reason are true and reason is the correct explanation of assertion.

Explanation

Assertion (A) is correct: The drift velocity of the free electrons is

$$\begin{aligned}\text v_\text d = \frac{\text{eE}}{\text m}\ \tau \\[1em] \text v_\text d = \frac{\text{eV}}{\text{ml}}\ \tau\end{aligned}$$

For a constant potential difference V, doubling the length l halves the drift velocity.

Reason (R) is also correct: The above expression shows that at constant potential difference the drift velocity is inversely proportional to the length of the conductor, which is exactly why doubling the length halves it. Hence the Reason explains the Assertion.

Therefore, both assertion and reason are true and reason is the correct explanation of assertion.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: The moment the switch is closed, an electric field is set up throughout the conductor almost instantaneously. The free electrons everywhere in the circuit begin to drift at once, so the current is established immediately and the bulb starts glowing without any perceptible delay.

Reason (R) is incorrect: The drift speed of the electrons in a metallic wire is in fact extremely small, of the order of 10^-4 m s^-1. The bulb glows instantly not because the electrons travel fast, but because the electric field is set up throughout the circuit at once.

Therefore, assertion is true but reason is false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: In the absence of an electric field the free electrons move randomly and their average velocity is zero, so there is no current. When an electric field is applied, each electron experiences a force opposite to the field, the electrons acquire a drift velocity and a current is established.

Reason (R) is incorrect: In the presence of an electric field the drift velocity of the electrons does not decrease; it increases from zero to a steady value given by v_d = (eE/m)τ.

Therefore, assertion is true but reason is false.

If assertion is true but reason is false.

Explanation

Assertion (A) is correct: The current through a conductor is

$\text I = \text{neAv}_\text d$

For a given conductor n, e and A are fixed, so the current is directly proportional to the drift velocity.

Reason (R) is incorrect: The same relation shows that the current increases with the drift velocity. It does not decrease.

Therefore, assertion is true but reason is false.

If both assertion and reason are false.

Explanation

Assertion (A) is incorrect: Ohm's law is obeyed by metallic conductors only over a limited range of temperature. It is not obeyed by semi-conductors, diodes, valves and electrolytes, which are non-ohmic conductors.

Reason (R) is also incorrect: Ohm's law is not a fundamental law of nature. It is purely an empirical relation which holds for certain materials under certain conditions.

Therefore, both assertion and reason are false.