Electric Resistance and Ohm's Law

metals

Reason — Ohm's law is not a universal law; it is true for metallic conductors only. In a metal the free electrons drift under the applied electric field and the ratio V/I remains constant so long as the temperature and other physical states of the conductor do not change.

is excited

Reason — A nerve carries signals in the form of a flow of charge. When an external electric current is passed through it, the charge flow stimulates the nerve, that is, the nerve gets excited.

$\frac{v}{2}$

Reason — The current in a wire in terms of the drift velocity is

$\text I = ne\text A v_d$

so that $v_d = \frac{\text I}{ne\text A}$.

For the first wire, $v = \frac{\text I}{ne\pi r^2}$.

For the second wire of the same metal (same n) the radius is doubled, so its area of cross-section becomes 4 times, and the current is 2I. Hence

$$\begin{aligned}v' = \frac{2\text I}{ne(4\pi r^2)} \\[1em] v' = \frac{1}{2}\left(\frac{\text I}{ne\pi r^2}\right) \\[1em] v' = \frac{v}{2}\end{aligned}$$

current only

Reason — For a steady current, the same charge crosses every cross-section of the conductor in a given time, so the current I is constant along the length of the conductor. The current density $j = \frac{\text I}{\text A}$ and the drift speed $v_d = \frac{\text I}{ne\text A}$ both depend on the area of cross-section A, which is not uniform. The electric field $\text E = \frac{j}{\sigma}$ therefore also varies. Hence only the current remains constant.

$\frac{l}{2}$ and 2A

Reason — The resistance of a conductor is

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

R is directly proportional to the length and inversely proportional to the area of cross-section. Hence the resistance is minimum when the length is made smallest and the area of cross-section largest, that is, for the length $\frac{l}{2}$ and the area 2A.

both have equal resistances

Reason — Let the side of the plate P be s and its thickness be t. The current flows along the side, so the length is s and the area of cross-section is s × t. Therefore

$$\begin{aligned}\text R_\text P = \rho\frac{s}{s \times t} \\[1em] \text R_\text P = \frac{\rho}{t}\end{aligned}$$

For Q the side is 2s and the thickness is the same t, so

$$\begin{aligned}\text R_\text Q = \rho\frac{2s}{2s \times t} \\[1em] \text R_\text Q = \frac{\rho}{t}\end{aligned}$$

The resistance depends only on the resistivity and the thickness, and not on the side. Hence both plates have equal resistances.

2 : 5

Reason — For a bulb of rated power P at the rated voltage V,

$\text P = \frac{\text V^2}{\text R} \quad \Rightarrow \quad \text R = \frac{\text V^2}{\text P}$

Since both bulbs have the same rated voltage, R is inversely proportional to P. Hence

$$\begin{aligned}\frac{\text R_{100}}{\text R_{40}} = \frac{\text P_{40}}{\text P_{100}} \\[1em] \frac{\text R_{100}}{\text R_{40}} = \frac{40}{100} \\[1em] \frac{\text R_{100}}{\text R_{40}} = \frac{2}{5}\end{aligned}$$

100 W filament is thicker

Reason — At the same supply voltage, $\text R = \frac{\text V^2}{\text P}$, so the 100 W bulb has the smaller resistance. For filaments of the same length and the same material,

$\text R = \rho\frac{l}{\text A} \quad \Rightarrow \quad \text R \propto \frac{1}{\text A}$

A smaller resistance therefore means a larger area of cross-section. Hence the filament of the 100 W bulb is thicker.

0.35 A

Reason — The resistance of the filament from its rating is

$$\begin{aligned}\text R = \frac{\text V^2}{\text P} \\[1em] \text R = \frac{(240)^2}{100} \\[1em] \text R = 576\ \Omega\end{aligned}$$

When the bulb is operated at 200 V, the current is

$$\begin{aligned}\text I = \frac{\text V}{\text R} \\[1em] \text I = \frac{200}{576} \\[1em] \text I = 0.35\ \text A\end{aligned}$$

The dim bulb

Reason — The bulbs in a house are connected in parallel, so the same potential difference V acts across both. The power consumed is

$\text P = \frac{\text V^2}{\text R}$

that is, $\text P \propto \frac{1}{\text R}$. The bulb which glows dimly consumes less power and therefore has the larger resistance.

both 1 and 2

Reason — Ohm's law fails whenever the V–I graph of the substance is not a straight line passing through the origin. This happens in two ways : the potential difference V may vary with the current I in a non-linear manner, or the relation between V and I may not be unique, that is, for the same current there may be more than one value of V. A GaAs semiconductor or a tunnel diode shows the second of these features. Hence Ohm's law fails in both the situations.

R < R₁ and R₂

Reason — For two resistances joined in parallel,

$\frac{1}{\text R} = \frac{1}{\text R_1} + \frac{1}{\text R_2}$

Since $\frac{1}{\text R}$ is greater than $\frac{1}{\text R_1}$ as well as $\frac{1}{\text R_2}$, the equivalent resistance R is smaller than each of R₁ and R₂. In parallel combination the equivalent resistance is always less than the least of the resistances joined.

10⁷

Reason — Let n' be the number of electrons per unit length of the beam. In one second, the electrons contained in a length v pass a given point, so the current is

$\text I = n' e v$

Substituting I = 1.6 × 10^-6 A, e = 1.6 × 10^-19 C and v = 10⁶ m/s,

$$\begin{aligned}n' = \frac{\text I}{ev} \\[1em] n' = \frac{1.6 \times 10^{-6}}{(1.6 \times 10^{-19}) \times 10^{6}} \\[1em] n' = 10^{7}\end{aligned}$$

Hence 1 metre of the beam contains 10⁷ electrons.

1.07 × 10^-3 A

Reason — The revolving electron crosses any point of its orbit once in each period T. Hence the charge flowing past that point per second, that is, the current, is

$\text I = \frac{e}{\text T}$

Substituting e = 1.6 × 10^-19 C and T = 1.5 × 10^-16 s,

$$\begin{aligned}\text I = \frac{1.6 \times 10^{-19}}{1.5 \times 10^{-16}} \\[1em] \text I = 1.07 \times 10^{-3}\ \text A\end{aligned}$$

v_d

Reason — The drift velocity of free electrons in terms of the potential difference and the length of the wire is

$v_d = \frac{e}{m}\tau\frac{\text V}{l}$

This expression contains only V and l, and does not contain the area of cross-section. Since V and l are unchanged, doubling the diameter leaves the drift velocity unaltered.

v_d ∝ E

Reason — The drift velocity acquired by the free electrons in an electric field E is

$v_d = \frac{e\text E}{m}\tau = \mu_e\text E$

where $\mu_e = \frac{e\tau}{m}$ is the mobility of electrons, a constant if the temperature remains constant. Thus v_d is directly proportional to E. This linear dependence leads to $j = \sigma\text E$, which is Ohm's law in vector form.

$\frac{\text I}{2}$

Reason — The current in terms of the drift velocity is

$\text I = ne\text A v_d = ne(\pi r^2)v$

For the second wire of the same material, the radius is $\frac{r}{2}$, so the area of cross-section becomes $\frac{\pi r^2}{4}$, and the drift velocity is 2v. Hence

$$\begin{aligned}\text I' = ne\left(\frac{\pi r^2}{4}\right)(2v) \\[1em] \text I' = \frac{1}{2}\left[ne(\pi r^2)v\right] \\[1em] \text I' = \frac{\text I}{2}\end{aligned}$$

reduces to $\frac{1}{3}$ of initial value

Reason — The drift velocity of free electrons in terms of the emf applied and the length of the conductor is

$v_d = \frac{e}{m}\tau\frac{\text E}{\text L}$

so that $v_d \propto \frac{1}{\text L}$ when E is kept constant. On stretching, the length becomes 3L, hence the drift velocity reduces to $\frac{1}{3}$ of its initial value.

Drift velocity alone

Reason — In the absence of an electric field the free electrons move randomly with large thermal velocities, but the average of these velocities over all the electrons is zero, so they carry no net current. When a potential difference is applied, the electrons acquire a small additional average velocity, the drift velocity, in a direction opposite to the field, and this alone gives the current $\text I = ne\text A v_d$.

maximum when the battery is connected across 1 cm × $\frac{1}{2}$ cm faces

Reason — For a given rod,

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

so the resistance is greatest when the length along the current is largest and the area of cross-section is smallest. When the battery is connected across the 1 cm × $\frac{1}{2}$ cm faces, the current flows along the 10 cm length through the smallest area of cross-section $\left(1 \times \frac{1}{2}\right)$ cm². Hence the resistance is maximum in this case.

increased by 2%

Reason — On stretching a wire, its volume remains constant, so the resistance varies as

$\text R = \rho\frac{l}{\text A} = \rho\frac{l^2}{\text V} \quad \Rightarrow \quad \text R \propto l^2$

Taking the fractional change,

$$\begin{aligned}\frac{\Delta\text R}{\text R} \times 100 = 2 \times \frac{\Delta l}{l} \times 100 \\[1em] \frac{\Delta\text R}{\text R} \times 100 = 2 \times 1% \\[1em] \frac{\Delta\text R}{\text R} \times 100 = 2%\end{aligned}$$

Hence the resistance increases by 2%.

I₁ = I₂, j₂ > j₁, v₂ > v₁

Reason — For a steady current the same charge crosses every cross-section in a given time, so the current is the same at the sections 1 and 2, that is, I₁ = I₂.

From the figure the wire tapers, so the area of cross-section at 2 is smaller than that at 1. Since

$j = \frac{\text I}{\text A} \quad \text{and} \quad v_d = \frac{\text I}{ne\text A}$

both the current density and the drift velocity are inversely proportional to the area of cross-section. Hence j₂ > j₁ and v₂ > v₁.

0 V

Reason — As seen in the figure, the battery B₂ has a plain connecting wire joined directly across its two terminals, that is, B₂ is short-circuited.

The whole current sent out by B₂ passes through this wire of zero resistance, and the potential difference between the two ends of a wire of zero resistance is

$$\begin{aligned}\text V = \text{IR} \\[1em] \text V = \text I \times 0 \\[1em] \text V = 0\end{aligned}$$

Hence the potential difference across the battery B₂ is zero.

conductivity

Reason — According to Ohm's law in vector form, the current density at a point is directly proportional to the electric field at that point,

$\vec{j} = \sigma\vec{\text E} \quad \Rightarrow \quad \sigma = \frac{j}{\text E}$

The constant of proportionality σ is the electrical conductivity (specific conductance) of the material, and it is independent of the electric field producing the current.

$\frac{\rho_1 + \rho_2}{2}$

Reason — Let each wire be of length l and area of cross-section A. In series the equivalent resistance is

$$\begin{aligned}\text R = \text R_1 + \text R_2 \\[1em] \text R = \rho_1\frac{l}{\text A} + \rho_2\frac{l}{\text A} \\[1em] \text R = (\rho_1 + \rho_2)\frac{l}{\text A}\end{aligned}$$

The combination is a single conductor of total length 2l and the same area A. If ρ is its equivalent resistivity,

$$\begin{aligned}\rho\frac{2l}{\text A} = (\rho_1 + \rho_2)\frac{l}{\text A} \\[1em] \rho = \frac{\rho_1 + \rho_2}{2}\end{aligned}$$

4

Reason — Let the two resistances be R₁ and R₂. Then

$\text S = \text R_1 + \text R_2 \qquad \text{and} \qquad \text P = \frac{\text R_1\text R_2}{\text R_1 + \text R_2}$

Hence

$$\begin{aligned}n = \frac{\text S}{\text P} \\[1em] n = \frac{(\text R_1 + \text R_2)^2}{\text R_1\text R_2}\end{aligned}$$

Writing $(\text R_1 + \text R_2)^2 = (\text R_1 - \text R_2)^2 + 4\text R_1\text R_2$,

$n = \frac{(\text R_1 - \text R_2)^2}{\text R_1\text R_2} + 4$

The first term is never negative, so n is least when R₁ = R₂. Hence the minimum possible value of n is 4.

25 W

Reason — The resistances of the two bulbs are

$$\begin{aligned}\text R_{25} = \frac{(220)^2}{25} \\[1em] \text R_{25} = 1936\ \Omega \\[1em] \text R_{100} = \frac{(220)^2}{100} \\[1em] \text R_{100} = 484\ \Omega\end{aligned}$$

The current in the series circuit is

$$\begin{aligned}\text I = \frac{440}{1936 + 484} \\[1em] \text I = \frac{440}{2420} \\[1em] \text I = \frac{2}{11}\ \text A \approx 0.18\ \text A\end{aligned}$$

The potential difference across the 25 W bulb is

$$\begin{aligned}\text V_{25} = \frac{2}{11} \times 1936 \\[1em] \text V_{25} = 352\ \text V\end{aligned}$$

This exceeds its rated 220 V, while the 100 W bulb gets only 88 V. Hence the 25 W bulb fuses.

cot 2θ

Reason — For a V–I graph, the resistance is the slope of V with respect to I.

From the figure the line for T₂ makes an angle θ with the V-axis and the line for T₁ makes an angle θ with the I-axis. Hence

$$\begin{aligned}\text R_2 = \tan(90^{\circ} - \theta) \\[1em] \text R_2 = \cot\theta \\[1em] \text R_1 = \tan\theta\end{aligned}$$

The change in resistance is proportional to the change in temperature, so

$$\begin{aligned}\text T_2 - \text T_1 \propto \text R_2 - \text R_1 \\[1em] = \cot\theta - \tan\theta \\[1em] = \frac{\cos\theta}{\sin\theta} - \frac{\sin\theta}{\cos\theta} \\[1em] = \frac{\cos^2\theta - \sin^2\theta}{\sin\theta\cos\theta} \\[1em] = \frac{\cos 2\theta}{\frac{1}{2}\sin 2\theta} \\[1em] = 2\cot 2\theta\end{aligned}$$

Hence (T₂ − T₁) is proportional to cot 2θ.

The number of free electrons in metals is of the order of 10²⁹ per metre³.

Explanation:
It is this very large free-electron density that enables a metal to carry a large current even though the charge and the drift speed of an electron are both extremely small.

The relation is

$\text I = ne\text A v_d$

where A is the area of cross-section of the wire, n is the free-electron density, that is, the number of free electrons per unit volume, and e is the magnitude of the electronic charge.

The dimensions of resistance are [M L² T^-3 A^-2].

Explanation:
This follows from

$$\begin{aligned}\text{ohm} = \frac{\text{volt}}{\text{ampere}} \\[1em] \text{ohm} = \frac{\text{newton-metre}}{\text{ampere}^2\text{-second}}\end{aligned}$$

Electric conductivity : The electric conductivity (specific conductance) of a material is the reciprocal of its resistivity, that is, $\sigma = \frac{1}{\rho}$.

Its SI unit is 'mho metre^-1' or ohm^-1 m^-1 (Ω^-1 m^-1), also called siemens per metre.

Note: The textbook's answer (reciprocal of resistance, unit mho) describes conductance, not conductivity.

In increasing order of electrical conductivity the metals are :

Iron, aluminium, copper, silver.

Explanation:
Silver has the lowest resistivity and therefore the highest conductivity, while iron has the highest resistivity among the four.

The relation is

$v_d = \frac{e\text V\tau}{ml}$

where m is the mass of the electron, τ the relaxation time and l the length of the wire. Thus the drift velocity is directly proportional to the potential difference applied across the ends of the wire.

The drift velocity increases.

Explanation:
The drift velocity is related to the electric field by $v_d = \frac{e\text E}{m}\tau$, so v_d is directly proportional to E. Hence an increase in the electric field increases the drift velocity of the electrons.

The resistance of a conductor depends upon the nature of the material, its length, its area of cross-section and its temperature. The relation is

$\text R = \frac{m}{ne^2\tau}\frac{l}{\text A} = \rho\frac{l}{\text A}$

that is, the resistance is directly proportional to the length and inversely proportional to the area of cross-section.

The specific resistance depends on the material only.

Explanation:
It is a characteristic property of the material of the conductor, $\rho = \frac{m}{ne^2\tau}$, and it does not depend upon the length, the area of cross-section or the resistance of the particular specimen.

The resistivity of a metal depends upon its temperature and strain.

Explanation:
The resistivity of a metal increases with a rise in temperature, and it also changes when the metal is strained.

The dimensions of specific resistance are [M L³ T^-3 A^-2].

Explanation:
They follow from $\rho = \frac{\text{RA}}{l}$, since the dimensions of resistance are [M L² T^-3 A^-2].

Silver is the best electrical conductor, because its specific resistance is the lowest of the three.

For a wire of given length and area of cross-section, $\text R = \rho\frac{l}{\text A}$, so the material of the least specific resistance offers the least resistance to the flow of current.

Metals have the greater temperature coefficient of resistance.

Explanation:
Alloys such as manganin and constantan have a very small temperature coefficient of resistance, so that their resistance is hardly affected by a change of temperature. This is why they are used for making standard resistances.

At absolute zero temperature pure germanium behaves as an insulator.

Explanation:
Germanium is a semiconductor and its free electrons are produced by thermal energy. At absolute zero no free electrons are available in it, so it cannot conduct electricity.

No, a wire carrying current is not charged.

Explanation:
An electric current is only a flow of electrons through the wire. As many electrons enter the wire at one end, the same number leave it at the other end. Hence the wire as a whole remains electrically neutral.

A large current is obtained because the number of free electrons in a metal is extremely large, being of the order of 10²⁸ per m³.

Since $\text I = ne\text A v_d$, the very large value of n compensates for the small values of e and v_d, and the current comes out to be large.

Yes, an electric field exists within the conductor.

Explanation:
A current flows in a conductor only when an electric field is established inside it, because it is this field which exerts a force on the free electrons and makes them drift. The field is related to the current density by $\vec{j} = \sigma\vec{\text E}$.

There is no change in the resistance.

Explanation:
The resistance of the wire is $\text R = \rho\frac{l}{\text A} = \rho\frac{l}{\pi r^2}$.

When the radius is halved, the area of cross-section becomes one-fourth, and the length is also reduced to one-fourth. Hence

$$\begin{aligned}\text R' = \rho\frac{\left(\frac{l}{4}\right)}{\left(\frac{\text A}{4}\right)} \\[1em] \text R' = \rho\frac{l}{\text A} \\[1em] \text R' = \text R\end{aligned}$$

The ratio of their resistances is 4 : 1.

Explanation:
For wires of the same material and the same length, $\text R \propto \frac{1}{\text A} \propto \frac{1}{d^2}$. Hence

$$\begin{aligned}\frac{\text R_1}{\text R_2} = \left(\frac{d_2}{d_1}\right)^2 \\[1em] \frac{\text R_1}{\text R_2} = \left(\frac{2}{1}\right)^2 \\[1em] \frac{\text R_1}{\text R_2} = \frac{4}{1}\end{aligned}$$

The ratio of the currents is 1 : 2.

Explanation:
For the same material and the same area of cross-section, $\text R \propto l$, so the resistances are in the ratio 2 : 1. Since the same potential difference is applied, $\text I = \frac{\text V}{\text R} \propto \frac{1}{\text R}$, and hence the currents are in the ratio 1 : 2.

The ratio of the currents is 2 : 1.

Explanation:
For the same material and the same length, $\text R \propto \frac{1}{\text A}$, so the resistances are in the ratio 1 : 2. For the same potential difference $\text I \propto \frac{1}{\text R}$, and hence the currents are in the ratio 2 : 1.

12 × 10³ Ω ± 20%

Explanation:
For the colour code, brown corresponds to the figure 1, red to the figure 2 and orange to the multiplier 10³. As there is no fourth band, the tolerance is 20%.

Kilowatt-hour is the practical unit of electrical energy. It is the electrical energy consumed when an appliance of 1 kilowatt power works for 1 hour.

It is commonly known as 1 unit of electricity.

One unit of electricity means that an electric appliance of 1000 watt power has been operated for 1 hour, that is, the electrical energy consumed is 1 kilowatt-hour.

1 kW-h = 3.6 × 10⁶ J

Explanation:

$$\begin{aligned}1\ \text{kW-h} = 1000\ \text W \times 3600\ \text s \\[1em] 1\ \text{kW-h} = 3.6 \times 10^6\ \text J\end{aligned}$$

Bulb A has the higher resistance.

Explanation:
Both bulbs are marked for the same voltage 220 V, and $\text R = \frac{\text V^2}{\text P}$, so R is inversely proportional to the rated power. Bulb A is of the lower wattage (40 W), hence it has the higher resistance.

The maximum safe voltage is 7 V.

Explanation:
For a resistor of resistance R which can safely dissipate a power P,

$$\begin{aligned}\text P = \frac{\text V^2}{\text R} \quad \Rightarrow \quad \text V = \sqrt{\text{PR}} \\[1em] \text V = \sqrt{0.5 \times 98} \\[1em] \text V = \sqrt{49} \\[1em] \text V = 7\ \text V\end{aligned}$$

3 × 10⁴ J

Explanation:
The energy liberated is

$$\begin{aligned}\text W = \text P t \\[1em] \text W = (1000\ \text W) \times (30\ \text s) \\[1em] \text W = 3 \times 10^4\ \text J\end{aligned}$$

No, Ohm's law is not true for all conductors.

Explanation:
It is true only for metallic conductors, and even for them only so long as the temperature and other physical states of the conductor do not change. Substances such as the vacuum diode and the semiconductor diode do not obey Ohm's law.

Copper wire and tungsten wire are examples of linear resistors.

Explanation:
Both are metallic conductors and give a straight-line V–I graph, that is, they obey Ohm's law. Carbon resistance and diode are non-ohmic.

Vacuum diode and semiconductor diode are conducting elements which do not obey Ohm's law.

Explanation:
For such elements the V–I graph is not a straight line, so the ratio V/I is not constant.

Yes, the formula V = I R is true for non-ohmic resistance also.

Explanation:
The relation V/I = R is merely the definition of the resistance of a conductor, whether the conductor obeys Ohm's law or not. Ohm's law is the additional statement that this ratio remains constant, which is true only for ohmic conductors.

The electric field lines are perpendicular to the surface of the sphere and there is no field inside it. This is the property of a conductor, so the material 'X' is a conductor.

For a conductor the resistivity increases with a rise in temperature, so the graph of resistivity versus temperature is as shown.

The carriers of electric current are :

(a) A metallic rod — free electrons.

(b) An electrolytic cell — positive and negative ions of the electrolyte.

(c) A hydrogen discharge tube — electrons and positive ions (protons) of hydrogen.

(d) A semiconductor — electrons and holes.

(e) A superconductor — electron pairs.

Free-electron model : According to this model, the valence electrons of the atoms of a metal are not bound to any particular atom but are free to move about within the whole volume of the metal. These are called free electrons and their number is of the order of 10²⁹ per m³.

In the absence of an electric field these free electrons move randomly in all directions with large thermal velocities, colliding with the positive metal ions. The average of their velocities is zero, so there is no net current.

When a potential difference is applied, an electric field is set up inside the conductor and each free electron acquires a small additional average velocity, the drift velocity, opposite to the field. This ordered drift of free electrons constitutes the electric current in the conductor.

(i) The number of free electrons passing through a cross-section of the wire in a time-interval Δt is

$\text N = n\text A v_d \Delta t$

(ii) The current in the wire is

$\text I = \frac{\text{N}e}{\Delta t} = ne\text A v_d$

(iii) The current density in the wire is

$j = \frac{\text I}{\text A} = ne v_d$

Electrical resistivity : The resistance of a wire is $\text R = \rho\frac{l}{\text A}$. If l = 1 m and A = 1 m², then R = ρ.

Thus the electrical resistivity (specific resistance) of a material is the resistance of a wire of that material of unit length and unit area of cross-section.

It is a characteristic property of the material, $\rho = \frac{m}{ne^2\tau}$, and its SI unit is ohm-metre (Ω-m).

The resistivity of a material is the resistance of a block of that material of length 1 metre and area of cross-section 1 metre².

Hence the statement means that a block of manganin of length 1 metre and area of cross-section 1 metre² has a resistance of 44 × 10^-8 Ω.

Specific conductance : The reciprocal of the specific resistance of a material is called its specific conductance (electrical conductivity),

$\sigma = \frac{1}{\rho}$

Its unit is (ohm-metre)^-1, that is, mho m^-1 or Ω^-1 m^-1.

Its dimensional formula is [M^-1 L^-3 T³ A²].

For a metallic conductor the resistivity increases with a rise in temperature.

The resistivity is $\rho = \frac{m}{ne^2\tau}$. On heating, the metal ions vibrate more vigorously, so the free electrons collide with them more frequently and the relaxation time τ decreases. Since ρ is inversely proportional to τ, the resistivity increases.

For a semiconductor the resistivity decreases rapidly with a rise in temperature, because the number of free electrons n increases greatly and this increase outweighs the decrease in τ.

The relaxation time decreases.

Explanation:
The resistance of a conductor is

$\text R = \frac{m}{ne^2\tau}\frac{l}{\text A}$

so that $\text R \propto \frac{1}{\tau}$. When the temperature of a good conductor increases, its resistance increases, and hence the relaxation time τ of the electrons decreases. Physically, the metal ions vibrate more vigorously and the electrons suffer collisions more frequently.

(i) Cu — positive

(ii) Si — negative

(iii) Ag — positive

(iv) Ge — negative

(v) C — negative

For metals (Cu, Ag) the resistance increases with a rise in temperature, so α is positive. For semiconductors and carbon (Si, Ge, C) the resistance decreases with a rise in temperature, so α is negative.

Silicon and germanium are semiconductors, and for a semiconductor the temperature coefficient of resistance is negative. With a rise in temperature the number of free electrons n increases greatly, so the resistance $\text R = \frac{ml}{ne^2\tau\text A}$ goes on decreasing.

Hence the graph of resistance R against temperature T is as shown.

In the absence of a potential difference the motion of the free electrons in a metal is purely random. They move in all directions with large thermal velocities, so the average of their velocities is zero and there is no net motion in any direction. Hence no charge is transported across any cross-section and there is no current.

When a potential difference (or electric field) is established across the conductor, each free electron acquires a drift velocity in a direction opposite to the field. This ordered drift of the free electrons constitutes the electric current.

(i) The drift velocity is $v_d = \frac{\text I}{ne\text A}$. When the radius is doubled, the area of cross-section becomes 4 times. For the same current in the same material, the drift velocity is reduced to one-fourth.

(ii) In iron the free-electron density n is smaller than that in copper. Since $v_d \propto \frac{1}{n}$ for the same current and the same radius, the drift velocity in the iron wire will increase.

No, it does not mean that all the free electrons are moving in the same direction.

Explanation:
The free electrons continue their random thermal motion in all directions with very large speeds. The drift velocity produced by the electric field is extremely small, of the order of 10^-5 m s^-1, and is merely superposed on these large random thermal velocities. It is only the average velocity of all the free electrons which is directed from the lower to the higher potential.

The electric field accelerates an electron only between two successive collisions. The electron frequently collides with the positive ions of the metal, and in each collision it loses the velocity it had gained since the previous collision.

Hence the velocity gained between two collisions is lost in the next collision, and the net acceleration averages out to zero. The result is that the electrons move with a small steady average velocity, the drift velocity, given by $v_d = \frac{e\text E}{m}\tau$.

(i) In the absence of an electric field — yes. The free electrons move with uniform velocity between two successive collisions, so their paths are straight lines.

(ii) In the presence of an electric field — no. Each electron is acted upon by a constant force $-e\vec{\text E} \space$between two collisions, so it moves with a constant acceleration and its path is in general curved.

When the switch is put into the 'ON' position, an electric impulse travels through the circuit with the speed of light, 3 × 10⁸ m s^-1. Hence an electric field is established almost instantly at every point of the circuit, however far the bulb may be from the switch.

As a result, the free electrons at every point of the circuit, including those already present in the filament, begin to drift at practically the same instant. So the current begins to flow and the bulb lights up almost immediately, even though the drift velocity of the electrons is very small.

(i) When V is halved : $\text E = \frac{\text V}{l}$, so E is also halved. The resistance $\text R\left(= \frac{\text V}{\text I}\right)$ remains unaffected, because the current I is also halved.

(ii) When l is halved : $\text E = \frac{\text V}{l}$, so E is doubled, and $\text R = \rho\frac{l}{\text A}$ is halved.

(iii) When d is doubled : E remains unaffected, since it does not depend on the diameter. The area of cross-section becomes 4 times, so $\text R\left(= \frac{\rho l}{\pi r^2}\right)$ becomes $\frac{\text R}{4}$.

The drift velocity is $v_d = \left(\frac{e\text V}{ml}\right)\tau$.

(i) When V is doubled : v_d is directly proportional to V, so the drift velocity is doubled.

(ii) When l is doubled : v_d is inversely proportional to l, so the drift velocity is halved.

(iii) When d is doubled : the area of cross-section becomes 4 times, so the resistance becomes one-fourth and the current becomes 4 times. Since $v_d = \frac{\text I}{ne\text A}$, with both I and A becoming 4 times, the drift velocity remains unchanged.

On stretching l to 3l the volume remains constant, so the area of cross-section is reduced to $\frac{\text A}{3}$.

(i) Resistance : $\text R = \rho\frac{l}{\text A}$ becomes $\rho\frac{3l}{\text A/3} = 9\text R$, that is, the resistance becomes 9 times.

(ii) Resistivity : ρ is a property of the material only, and the material is unchanged. Hence the resistivity remains unchanged.

(iii) Drift speed : Since R becomes 9 times and V is constant, the current $\text I = \frac{\text V}{\text R}$ reduces to $\frac{1}{9}$ of its value. Now $v_d = \frac{\text I}{ne\text A}$, and A is reduced to $\frac{\text A}{3}$, so the drift speed becomes one-third of its initial value.

On stretching, the volume of the wire remains constant. If l and A are the initial and l' and A' the final length and area of cross-section, then

$\text A l = \text A' l'$

When the radius is halved, the area of cross-section becomes one-fourth, so $\frac{\text A'}{\text A} = \frac{1}{4}$ and therefore $\frac{l'}{l} = \frac{\text A}{\text A'} = 4$.

If R' is the new resistance, then

$$\begin{aligned}\frac{\text R'}{\text R} = \frac{\rho(l'/\text A')}{\rho(l/\text A)} \\[1em] \frac{\text R'}{\text R} = \frac{l'}{l} \times \frac{\text A}{\text A'} \\[1em] \frac{\text R'}{\text R} = 4 \times 4 \\[1em] \frac{\text R'}{\text R} = 16\end{aligned}$$

Hence, the new resistance is R' = 16 R.

As soon as the current flows, heat is produced in the wire and its temperature rises. For a metallic wire the resistance increases with a rise in temperature, so the current $\left(\text I = \frac{\text V}{\text R}\right)$ decreases.

As the temperature of the wire rises, the rate of loss of heat by radiation also increases. When the rate of heat radiated becomes equal to the rate at which electrical energy is dissipated in the wire, the temperature of the wire becomes constant. Hence the resistance and the current also become steady at a lower value.

The drift velocity of the free electrons decreases.

Explanation:
On being heated the resistance of the wire increases, so for the same applied potential difference the current $\left(\text I = \frac{\text V}{\text R}\right)$ decreases. Since

$v_d = \frac{\text I}{ne\text A}$

a decrease in the current decreases the drift velocity of the free electrons.

Using the colour code, the first two bands give the significant figures, the third band gives the multiplier and the fourth band gives the tolerance (gold 5%, silver 10%, no fourth band 20%).

(i) Yellow, Violet, Brown, Gold — 4, 7, multiplier 10¹, tolerance 5%

$$\begin{aligned}\text R = 47 \times 10^1\ \Omega \pm 5% \\[1em] \text R = 470\ \Omega \pm 5%\end{aligned}$$

(ii) Yellow, Red, Orange, Silver — 4, 2, multiplier 10³, tolerance 10%

$\text R = 42 \times 10^3\ \Omega \pm 10%$

(iii) Green, Violet, Red — 5, 7, multiplier 10², no fourth band so tolerance 20%

$$\begin{aligned}\text R = 57 \times 10^2\ \Omega \pm 20% \\[1em] \text R = 5700\ \Omega \pm 20%\end{aligned}$$

(iv) Red, Yellow, Orange — 2, 4, multiplier 10³, no fourth band so tolerance 20%

$\text R = 24 \times 10^3\ \Omega \pm 20%$

(i) 47 kΩ = 47 × 10³ Ω

The significant figures are 4 and 7, so the first two bands are Yellow and Violet, and the multiplier 10³ is Orange.

Colour sequence : Yellow, Violet, Orange.

(ii) 74 kΩ = 74 × 10³ Ω

The significant figures are 7 and 4, so the first two bands are Violet and Yellow, and the multiplier 10³ is Orange.

Colour sequence : Violet, Yellow, Orange.

In series the same current I flows through all the three bulbs, so the potential difference across a bulb is proportional to its resistance,

$\text V = \text{IR}$

The resistance of a bulb is $\text R = \frac{\text V^2}{\text P}$, so the bulb of the lowest wattage has the highest resistance.

Hence the PD is highest across the 40 W bulb and lowest across the 100 W bulb.

No, the bulb does not necessarily consume greater power.

Explanation:
The steady temperature attained by a resistor does not depend upon the power consumed alone. It also depends upon its surface area, emittivity and similar factors, which decide the rate at which the resistor loses energy by radiation.

The tungsten filament of the bulb has a very small surface area, so it loses heat slowly and attains a much higher temperature even while consuming less power than the nichrome element.

Both wires are given the same voltage V, so the heat produced in time t is

$\text W = \frac{\text V^2 t}{\text R}$

that is, $\text W \propto \frac{1}{\text R}$.

For wires of the same length and same material, $\text R \propto \frac{1}{\text A}$, so $\text W \propto \text A$.

Since the areas of cross-section are in the ratio 1 : 4, the ratio of the heat produced is 1 : 4.

By Ohm's law, V = I R. If the current I is kept constant, then

$\text V \propto \text R$

Hence the graph between voltage and resistance is a straight line passing through the origin.

Ohmic resistance : If the temperature and other physical states of a conductor remain unchanged, and its resistance remains constant whatever be the potential difference applied between its ends, then it is called ohmic resistance. Its V–I graph is a straight line.

Examples : metallic wires, mercury.

Non-ohmic resistance : If the resistance changes with a change in the potential difference applied, then it is called non-ohmic resistance. Its V–I graph is not a straight line.

Examples : diode valve, thermistor.

No, a light bulb is not an ohmic resistor.

An ohmic resistor is one whose resistance does not change over a fairly wide range of applied voltages and currents. In a light bulb the filament becomes very hot as the voltage is raised, and since the resistance of a metal increases with a rise in temperature, the resistance of the filament changes considerably between zero voltage and its rated voltage. Hence the bulb is non-ohmic.

The resistivity of copper is less than that of nichrome, so for the same length and same radius the copper wire has the smaller resistance.

In series : the same current flows through both wires, and the rate of production of heat is P = I²R. The nichrome wire, having the larger resistance, generates heat at a higher rate and hence the nichrome wire glows first.

In parallel : the same voltage acts across both wires, and the rate of production of heat is $\text P = \frac{\text V^2}{\text R}$. The copper wire, having the lower resistance, generates heat at a higher rate and hence the copper wire glows first.

If connected separately to the same line, the two bulbs would glow equally because they are of equal wattage.

When they are joined in series, the resistance of the circuit increases and so the current in each bulb decreases. Hence both glow less brightly than before and the temperature of their filaments is lowered.

Due to this lowering of temperature, the resistance of the carbon filament slightly increases (carbon has a negative temperature coefficient of resistance) while that of the metallic filament decreases. Since the same current flows through both and P = I²R, the carbon filament will glow more.

The length of the element should be decreased to $\frac{2}{3}$ of its initial value.

With a given DC supply of V volt, the heat produced in the element of resistance R in time t is

$\text H = \frac{\text V^2 t}{\text R}\ \text J$

The same quantity of heat is needed to boil the water, so for a smaller time t the resistance R must be reduced in the same ratio. To reduce the time from 15 min to 10 min, the resistance must become $\frac{10}{15} = \frac{2}{3}$ of its initial value.

Since $\text R \propto l$ for the same wire, the length of the heating element should be decreased to $\frac{2}{3}$ of its initial value.

In series : The 100 W bulb has a lower resistance than the 60 W bulb. On replacing the 60 W bulb by the 100 W bulb, the total resistance of the series circuit decreases and so the current increases. The heat produced by the heater of resistance R is $\text H = \frac{\text I^2\text R t}{\text J}$, so the heater will produce more heat.

In parallel : When the heater and the bulb are joined in parallel, the potential difference across each is equal to the mains voltage, whatever be the resistance of the bulb. Hence on replacing the bulb the current through the heater remains unchanged and it will produce the same heat as before.

When the accumulator of emf E and internal resistance r is connected to R₁ and then to R₂, the currents are

$\text I_1 = \frac{\text E}{\text R_1 + r}, \qquad \text I_2 = \frac{\text E}{\text R_2 + r}$

The heat dissipated in time t is $\text H = \frac{\text I^2\text R t}{\text J}$. For the heat to be the same in both cases,

$$\begin{aligned}\text I_1^2\text R_1 = \text I_2^2\text R_2 \\[1em] \frac{\text E^2\text R_1}{(\text R_1 + r)^2} = \frac{\text E^2\text R_2}{(\text R_2 + r)^2}\end{aligned}$$

Taking the square root of both sides,

$$\begin{aligned}\frac{\sqrt{\text R_1}}{\text R_1 + r} = \frac{\sqrt{\text R_2}}{\text R_2 + r} \\[1em] \sqrt{\text R_1}\thinspace\text R_2 + r\sqrt{\text R_1} = \sqrt{\text R_2}\thinspace\text R_1 + r\sqrt{\text R_2} \\[1em] \sqrt{\text R_1\text R_2}\left(\sqrt{\text R_2} - \sqrt{\text R_1}\right) = r\left(\sqrt{\text R_2} - \sqrt{\text R_1}\right)\end{aligned}$$

Hence, $r = \sqrt{\text R_1\text R_2}$.

(i) In parallel : The potential difference V across both the bulbs is the same. The rate of energy production is

$\text P = \frac{\text V^2}{\text R}$

so P is higher for the bulb of lower resistance. The bulb of higher wattage has the lower resistance, hence the 100 W bulb glows brighter.

On switching off one bulb, the potential difference across the other is unchanged, so its brightness is unaffected. Since one source of light has been removed, the total light in the room will decrease.

(ii) In series : The same current I flows through both the bulbs, and the rate of energy production is

$\text P = \text I^2\text R$

so P is higher for the bulb of higher resistance. The bulb of lower wattage has the higher resistance, hence the 60 W bulb will glow brighter.

If one bulb is taken out and the other is left across the mains, the total resistance falls, the current rises, and the light in the room increases.

In the given circuit the bulb B₁ is in series with the parallel combination of B₂ and B₃.

On removing B₃, the parallel combination is lost and the equivalent resistance of the circuit increases. Hence the main current, which flows through B₁, decreases, and so B₁ becomes dim.

Earlier the current through the parallel combination was shared between B₂ and B₃. Now the whole of the current passes through B₂ alone, so the current in B₂ increases and B₂ glows more bright.

(i) The drift velocity in terms of the applied potential difference is

$v_d = \frac{e\tau}{m}\frac{\text V}{l} \quad \Rightarrow \quad \frac{v_d}{\text V} = \frac{e\tau}{ml}$

Hence the slope of the line represents $\frac{e\tau}{ml}$, that is, the drift velocity acquired per unit potential difference.

(ii) From the above, the slope is inversely proportional to the length of the wire,

$\text{slope} \propto \frac{1}{l}$

From the figure the line B has the smaller slope, hence the wire B is longer.

(a) Magnitude of the electric field : The field within a rod is $\text E = \frac{\text V}{l}$. Hence

$\text E_1 = \frac{\text V}{\text L}, \qquad \text E_2 = \frac{\text V}{2\text L}, \qquad \text E_3 = \frac{2\text V}{3\text L}$

so that E₁ > E₃ > E₂.

(b) Current density : By Ohm's law in vector form, j = σE, and all the rods are of copper, so σ is the same for all. Hence j is proportional to E, and

j₁ > j₃ > j₂.

(c) Drift speed : Since $v_d = \frac{e\text E}{m}\tau$, the drift speed is also proportional to E. Hence

$v_{d_1}$ > $v_{d_3}$ > $v_{d_2}$.

In each case the diameter does not enter the result, because the electric field, the current density and the drift speed all depend only on the potential difference and the length of the rod.

Consider a conductor of area of cross-section A in which the free-electron density is n, that is, there are n free electrons per unit volume. Let each electron have a charge e and let the free electrons drift with a drift velocity v_d.

In a small time-interval Δt every free electron covers a distance v_dΔt. Hence all the free electrons contained in a cylinder of length v_dΔt and area of cross-section A cross the given cross-section in this interval.

Volume of this cylinder = A v_d Δt

Number of free electrons in it, N = n A v_d Δt

Charge crossing the cross-section in time Δt is

$$\begin{aligned}\Delta \text q = \text Ne \\[1em] \Delta \text q = ne\text A v_d \Delta t\end{aligned}$$

Therefore the current in the conductor is

$$\begin{aligned}\text I = \frac{\Delta \text q}{\Delta t} \\[1em] \text I = ne\text A v_d\end{aligned}$$

The current density is defined as the current per unit area of cross-section,

$j = \frac{\text I}{\text A}$

Hence

$$\begin{aligned}j = \frac{ne\text A v_d}{\text A} \\[1em] j = ne v_d\end{aligned}$$

Thus, the current density in a current-carrying conductor is j = n e v_d.

Specific resistance (resistivity) : The resistance of a wire of a given material is directly proportional to its length l and inversely proportional to its area of cross-section A,

$\text R \propto \frac{l}{\text A} \quad \Rightarrow \quad \text R = \rho\frac{l}{\text A}$

The constant of proportionality ρ is called the specific resistance or resistivity of the material.

If l = 1 m and A = 1 m², then R = ρ. Hence the specific resistance of a material is the resistance of a wire of that material of unit length and unit area of cross-section.

Unit : From $\rho = \frac{\text{RA}}{l}$, the SI unit of resistivity is ohm-metre (Ω-m).

Dependence of resistance on the dimensions : From the proof of Ohm's law, the resistance of a wire of length l and area of cross-section A is

$\text R = \frac{m}{ne^2\tau}\frac{l}{\text A}$

In this expression $\frac{m}{ne^2\tau}$ is the resistivity ρ, which is a characteristic property of the material and does not depend on the dimensions of the specimen. Hence for a conductor made of a given material

$\text R \propto \frac{l}{\text A}$

that is, the resistance of a conductor is directly proportional to its length and inversely proportional to its area of cross-section.

In the equation $\vec{\text j} = \sigma\vec{\text E}$:

• $\vec{\text j} \space$is the current density, that is, the current flowing per unit area of cross-section held normal to the direction of flow of charge. Its SI unit is A m^-2.
• σ is the specific conductance (electrical conductivity) of the material, the reciprocal of its resistivity. Its SI unit is (Ω m)^-1.
• $\vec{\text E} \space$is the electric field intensity established inside the conductor. Its SI unit is V m^-1.

Equality of units :

Unit of the left-hand side = A m^-2

Unit of the right-hand side

$$\begin{aligned}= (\Omega\ \text m)^{-1} \times (\text V\ \text m^{-1}) \\[1em] = \frac{\text V}{\Omega}\ \text m^{-2}\end{aligned}$$

But by Ohm's law $\frac{\text{volt}}{\text{ohm}} = \text{ampere}$. Hence

$= \text A\ \text m^{-2}$

Thus the combination of units on the two sides is identical.

Resistivity : The resistivity of a conductor is the resistance of a wire of that material of unit length and unit area of cross-section, $\rho = \frac{\text{RA}}{l}$. In terms of the free-electron model,

$\rho = \frac{m}{ne^2\tau}$

(i) Variation for a metallic wire : When a metallic wire is heated, the positive metal ions vibrate more vigorously about their mean positions. The free electrons therefore collide with them more frequently and the relaxation time τ decreases. In a metal the free-electron density n is already very large and does not change appreciably with temperature. Since $\rho \propto \frac{1}{\tau}$, the resistivity and hence the resistance of the metallic wire increases with a rise in temperature. The temperature coefficient of resistance of a metal is therefore positive.

(ii) Variation for a piece of carbon : Carbon is a semiconductor. In it the number of free electrons n is very small at ordinary temperatures, but on heating a large number of additional free electrons are produced, so that n increases greatly. Although the relaxation time τ also decreases, the increase in n is much larger than the decrease in τ. Hence the net effect is that the resistivity, and so the resistance, of carbon decreases with a rise in temperature. The temperature coefficient of resistance of carbon is therefore negative.

Energy transfer in an electric circuit : When a potential difference V is applied across the ends of a conductor, the free electrons drift through it. In this drift the electrons repeatedly collide with the positive ions of the conductor and give up to them the kinetic energy gained from the electric field. Hence the electrical energy supplied by the source is transferred to the conductor and appears as heat in it.

Derivation : Let a charge q be carried through the conductor across which a potential difference V is maintained. Since the potential difference is the work done per unit charge, the work done by the source is

$\text W = \text{qV}$

If a steady current I flows for a time t, then the charge transferred is

$\text q = \text I t$

Therefore

$\text W = \text{VI}t$

By Ohm's law, V = I R. Substituting for V,

$$\begin{aligned}\text W = (\text{IR})\text I t \\[1em] \text W = \text I^2\text R t\end{aligned}$$

Again, substituting $\text I = \frac{\text V}{\text R}$,

$$\begin{aligned}\text W = \text V\left(\frac{\text V}{\text R}\right)t \\[1em] \text W = \frac{\text V^2 t}{\text R}\end{aligned}$$

Hence, W = VI t = I²Rt = $\frac{\text V^2 t}{\text R}$, where W is in joule when V is in volt, I in ampere, R in ohm and t in second.

Electric power : The rate at which electrical energy is consumed in an electric circuit is called electric power.

$\text P = \frac{\text W}{t}$

Expressions for electric power : If a steady current I flows for a time t across a potential difference V, the electrical energy consumed is W = VI t. Hence

$$\begin{aligned}\text P = \frac{\text{VI}t}{t} \\[1em] \text P = \text{VI}\end{aligned}$$

Using Ohm's law V = I R,

$$\begin{aligned}\text P = (\text{IR})\text I \\[1em] \text P = \text I^2\text R\end{aligned}$$

and substituting $\text I = \frac{\text V}{\text R}$,

$$\begin{aligned}\text P = \text V\left(\frac{\text V}{\text R}\right) \\[1em] \text P = \frac{\text V^2}{\text R}\end{aligned}$$

Thus, P = VI = I²R = $\frac{\text V^2}{\text R}$.

The SI unit of electric power is the watt (W), where 1 watt = 1 volt × 1 ampere = 1 joule per second. Its bigger unit is the kilowatt, 1 kW = 1000 W.

Ohmic conductors : Those conductors whose resistance remains constant whatever be the potential difference applied between their ends, the temperature and other physical states remaining unchanged, are called ohmic conductors. They obey Ohm's law and their V–I graph is a straight line passing through the origin.

Example : a metallic wire such as a manganin wire.

Non-ohmic conductors : Those conductors whose resistance changes with a change in the potential difference applied are called non-ohmic conductors. They do not obey Ohm's law and their V–I graph is not a straight line.

Example : a diode valve (vacuum-tube diode).

V–I characteristics :

For the ohmic conductor the graph is a straight line, showing that the ratio V/I is constant. For the non-ohmic conductor the graph is curved, showing that the ratio V/I changes with the applied potential difference.

(i) The student's response is incorrect.

The additional cell should have been connected in series with the existing cell. When an identical cell is connected in parallel with the existing cell, the equivalent emf of the combination remains the same as that of a single cell. Hence the potential difference across the circuit is not increased.

(ii) The student's response is correct.

The examiner wants the current to be decreased without changing the applied voltage. By Ohm's law $\text I = \frac{\text V}{\text R}$, so at constant V the current can be decreased only by increasing the resistance. Increasing the resistance through the rheostat does exactly this.

(iii) The student's response is correct.

The resistance of a wire is $\text R = \rho\frac{l}{\text A}$, so $\text R \propto \frac{1}{\text A}$ for a wire of the same material and the same length. A thick wire has a larger area of cross-section, hence a smaller resistance.

Drift velocity : Drift velocity is the average velocity with which the free electrons get drifted towards the positive end of the conductor under the influence of an externally applied electric field. It is given by

$v_d = \frac{e\text E}{m}\tau$

and is of the order of 10^-5 m s^-1.

Time of relaxation : Relaxation time is the average time interval between two successive collisions of a free electron with the metal ions. It is the ratio of the mean free path of an electron to the rms velocity of the electron,

$\tau = \frac{\lambda}{v_\text{rms}}$

Its value is of the order of 10^-14 s.

Current density : Current density is the current flowing per unit area of cross-section held normal to the direction of flow of charge,

$j = \frac{\text I}{\text A} = ne v_d$

Its SI unit is A m^-2.

Why current is a scalar : Although an electric current has both magnitude and direction, it is treated as a scalar because currents do not add according to the law of vector addition. When two currents I₁ and I₂ meet at a point at any angle θ, the resulting current is simply I₁ + I₂, whatever the value of θ; currents add as scalars, not by the parallelogram law. Current density, on the other hand, is a true vector, its direction being the direction of flow of positive charge at a point, and it obeys the laws of vector addition.

Ohm's law : If there is no change in the physical state of a conductor (such as temperature etc.), then the ratio of the potential difference applied at its ends and the current flowing through it is constant.

$\frac{\text V}{\text I} = \text R = \text{constant}$

Derivation on the basis of drift velocity : Consider a conductor of length l and area of cross-section A having n free electrons per unit volume. Let a potential difference V be applied across its ends.

The electric field set up inside the conductor is

$\text E = \frac{\text V}{l}$

Each free electron of charge e and mass m experiences a force eE, so its acceleration is

$a = \frac{e\text E}{m}$

The free electrons are repeatedly accelerated by this field and then suffer collisions with the ions of the metallic substance, losing in each collision the velocity gained since the previous collision. If τ is the relaxation time, that is, the average time between two successive collisions, then the average velocity acquired, the drift velocity, is

$$\begin{aligned}v_d = a\tau \\[1em] v_d = \frac{e\text E}{m}\tau \\[1em] v_d = \frac{e\text V\tau}{ml}\end{aligned}$$

The current in the conductor is

$$\begin{aligned}\text I = ne\text A v_d \\[1em] \text I = ne\text A\left(\frac{e\text V\tau}{ml}\right) \\[1em] \text I = \frac{ne^2\tau\text A}{ml}\text V\end{aligned}$$

Therefore

$\frac{\text V}{\text I} = \frac{m}{ne^2\tau}\frac{l}{\text A}$

For a given conductor at a constant temperature, m, n, e, τ, l and A are all constants. Hence

$\frac{\text V}{\text I} = \text{constant} = \text R$

which is Ohm's law, and

$\text R = \frac{m}{ne^2\tau}\frac{l}{\text A}$

Vector form of Ohm's law : The current density in a conductor is directly proportional to the electric field established inside it,

$\vec{j} = \sigma\vec{\text E}$

where σ is the specific conductance of the material. Thus the ratio of the current density to the electric field is a constant σ, which is independent of the electric field producing the current, and the direction of $\vec{j} \space$coincides with that of $\vec{\text E} \space$.

Is the law universal? No, Ohm's law is not a universal law. It holds only for metallic conductors, and even for them only so long as the temperature and other physical states of the conductor remain unchanged. Many substances such as the vacuum diode, the semiconductor diode and the thermistor do not obey it.

Ohmic and non-ohmic nature :

For an ohmic conductor, for example a metallic wire of manganin or constantan, the V–I graph is a straight line passing through the origin, showing that the resistance is constant.

For a non-ohmic conductor, for example a semiconductor diode or a thermistor, the V–I graph is not a straight line, showing that the resistance changes with the applied potential difference.

Resistivity : The resistivity (specific resistance) of a material is the resistance of a wire of that material of unit length and unit area of cross-section,

$\rho = \frac{\text{RA}}{l} = \frac{m}{ne^2\tau}$

where n is the free-electron density and τ the relaxation time.

(i) Conductors : When the temperature of a metal is raised, its ions vibrate more vigorously, so the free electrons collide with them more frequently and the relaxation time τ decreases. The free-electron density n is very large and remains practically unchanged. Since $\rho \propto \frac{1}{\tau}$, the resistivity of a conductor increases with a rise in temperature, that is, its temperature coefficient of resistance is positive. The variation is nearly linear over a moderate range of temperature.

(ii) Semi-conductors : In semiconductors such as silicon, germanium and carbon, the value of n is very small compared to that in metals, but with a rise in temperature the number of free electrons increases greatly. Although τ also decreases, the increase in n is much larger than the decrease in τ. Hence the resistivity of a semiconductor decreases rapidly with a rise in temperature, that is, its temperature coefficient of resistance is negative.

(iii) Alloys : Alloys such as nichrome, manganin and constantan have a high resistivity and a very small temperature coefficient of resistance. Their resistivity increases only very slightly with a rise in temperature, so their resistance is not much affected by a change of temperature. For this reason they are used for making standard resistances and heating elements.

Series combination : Let three resistances R₁, R₂ and R₃ be joined end to end and let the same current I flow through all of them. If V₁, V₂ and V₃ are the potential differences across them, then

$\text V_1 = \text{IR}_1, \qquad \text V_2 = \text{IR}_2, \qquad \text V_3 = \text{IR}_3$

The total potential difference is

$$\begin{aligned}\text V = \text V_1 + \text V_2 + \text V_3 \\[1em] \text V = \text I(\text R_1 + \text R_2 + \text R_3)\end{aligned}$$

If R is the equivalent resistance, then V = I R. Comparing,

$\text R = \text R_1 + \text R_2 + \text R_3$

Since each of R₁, R₂, R₃ is positive, R is greater than even the greatest of the resistances connected. Hence the effective resistance in series combination increases.

Parallel combination : Let the same three resistances be joined between two points across which the potential difference is V. The currents in them are

$\text I_1 = \frac{\text V}{\text R_1}, \qquad \text I_2 = \frac{\text V}{\text R_2}, \qquad \text I_3 = \frac{\text V}{\text R_3}$

The main current is

$$\begin{aligned}\text I = \text I_1 + \text I_2 + \text I_3 \\[1em] \text I = \text V\left(\frac{1}{\text R_1} + \frac{1}{\text R_2} + \frac{1}{\text R_3}\right)\end{aligned}$$

If R is the equivalent resistance, then $\text I = \frac{\text V}{\text R}$. Comparing,

$\frac{1}{\text R} = \frac{1}{\text R_1} + \frac{1}{\text R_2} + \frac{1}{\text R_3}$

Here $\frac{1}{\text R}$ is greater than each of $\frac{1}{\text R_1}$, $\frac{1}{\text R_2}$ and $\frac{1}{\text R_3}$, so R is less than the least of the resistances joined. Hence the effective resistance in parallel combination decreases.

Iron and copper wires in series : The two wires have the same length and the same area of cross-section, but the resistivity of iron is much greater than that of copper. Hence, from $\text R = \rho\frac{l}{\text A}$, the iron wire has a much larger resistance than the copper wire.

In series the same current I flows through both, and the rate of production of heat is P = I²R. Therefore heat is produced at a far greater rate in the iron wire than in the copper wire. The iron wire becomes white hot and burns out, while the copper wire produces so little heat that it does not even glow visibly.

(i) When the motor starts, its coil is at rest and no back emf is developed in it, so the resistance offered to the source is only the small resistance of the coil and the motor takes a large current. As the motor picks up speed, a back emf is developed in its coil which opposes the applied potential difference. Hence the current decreases and finally becomes steady when the motor attains its constant speed.

(ii) In a storage battery on charging, the electrical energy is stored in the form of chemical energy.

(iii) The commercial unit of electrical energy is the kilowatt-hour (kW-h), commonly called 1 unit, where 1 kW-h = 3.6 × 10⁶ J.

(iv) The heat produced in a resistance R in time t when a current I flows through it is

$\text H = \text I^2\text R t$

(v)

Given,

• Emf of the battery, E = 3.0 V
• Power dissipated in the resistor, P = 0.5 W
• Terminal voltage of the battery, V = I R = 2.5 V

The power dissipated in the resistor is

$\text P = \text I^2\text R = 0.5 \qquad \text{...(i)}$

and the terminal voltage is

$\text{IR} = 2.5 \qquad \text{...(ii)}$

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

$$\begin{aligned}\frac{\text I^2\text R}{\text{IR}} = \frac{0.5}{2.5} \\[1em] \text I = 0.2\ \text A\end{aligned}$$

Hence, the current in the circuit is 0.2 A.

(i) The drift speed of the electrons is $v_d = \frac{e\text E}{m}\tau$. Hence it depends upon

(a) the electric field established in the conductor, that is, the potential difference applied across its ends, and

(b) the relaxation time of the free electrons, which depends upon the material of the conductor and its temperature.

(ii) The current in a conductor is I = n e A v_d. Although the charge e and the drift speed v_d of an electron are both extremely small, the number of free electrons per unit volume n in a conductor is very large, being of the order of 10²⁸ per m³. This very large value of n compensates for the small values of e and v_d, and hence we get a large current in a conductor.

(iii) When the switch is closed, an electric impulse travels through the whole circuit with the speed of light, 3 × 10⁸ m s^-1. Hence an electric field is established almost instantly at every point of the circuit, and the free electrons already present everywhere in the circuit, including those in the filament of the bulb, start drifting at practically the same instant. So the current is set up throughout the circuit at once and the bulb glows immediately, in spite of the very small drift speed.

(iv) Only the current is constant along the conductor. For a steady current the same charge crosses every cross-section in a given time. The current density $\left(j = \frac{\text I}{\text A}\right)$, the electric field $\left(\text E = \frac{j}{\sigma}\right)$ and the drift speed $\left(v_d = \frac{\text I}{ne\text A}\right)$ all depend on the area of cross-section, which is not uniform.

(v) The relation between the current density and the drift velocity is

$\vec{\text j} = -ne\vec{\text v}_d$

where n is the free-electron density in the conductor. The negative sign shows that the current density is directed opposite to the drift velocity of the electrons, because the electrons carry a negative charge.

Given,

• Current, I = 20 µA = 20 × 10^-6 A
• Time, t = 30 s
• Electronic charge, e = 1.6 × 10^-19 C

The electric current is the rate of flow of charge,

$\text I = \frac{\text q}{\text t} \quad \Rightarrow \quad \text q = \text{It}$

Substituting the values,

$$\begin{aligned}\text q = (20 \times 10^{-6}) \times 30 \\[1em] \text q = 6 \times 10^{-4}\ \text C\end{aligned}$$

If N is the number of free electrons transferred, then q = Ne, so

$$\begin{aligned}\text N = \frac{\text q}{\text e} \\[1em] \text N = \frac{6 \times 10^{-4}}{1.6 \times 10^{-19}} \\[1em] \text N = 3.75 \times 10^{15}\end{aligned}$$

Hence, the charge transferred is 6 × 10^-4 C and the number of free electrons transferred is 3.75 × 10¹⁵.

Given, the current–time graph of the wire. The time axis of the graph is in milliseconds, so the intervals are 0 to 5 ms and 0 to 9 ms.

The charge passing through the wire is $\text q = \int \text I\thinspace\text{dt}$, which is equal to the area under the current–time graph.

(i) From 0s to 5s :

From 0 to 3 ms the current rises uniformly from 0 to 3 A, and from 3 ms to 5 ms it stays constant at 3 A.

$$\begin{aligned}\text q_1 = \left(\frac{1}{2} \times 3 \times 3\right) + (3 \times 2) \\[1em] \text q_1 = 4.5 + 6 \\[1em] \text q_1 = 10.5\ \text{mC}\end{aligned}$$

(ii) From 0 to 9s :

From 5 ms to 6 ms the current falls uniformly from 3 A to 2 A, from 6 ms to 8 ms it stays constant at 2 A, and from 8 ms to 9 ms it falls uniformly from 2 A to 0.

$$\begin{aligned}\text q_2 = 10.5 + \left[\frac{1}{2}(3 + 2) \times 1\right] + (2 \times 2) + \left(\frac{1}{2} \times 2 \times 1\right) \\[1em] \text q_2 = 10.5 + 2.5 + 4 + 1 \\[1em] \text q_2 = 18\ \text{mC}\end{aligned}$$

Hence, the charge passing through the wire is 10.5 mC in the first case and 18 mC in the second case.

Given,

• Current, i = 4 + 2t (i in ampere, t in second)
• Time interval, from t = 2 s to t = 6 s

The current is the rate of flow of charge,

$\text i = \frac{\text{dq}}{\text{dt}} \quad \Rightarrow \quad \text{dq} = \text i\thinspace\text{dt}$

Integrating between the given limits,

$$\begin{aligned}\text q = \int_{2}^{6} (4 + 2\text t)\thinspace\text{dt} \\[1em] \text q = \left[4\text t + \frac{2\text t^2}{2}\right]_{2}^{6} \\[1em] \text q = \left[4\text t + \text t^2\right]_{2}^{6} \\[1em] \text q = \left[(4 \times 6) + 6^2\right] - \left[(4 \times 2) + 2^2\right] \\[1em] \text q = (24 + 36) - (8 + 4) \\[1em] \text q = 60 - 12 \\[1em] \text q = 48\ \text C\end{aligned}$$

Hence, the quantity of charge passed is 48 C.

Given,

• Area of cross-section, A = 1.0 × 10^-7 m²
• Free-electron density, n = 2.0 × 10²⁸ m^-3
• Current, I = 3.2 A
• Electronic charge, e = 1.6 × 10^-19 C

The current in terms of the drift velocity is

$\text I = \text{neA}\text v_\text d \quad \Rightarrow \quad \text v_\text d = \frac{\text I}{\text{neA}}$

Substituting the values,

$$\begin{aligned}\text v_\text d = \frac{3.2}{(2.0 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (1.0 \times 10^{-7})} \\[1em] \text v_\text d = \frac{3.2}{3.2 \times 10^{2}} \\[1em] \text v_\text d = 1.0 \times 10^{-2}\ \text{m s}^{-1}\end{aligned}$$

Hence, the drift velocity of the free electrons is 1.0 × 10^-2 m s^-1.

Given,

• Charge, q = 90 C
• Time, t = 75 minutes = 75 × 60 = 4500 s
• Diameter of the wire, d = 1.0 mm, so radius r = 0.5 mm = 0.5 × 10^-3 m
• Free-electron density, n = 5.8 × 10²⁸ m^-3
• Electronic charge, e = 1.6 × 10^-19 C

Current in the wire :

$$\begin{aligned}\text I = \frac{\text q}{\text t} \\[1em] \text I = \frac{90}{4500} \\[1em] \text I = 0.02\ \text A\end{aligned}$$

Area of cross-section :

$$\begin{aligned}\text A = \pi\text r^2 \\[1em] \text A = 3.14 \times (0.5 \times 10^{-3})^2 \\[1em] \text A = 7.85 \times 10^{-7}\ \text m^2\end{aligned}$$

Drift velocity : From I = n e A v_d,

$$\begin{aligned}\text v_\text d = \frac{\text I}{\text{neA}} \\[1em] \text v_\text d = \frac{0.02}{(5.8 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (7.85 \times 10^{-7})} \\[1em] \text v_\text d = \frac{0.02}{7.28 \times 10^{3}} \\[1em] \text v_\text d = 2.7 \times 10^{-6}\ \text{m s}^{-1}\end{aligned}$$

Hence, the current is 0.02 A and the drift velocity of the free electrons is 2.7 × 10^-6 m s^-1.

Given,

• Area of cross-section, A = 1.0 × 10^-7 m²
• Current, I = 1.5 A
• Density of copper, d = 9.0 × 10³ kg m^-3
• Atomic mass of copper = 63.5, that is, 63.5 × 10^-3 kg mol^-1
• Avogadro's number, N = 6.0 × 10²³ mol^-1

Free-electron density : Each copper atom contributes one conduction electron, so n is equal to the number of copper atoms per unit volume,

$$\begin{aligned}\text n = \frac{\text{density}}{\text{atomic mass}} \times \text N \\[1em] \text n = \frac{9.0 \times 10^{3}}{63.5 \times 10^{-3}} \times (6.0 \times 10^{23}) \\[1em] \text n = (1.417 \times 10^{5}) \times (6.0 \times 10^{23}) \\[1em] \text n = 8.5 \times 10^{28}\ \text m^{-3}\end{aligned}$$

Drift speed : From I = n e A v_d,

$$\begin{aligned}\text v_\text d = \frac{\text I}{\text{neA}} \\[1em] \text v_\text d = \frac{1.5}{(8.5 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (1.0 \times 10^{-7})} \\[1em] \text v_\text d = \frac{1.5}{1.36 \times 10^{3}} \\[1em] \text v_\text d = 1.1 \times 10^{-3}\ \text{m s}^{-1}\end{aligned}$$

Hence, the average drift speed of the conduction electrons is 1.1 × 10^-3 m s^-1.

Given,

• Length of the wire, l = 3.0 m
• Current, I = 3.0 A
• Area of cross-section, A = 2.0 × 10^-6 m²
• Free-electron density, n = 8.5 × 10²⁸ m^-3
• Electronic charge, e = 1.6 × 10^-19 C

Drift velocity : From I = n e A v_d,

$$\begin{aligned}\text v_\text d = \frac{\text I}{\text{neA}} \\[1em] \text v_\text d = \frac{3.0}{(8.5 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (2.0 \times 10^{-6})} \\[1em] \text v_\text d = \frac{3.0}{2.72 \times 10^{4}} \\[1em] \text v_\text d = 1.103 \times 10^{-4}\ \text{m s}^{-1}\end{aligned}$$

Time taken to drift the whole length :

$$\begin{aligned}\text t = \frac{\text l}{\text v_\text d} \\[1em] \text t = \frac{3.0}{1.103 \times 10^{-4}} \\[1em] \text t = 2.72 \times 10^{4}\ \text s \\[1em] \text t = \frac{2.72 \times 10^{4}}{3600}\ \text h \\[1em] \text t = 7.56\ \text h \approx 7.6\ \text h\end{aligned}$$

Hence, an electron takes about 7.6 hours to drift from one end of the wire to the other.

Note: The textbook gives 7.5 h, which is 7.56 h truncated rather than rounded.

Given,

• Current density, j = 2.5 A-m^-2
• Electric field, E = 15 V-m^-1

By Ohm's law in vector form, $\text j = \sigma\text E = \frac{\text E}{\rho}$.

Therefore the resistivity is

$\rho = \frac{\text E}{\text j}$

Substituting the values,

$$\begin{aligned}\rho = \frac{15}{2.5} \\[1em] \rho = 6\ \Omega\text{-m}\end{aligned}$$

Hence, the resistivity of the conductor is 6 Ω-m.

Given,

• Length of the wire, l = 15 m
• Area of cross-section, A = 6.0 × 10^-7 m²
• Resistance, R = 5.0 Ω

The resistance of a wire is

$\text R = \rho\frac{\text l}{\text A} \quad \Rightarrow \quad \rho = \frac{\text{RA}}{\text l}$

Substituting the values,

$$\begin{aligned}\rho = \frac{5.0 \times (6.0 \times 10^{-7})}{15} \\[1em] \rho = \frac{3.0 \times 10^{-6}}{15} \\[1em] \rho = 2.0 \times 10^{-7}\ \Omega\text{-m}\end{aligned}$$

The specific conductance is the reciprocal of the specific resistance,

$$\begin{aligned}\sigma = \frac{1}{\rho} \\[1em] \sigma = \frac{1}{2.0 \times 10^{-7}} \\[1em] \sigma = 5.0 \times 10^{6}\ (\Omega\text{-m})^{-1}\end{aligned}$$

Hence, the specific resistance is 2.0 × 10^-7 Ω-m and the specific conductance is 5.0 × 10⁶ (Ω-m)^-1.

Given,

• Diameter of the wire, d = 0.46 mm = 0.046 cm, so radius r = 0.023 cm
• Specific resistance, ρ = 50 × 10^-6 ohm-cm
• Resistance required, R = 10 ohm

Area of cross-section :

$$\begin{aligned}\text A = \pi\text r^2 \\[1em] \text A = 3.14 \times (0.023)^2 \\[1em] \text A = 1.66 \times 10^{-3}\ \text{cm}^2\end{aligned}$$

Length required : From $\text R = \rho\frac{\text l}{\text A}$,

$$\begin{aligned}\text l = \frac{\text{RA}}{\rho} \\[1em] \text l = \frac{10 \times (1.66 \times 10^{-3})}{50 \times 10^{-6}} \\[1em] \text l = \frac{1.66 \times 10^{-2}}{5 \times 10^{-5}} \\[1em] \text l = 332\ \text{cm}\end{aligned}$$

Hence, a wire of length 332 cm will be required.

Given,

• Mass of copper, m = 50 g = 0.05 kg
• Length of the wire, l = 50 m
• Density of copper, d = 8.9 × 10³ kg m^-3
• Specific resistance, ρ = 1.7 × 10^-8 Ω-m

Volume of the wire :

$$\begin{aligned}\text V = \frac{\text m}{\text d} \\[1em] \text V = \frac{0.05}{8.9 \times 10^{3}} \\[1em] \text V = 5.62 \times 10^{-6}\ \text m^3\end{aligned}$$

Area of cross-section : Since V = A l,

$$\begin{aligned}\text A = \frac{\text V}{\text l} \\[1em] \text A = \frac{5.62 \times 10^{-6}}{50} \\[1em] \text A = 1.124 \times 10^{-7}\ \text m^2\end{aligned}$$

Resistance of the wire :

$$\begin{aligned}\text R = \rho\frac{\text l}{\text A} \\[1em] \text R = \frac{(1.7 \times 10^{-8}) \times 50}{1.124 \times 10^{-7}} \\[1em] \text R = \frac{8.5 \times 10^{-7}}{1.124 \times 10^{-7}} \\[1em] \text R = 7.6\ \Omega\end{aligned}$$

Hence, the resistance of the wire is 7.6 Ω.

Given,

• The wires A and B are of copper and of equal masses
• l_A = 2 l_B
• R_A = 160 Ω

Since the two wires are of the same material and of equal masses, their volumes are also equal. For a wire of volume V,

$$\begin{aligned}\text R = \rho\frac{\text l}{\text A} \\[1em] \text R = \rho\frac{\text l}{\left(\frac{\text V}{\text l}\right)} \\[1em] \text R = \frac{\rho\text l^2}{\text V}\end{aligned}$$

Hence for wires of the same material and the same volume,

$$\begin{aligned}\text R \propto \text l^2 \quad \Rightarrow \quad \frac{\text R_\text A}{\text R_\text B} = \left(\frac{\text l_\text A}{\text l_\text B}\right)^2 \\[1em] \frac{\text R_\text A}{\text R_\text B} = \left(\frac{2}{1}\right)^2 \\[1em] \frac{\text R_\text A}{\text R_\text B} = 4\end{aligned}$$

Therefore

$$\begin{aligned}\text R_\text B = \frac{\text R_\text A}{4} \\[1em] \text R_\text B = \frac{160}{4} \\[1em] \text R_\text B = 40\ \Omega\end{aligned}$$

Hence, the resistance of the wire B is 40 Ω.

Given,

• Potential difference, V = 200 V
• Current, I = 0.2 A
• Length of the filament, l = 22 cm
• Specific resistance, ρ = 7 × 10^-5 Ω-cm
• $\pi = \frac{22}{7}$

Resistance of the filament :

$$\begin{aligned}\text R = \frac{\text V}{\text I} \\[1em] \text R = \frac{200}{0.2} \\[1em] \text R = 1000\ \Omega\end{aligned}$$

Area of cross-section : From $\text R = \rho\frac{\text l}{\text A}$,

$$\begin{aligned}\text A = \frac{\rho\text l}{\text R} \\[1em] \text A = \frac{(7 \times 10^{-5}) \times 22}{1000} \\[1em] \text A = 1.54 \times 10^{-6}\ \text{cm}^2\end{aligned}$$

Radius of the filament : Since A = πr²,

$$\begin{aligned}\text r^2 = \frac{\text A}{\pi} \\[1em] \text r^2 = \frac{1.54 \times 10^{-6} \times 7}{22} \\[1em] \text r^2 = 4.9 \times 10^{-7}\ \text{cm}^2 \\[1em] \text r = \sqrt{4.9 \times 10^{-7}} \\[1em] \text r = 7 \times 10^{-4}\ \text{cm}\end{aligned}$$

Hence, the radius of the filament is 7 × 10^-4 cm.

Given,

• Dimensions of the block, 1.2 cm × 1.2 cm × 15 cm
• Specific resistance of iron, ρ = 9.68 × 10^-8 Ω m

(a) Between the two square ends :

Here the length along the current is l = 15 cm = 0.15 m and the area of cross-section is

$$\begin{aligned}\text A = 1.2\ \text{cm} \times 1.2\ \text{cm} \\[1em] \text A = 1.44\ \text{cm}^2 \\[1em] \text A = 1.44 \times 10^{-4}\ \text m^2 \\[1em] \text R = \rho\frac{\text l}{\text A} \\[1em] \text R = \frac{(9.68 \times 10^{-8}) \times 0.15}{1.44 \times 10^{-4}} \\[1em] \text R = \frac{1.452 \times 10^{-8}}{1.44 \times 10^{-4}} \\[1em] \text R = 1.0 \times 10^{-4}\ \Omega \\[1em] \text R = 100\ \mu\Omega\end{aligned}$$

(b) Between the two opposite rectangular faces :

Here the length along the current is l = 1.2 cm = 0.012 m and the area of cross-section is

$$\begin{aligned}\text A = 15\ \text{cm} \times 1.2\ \text{cm} \\[1em] \text A = 18\ \text{cm}^2 \\[1em] \text A = 18 \times 10^{-4}\ \text m^2 \\[1em] \text R = \frac{(9.68 \times 10^{-8}) \times 0.012}{18 \times 10^{-4}} \\[1em] \text R = \frac{1.1616 \times 10^{-9}}{18 \times 10^{-4}} \\[1em] \text R = 0.65 \times 10^{-6}\ \Omega \\[1em] \text R = 0.65\ \mu\Omega\end{aligned}$$

Hence, the resistance is 100 µΩ between the square ends and 0.65 µΩ between the opposite rectangular faces.

Given,

• l₁ = 50 cm, d₁ = 0.2 mm, R₁ = 0.8 Ω
• l₂ = 75 cm, d₂ = 0.1 mm, R₂ = 1.5 Ω

The specific resistance of a wire is

$$\begin{aligned}\rho = \frac{\text{RA}}{\text l} \\[1em] \rho = \frac{\text R\pi\text d^2}{4\text l}\end{aligned}$$

Hence, for the two wires,

$$\begin{aligned}\frac{\rho_1}{\rho_2} = \frac{\text R_1\text d_1^2}{\text l_1} \times \frac{\text l_2}{\text R_2\text d_2^2} \\[1em] \frac{\rho_1}{\rho_2} = \frac{\text R_1}{\text R_2} \times \left(\frac{\text d_1}{\text d_2}\right)^2 \times \frac{\text l_2}{\text l_1}\end{aligned}$$

Substituting the values,

$$\begin{aligned}\frac{\rho_1}{\rho_2} = \frac{0.8}{1.5} \times \left(\frac{0.2}{0.1}\right)^2 \times \frac{75}{50} \\[1em] \frac{\rho_1}{\rho_2} = \frac{0.8}{1.5} \times 4 \times 1.5 \\[1em] \frac{\rho_1}{\rho_2} = 3.2 \\[1em] \frac{\rho_1}{\rho_2} = \frac{16}{5}\end{aligned}$$

Hence, the ratio of the specific resistances of the materials is 16 : 5.

Given,

• Initial resistance, R = 10 Ω
• The length becomes three times, that is, l' = 3l, so n = 3

On drawing the wire, its volume remains constant, so if the length becomes n times, the area of cross-section becomes $\frac{1}{n}$ times. Hence

$$\begin{aligned}\text R' = \rho\frac{\text l'}{\text A'} \\[1em] \text R' = \rho\frac{n\text l}{\left(\frac{\text A}{n}\right)} \\[1em] \text R' = n^2\left(\rho\frac{\text l}{\text A}\right) \\[1em] \text R' = n^2\text R\end{aligned}$$

Substituting n = 3 and R = 10 Ω,

$$\begin{aligned}\text R' = 3^2 \times 10 \\[1em] \text R' = 9 \times 10 \\[1em] \text R' = 90\ \Omega\end{aligned}$$

Hence, the new resistance of the wire is 90 Ω.

Given,

• Two pieces of copper of equal mass (1 kg each)
• Thickness (diameter) of the first wire, d₁ = 1 mm
• Thickness (diameter) of the second wire, d₂ = 2 mm

Since both pieces are of the same material and of equal masses, their volumes V are equal. For such a wire,

$\text R = \rho\frac{\text l}{\text A} \quad \text{and} \quad \text l = \frac{\text V}{\text A}$

Therefore

$\text R = \frac{\rho\text V}{\text A^2} \quad \Rightarrow \quad \text R \propto \frac{1}{\text A^2} \propto \frac{1}{\text d^4}$

Hence

$$\begin{aligned}\frac{\text R_1}{\text R_2} = \left(\frac{\text d_2}{\text d_1}\right)^4 \\[1em] \frac{\text R_1}{\text R_2} = \left(\frac{2}{1}\right)^4 \\[1em] \frac{\text R_1}{\text R_2} = 16\end{aligned}$$

Hence, the resistances of the two wires are in the ratio 16 : 1.

Given,

• Resistance at 0°C, R₀ = 1.25 Ω
• Resistance at t°C, R_t = 2 × 1.25 = 2.50 Ω
• Temperature coefficient of resistance of silver, α = 0.00375°C^-1

The resistance of a conductor at t°C is

$\text R_\text t = \text R_0(1 + \alpha\text t)$

Substituting the values,

$$\begin{aligned}2.50 = 1.25(1 + 0.00375\thinspace\text t) \\[1em] 2 = 1 + 0.00375\thinspace\text t \\[1em] \text t = \frac{1}{0.00375} \\[1em] \text t = 266.7 \approx 267^{\circ}\text C\end{aligned}$$

Yes, the temperature will be the same for all conductors of silver. The temperature coefficient of resistance α is a property of the material and does not depend on the dimensions of the specimen. Since the resistance is to be doubled in every case, the relation gives the same temperature for every silver conductor.

Hence, the wire must be heated up to 267°C, and the temperature will be the same for all conductors of silver.

Given,

• Resistance at 0°C, R₀ = 3.00 Ω
• Resistance at 100°C, R₁₀₀ = 3.75 Ω
• Resistance at the unknown temperature t°C, R_t = 3.15 Ω

The resistance at t°C is given by R_t = R₀(1 + αt).

Temperature coefficient of resistance :

$$\begin{aligned}3.75 = 3.00(1 + \alpha \times 100) \\[1em] 1.25 = 1 + 100\alpha \quad \Rightarrow \quad \alpha = 0.0025^{\circ}\text C^{-1}\end{aligned}$$

Unknown temperature :

$$\begin{aligned}3.15 = 3.00(1 + 0.0025\thinspace\text t) \\[1em] 1.05 = 1 + 0.0025\thinspace\text t \\[1em] \text t = \frac{0.05}{0.0025} \\[1em] \text t = 20^{\circ}\text C\end{aligned}$$

Hence, the unknown temperature is 20°C.

Given,

• Resistance of the carbon filament at 0°C, R_C = 104 Ω
• Temperature coefficient of resistance of carbon, α_C = − 0.0003°C^-1
• Temperature coefficient of resistance of iron, α_Fe = + 0.0052°C^-1

Let R_Fe be the resistance of the iron filament at 0°C. In series the combined resistance at t°C is

$$\begin{aligned}\text R = \text R_\text C(1 + \alpha_\text C\text t) + \text R_{\text{Fe}}(1 + \alpha_{\text{Fe}}\text t) \\[1em] \text R = (\text R_\text C + \text R_{\text{Fe}}) + \text t(\text R_\text C\alpha_\text C + \text R_{\text{Fe}}\alpha_{\text{Fe}})\end{aligned}$$

For the combined resistance to be independent of temperature, the coefficient of t must be zero,

$\text R_\text C\alpha_\text C + \text R_{\text{Fe}}\alpha_{\text{Fe}} = 0$

Substituting the values,

$$\begin{aligned}104 \times (-0.0003) + \text R_{\text{Fe}} \times (0.0052) = 0 \\[1em] \text R_{\text{Fe}} = \frac{0.0312}{0.0052} \\[1em] \text R_{\text{Fe}} = 6.00\ \Omega\end{aligned}$$

Hence, an iron filament of resistance 6.00 ohm at 0°C should be connected in series.

Given,

• Resistance at 0°C, R₀ = 3.0 Ω
• Resistance at 150°C, R_t = 4.8 Ω
• Temperature, t = 150°C

The resistance at t°C is

$\text R_\text t = \text R_0(1 + \alpha\text t)$

Substituting the values,

$$\begin{aligned}4.8 = 3.0(1 + \alpha \times 150) \\[1em] 1.6 = 1 + 150\alpha \\[1em] \alpha = \frac{0.6}{150} \\[1em] \alpha = 4 \times 10^{-3}\ ^{\circ}\text C^{-1}\end{aligned}$$

Hence, the temperature coefficient of resistance of the material is 4 × 10^-3 °C^-1.

Given, three resistances of 2 Ω, 3 Ω and 6 Ω, and the required equivalent resistance is 4 Ω.

Connect the 3 Ω and the 6 Ω resistances in parallel, and join the 2 Ω resistance in series with this combination.

The equivalent resistance of the parallel combination is

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

This is in series with the 2 Ω resistance, so the equivalent resistance of the whole combination is

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

Hence, the 2 Ω resistance should be connected in series with the parallel combination of the 3 Ω and 6 Ω resistances.

Given, three resistances of 3 Ω each, and the required effective resistance is 2 Ω.

Connect two of the 3 Ω resistances in series and join the third 3 Ω resistance in parallel with this combination.

The resistance of the two in series is

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

This is joined in parallel with the third resistance of 3 Ω,

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

Hence, two resistances should be joined in series and the third in parallel with them.

Given,

• Resistance in series, R₁ = 5 Ω
• n resistors each of 6 Ω joined in parallel
• Equivalent resistance, R = 7 Ω

For n equal resistances of 6 Ω each in parallel, the equivalent resistance is

$\text R_\text P = \frac{6}{n}$

This is joined in series with the 5 Ω resistor, so

$\text R = 5 + \frac{6}{n}$

Substituting R = 7 Ω,

$$\begin{aligned}7 = 5 + \frac{6}{n} \quad \Rightarrow \quad \frac{6}{n} = 2 \\[1em] n = 3\end{aligned}$$

Hence, the number of resistors joined in parallel is n = 3.

Given,

• R_AB = 40 Ω, R_BC = 60 Ω, R_CA = 100 Ω
• Equivalent resistance is required between A and B

Between the points A and B there are two paths in parallel :

(i) the direct side AB, of resistance 40 Ω, and

(ii) the path A → C → B, consisting of CA and BC in series,

$$\begin{aligned}\text R' = 100 + 60 \\[1em] \text R' = 160\ \Omega\end{aligned}$$

These two are in parallel between A and B, so

$$\begin{aligned}\text R = \frac{40 \times 160}{40 + 160} \\[1em] \text R = \frac{6400}{200} \\[1em] \text R = 32\ \Omega\end{aligned}$$

Hence, the equivalent resistance between the points A and B is 32 Ω.

Given, four resistors each of 10 Ω joined in the form of a square, and the equivalent resistance is required between two diagonally-opposite corners.

Let the square be ABCD and let the equivalent resistance be required between the diagonally-opposite corners A and C. There are two paths in parallel between A and C :

(i) the path A → B → C, of resistance

$10 + 10 = 20\ \Omega$

(ii) the path A → D → C, of resistance

$10 + 10 = 20\ \Omega$

These two branches of 20 Ω each are in parallel between A and C, so

$$\begin{aligned}\frac{1}{\text R} = \frac{1}{20} + \frac{1}{20} \\[1em] \frac{1}{\text R} = \frac{2}{20} \\[1em] \text R = 10\ \Omega\end{aligned}$$

Hence, the equivalent resistance between the two diagonally-opposite corners is 10 Ω.

(a) From the figure, A is joined to the upper rail and B to the lower rail by plain connecting wires. Hence all the three 10 Ω resistances are connected between the same two points A and B, that is, they are in parallel.

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

(b) Let the junctions be marked P (after the first 10 Ω from A), Q (the next junction) and B.

Between Q and B two resistances of 10 Ω each are joined in parallel (the lower 10 Ω and the upper 10 Ω),

$$\begin{aligned}\text R_1 = \frac{10 \times 10}{10 + 10} \\[1em] \text R_1 = 5\ \Omega\end{aligned}$$

Between P and B there are now two paths in parallel : the path P → Q → B of resistance

$10 + 5 = 15\ \Omega$

and the upper 10 Ω joined directly from P to B. Hence

$$\begin{aligned}\text R_2 = \frac{15 \times 10}{15 + 10} \\[1em] \text R_2 = \frac{150}{25} \\[1em] \text R_2 = 6\ \Omega\end{aligned}$$

Finally, the 10 Ω between A and P is in series with this,

$$\begin{aligned}\text R = 10 + 6 \\[1em] \text R = 16\ \Omega\end{aligned}$$

Hence, the equivalent resistance is $\frac{10}{3}$ Ω in the network (a) and 16 Ω in the network (b).

(a) The network is a Wheatstone bridge with the 6 Ω and 9 Ω in the upper two arms, the 4 Ω and 6 Ω in the lower two arms, and the 7 Ω as the bridge arm.

In a Wheatstone bridge, if the ratio of the resistances in the two arms on one side equals that on the other side ($\frac{\text P}{\text Q} = \frac{\text R}{\text S}$), the bridge is balanced and no current flows through the bridge arm.

The bridge is balanced, because

$\frac{6}{9} = \frac{2}{3} \qquad \text{and} \qquad \frac{4}{6} = \frac{2}{3}$

Hence no current flows through the 7 Ω arm and it may be ignored. The upper branch (6 + 9 = 15 Ω) and the lower branch (4 + 6 = 10 Ω) are then in parallel between A and B,

$$\begin{aligned}\text R = \frac{15 \times 10}{15 + 10} \\[1em] \text R = \frac{150}{25} \\[1em] \text R = 6\ \Omega\end{aligned}$$

(b) Here also the arrangement is a balanced Wheatstone bridge. The arms are AC = 1 Ω, CB = 2 Ω, AD = 2 Ω and DB = 4 Ω, with the 10 Ω between C and D as the bridge arm.

$$\begin{aligned}\frac{\text R_\text{AC}}{\text R_\text{CB}} = \frac{1}{2} \\[1em] \frac{\text R_\text{AD}}{\text R_\text{DB}} = \frac{2}{4} \\[1em] \frac{\text R_\text{AD}}{\text R_\text{DB}} = \frac{1}{2}\end{aligned}$$

so the bridge is balanced and no current flows through the 10 Ω. The two branches in parallel are then

$$\begin{aligned}1 + 2 = 3\ \Omega \qquad \text{and} \qquad 2 + 4 = 6\ \Omega \\[1em] \text R = \frac{3 \times 6}{3 + 6} \\[1em] \text R = \frac{18}{9} \\[1em] \text R = 2\ \Omega\end{aligned}$$

(c) In this network the connecting wire joins A to D, and another connecting wire joins C to B. Hence A and D are the same point, and C and B are the same point.

Therefore each of the three 6 Ω resistances is connected between A and B, that is, all three are in parallel,

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

Hence, the equivalent resistance is 6 Ω in (a), 2 Ω in (b) and 2 Ω in (c).

Given,

• Resistance of the wire = 0.1 Ω/cm
• Side of the square, 10 cm
• Emf of the battery, V = 2 V

Resistance of each side :

$$\begin{aligned}\text R = 0.1 \times 10 \\[1em] \text R = 1\ \Omega\end{aligned}$$

Equivalent resistance between A and C : The arms AB and BC are equal (1 Ω each) and the arms AD and DC are also equal (1 Ω each). Hence the points B and D are at the same potential and no current flows through the wire joining B and D, whatever be its resistance.

Therefore the two branches A → B → C and A → D → C, each of resistance

$1 + 1 = 2\ \Omega$

are in parallel between A and C,

$$\begin{aligned}\text R_{\text{AC}} = \frac{2 \times 2}{2 + 2} \\[1em] \text R_{\text{AC}} = 1.0\ \Omega\end{aligned}$$

Current drawn from the battery :

$$\begin{aligned}\text I = \frac{\text V}{\text R_{\text{AC}}} \\[1em] \text I = \frac{2}{1.0} \\[1em] \text I = 2.0\ \text A\end{aligned}$$

Reading of the ammeter across B and D : Since B and D are at the same potential, no current passes between them. Hence the ammeter reads zero.

Hence, the equivalent resistance is 1.0 Ω, the current drawn is 2.0 A and the ammeter reads zero.

Given,

• Rating of the heater, 250 V-1000 W
• Time, t = 1 minute = 60 s

Current drawn by the heater :

$$\begin{aligned}\text I = \frac{\text P}{\text V} \\[1em] \text I = \frac{1000}{250} \\[1em] \text I = 4\ \text A\end{aligned}$$

Charge flowing in one minute :

$$\begin{aligned}\text q = \text{It} \\[1em] \text q = 4 \times 60 \\[1em] \text q = 240\ \text C\end{aligned}$$

Hence, the charge flowing through the heater in one minute is 240 C.

Given,

• Power of the heater element, P = 2 kW = 2000 W
• Resistance when hot, R = 20 Ω

The power consumed in terms of the voltage drop across the element is

$\text P = \frac{\text V^2}{\text R} \quad \Rightarrow \quad \text V = \sqrt{\text{PR}}$

Substituting the values,

$$\begin{aligned}\text V = \sqrt{2000 \times 20} \\[1em] \text V = \sqrt{40000} \\[1em] \text V = 200\ \text V\end{aligned}$$

Hence, the voltage drop across the heater element is 200 V.

Given,

• Rating of the heating element, 210 V, 630 W
• Supply voltage, V = 210 V

Current drawn :

$$\begin{aligned}\text I = \frac{\text P}{\text V} \\[1em] \text I = \frac{630}{210} \\[1em] \text I = 3.0\ \text A\end{aligned}$$

Resistance of the element :

$$\begin{aligned}\text R = \frac{\text V}{\text I} \\[1em] \text R = \frac{210}{3.0} \\[1em] \text R = 70\ \Omega\end{aligned}$$

Hence, the current drawn is 3.0 A and the resistance of the element is 70 Ω.

Given,

• Power of the motor, P = 1 HP = 746 W
• Supply voltage, V = 220 V

The power consumed is P = VI, so

$\text I = \frac{\text P}{\text V}$

Substituting the values,

$$\begin{aligned}\text I = \frac{746}{220} \\[1em] \text I = 3.39\ \text A\end{aligned}$$

Hence, the current in the motor is 3.39 A.

Given,

• Resistance of the press, R = 44 Ω
• Current, I = 5.0 A

The power consumed by a resistor carrying a current I is

$\text P = \text I^2\text R$

Substituting the values,

$$\begin{aligned}\text P = (5.0)^2 \times 44 \\[1em] \text P = 25 \times 44 \\[1em] \text P = 1100\ \text W\end{aligned}$$

Hence, the power consumed by the press is 1100 W.

Given,

• Current, I = 4.0 A
• Power of the heating-wire, P = 100 W

The power consumed is P = I²R, so

$\text R = \frac{\text P}{\text I^2}$

Substituting the values,

$$\begin{aligned}\text R = \frac{100}{(4.0)^2} \\[1em] \text R = \frac{100}{16} \\[1em] \text R = 6.25\ \Omega\end{aligned}$$

Hence, the resistance of the wire is 6.25 Ω.

Given,

• Power of the first bulb, P₁ = 40 W
• Power of the second bulb, P₂ = 100 W
• Both are designed for the same supply line of V = 220 V

The resistance of a bulb from its rating is

$\text R = \frac{\text V^2}{\text P}$

Since both bulbs work at the same voltage, R is inversely proportional to P. Hence

$$\begin{aligned}\frac{\text R_1}{\text R_2} = \frac{\text P_2}{\text P_1} \\[1em] \frac{\text R_1}{\text R_2} = \frac{100}{40} \\[1em] \frac{\text R_1}{\text R_2} = \frac{5}{2}\end{aligned}$$

Hence, the ratio of the resistances of the 40 W bulb and the 100 W bulb is 5 : 2.

Given,

• Rating of the bulb, 100 W-220 V
• Length of the filament, l = 0.1 m
• Radius of the filament, r = 2.5 × 10^-5 m

Resistance of the filament when glowing :

$$\begin{aligned}\text R = \frac{\text V^2}{\text P} \\[1em] \text R = \frac{(220)^2}{100} \\[1em] \text R = \frac{48400}{100} \\[1em] \text R = 484\ \Omega\end{aligned}$$

Area of cross-section of the filament :

$$\begin{aligned}\text A = \pi\text r^2 \\[1em] \text A = 3.14 \times (2.5 \times 10^{-5})^2 \\[1em] \text A = 1.96 \times 10^{-9}\ \text m^2\end{aligned}$$

Specific resistance : From $\text R = \rho\frac{\text l}{\text A}$,

$$\begin{aligned}\rho = \frac{\text{RA}}{\text l} \\[1em] \rho = \frac{484 \times (1.96 \times 10^{-9})}{0.1} \\[1em] \rho = \frac{9.5 \times 10^{-7}}{0.1} \\[1em] \rho = 9.5 \times 10^{-6}\ \Omega\text{-m}\end{aligned}$$

Hence, the resistance of the filament is 484 Ω and the specific resistance of its material is 9.5 × 10^-6 Ω-m.

Given,

• Rating of the bulb, 240 V-200 W
• Resistance when glowing, R = 10 R₀, where R₀ is the resistance at 0°C
• Temperature of the glowing filament, t = 2000°C

Resistance of the bulb when glowing :

$$\begin{aligned}\text R = \frac{\text V^2}{\text P} \\[1em] \text R = \frac{(240)^2}{200} \\[1em] \text R = \frac{57600}{200} \\[1em] \text R = 288\ \Omega\end{aligned}$$

Resistance at 0°C :

$$\begin{aligned}\text R_0 = \frac{\text R}{10} \\[1em] \text R_0 = \frac{288}{10} \\[1em] \text R_0 = 28.8\ \Omega\end{aligned}$$

Temperature coefficient of resistance : Using R_t = R₀(1 + αt),

$$\begin{aligned}288 = 28.8(1 + \alpha \times 2000) \\[1em] 10 = 1 + 2000\alpha \\[1em] \alpha = \frac{9}{2000} \\[1em] \alpha = 4.5 \times 10^{-3}\ ^{\circ}\text C^{-1}\end{aligned}$$

Hence, the resistance of the glowing bulb is 288 Ω and the temperature coefficient of resistance of the filament is 4.5 × 10^-3 °C^-1.

Given,

• Rating of the lamp, 15 V-20 W
• Emf of the DC source, E = 25 V
• A resistance R is joined in series with the lamp

(i) Current in the circuit : For full illumination the lamp works at its rated values, so

$$\begin{aligned}\text I = \frac{\text P}{\text V} \\[1em] \text I = \frac{20}{15} \\[1em] \text I = \frac{4}{3}\ \text A\end{aligned}$$

(ii) Resistance of the lamp :

$$\begin{aligned}\text R_\text L = \frac{\text V}{\text I} \\[1em] \text R_\text L = \frac{15}{\left(\frac{4}{3}\right)} \\[1em] \text R_\text L = \frac{45}{4}\ \Omega\end{aligned}$$

(iii) Resistance R : The potential difference across R is the remaining part of the source voltage,

$$\begin{aligned}\text V_\text R = 25 - 15 \\[1em] \text V_\text R = 10\ \text V \\[1em] \text R = \frac{\text V_\text R}{\text I} \\[1em] \text R = \frac{10}{\left(\frac{4}{3}\right)} \\[1em] \text R = 7.5\ \Omega\end{aligned}$$

Hence, the current is $\frac{4}{3}$ A, the resistance of the lamp is $\frac{45}{4}$ Ω and the resistance R is 7.5 Ω.

Given,

• Line voltage, V = 220 V
• Appliances : a 60 W bulb, a 1000 W heater and a 40 W radio

All the three appliances are joined in parallel, so a potential difference of 220 V acts across each of them.

(a) Current drawn by the heater :

$$\begin{aligned}\text I_\text H = \frac{\text P_\text H}{\text V} \\[1em] \text I_\text H = \frac{1000}{220} \\[1em] \text I_\text H = \frac{50}{11}\ \text A \\[1em] \text I_\text H = 4.55\ \text A\end{aligned}$$

(b) Current through the fuse : The fuse carries the total current drawn by all the three appliances. The total power consumed is

$$\begin{aligned}\text P = 60 + 1000 + 40 \\[1em] \text P = 1100\ \text W \\[1em] \text I = \frac{\text P}{\text V} \\[1em] \text I = \frac{1100}{220} \\[1em] \text I = 5.0\ \text A\end{aligned}$$

Hence, the current drawn by the heater is $\frac{50}{11}$ A and the current passing through the fuse is 5.0 A.

Given,

• Rating of the bulb, 30 W-6 V
• Supply voltage, V = 120 V

Current required for the bulb to glow properly :

$$\begin{aligned}\text I = \frac{\text P}{\text V} \\[1em] \text I = \frac{30}{6} \\[1em] \text I = 5\ \text A\end{aligned}$$

The bulb needs only 6 V across it, whereas the supply is of 120 V. Hence a resistance must be joined in series with the bulb so that the excess voltage is dropped across it.

Potential difference to be dropped across the series resistance :

$$\begin{aligned}\text V_\text R = 120 - 6 \\[1em] \text V_\text R = 114\ \text V\end{aligned}$$

Resistance required :

$$\begin{aligned}\text R = \frac{\text V_\text R}{\text I} \\[1em] \text R = \frac{114}{5} \\[1em] \text R = 22.8\ \Omega\end{aligned}$$

Hence, a resistance wire of 22.8 Ω should be connected in series with the bulb.

Given,

• Distance between the cities, l = 150 km
• Potential-drop in the wires = 8 V/km
• Average resistance of the wires = 0.5 Ω/km

(i) Current flowing in the wires : Consider 1 km length of the wire. Across it the potential drop is 8 V and its resistance is 0.5 Ω. By Ohm's law,

$$\begin{aligned}\text I = \frac{\text V}{\text R} \\[1em] \text I = \frac{8}{0.5} \\[1em] \text I = 16\ \text A\end{aligned}$$

Since the same current flows through the whole line, the current in the wires is 16 A.

(ii) Loss of electric power in the wires :

Total resistance of the wires,

$$\begin{aligned}\text R = 0.5 \times 150 \\[1em] \text R = 75\ \Omega \\[1em] \text P = \text I^2\text R \\[1em] \text P = (16)^2 \times 75 \\[1em] \text P = 256 \times 75 \\[1em] \text P = 19200\ \text W \\[1em] \text P = 19.2\ \text{kW}\end{aligned}$$

Hence, the current flowing in the wires is 16 A and the loss of electric power in them is 19.2 kW.

Given,

• Rating of the heater, 500 W at 115 V
• Reduced line voltage, V' = 110 V

The resistance of the heating element is taken to be unchanged, so the power liberated is

$\text P = \frac{\text V^2}{\text R} \quad \Rightarrow \quad \text P \propto \text V^2$

Hence

$$\begin{aligned}\frac{\text P'}{\text P} = \left(\frac{\text V'}{\text V}\right)^2 \\[1em] \frac{\text P'}{\text P} = \left(\frac{110}{115}\right)^2 \\[1em] \frac{\text P'}{\text P} = (0.9565)^2 \\[1em] \frac{\text P'}{\text P} = 0.915 \\[1em] \text P' = 0.915 \times 500 \\[1em] \text P' = 457.5\ \text W\end{aligned}$$

Percentage loss of energy liberated :

$$\begin{aligned}\text{Percentage loss} = \frac{\text P - \text P'}{\text P} \times 100 \\[1em] = \frac{500 - 457.5}{500} \times 100 \\[1em] = \frac{42.5}{500} \times 100 \\[1em] \text{Percentage loss} = 8.5%\end{aligned}$$

Hence, the percentage loss of energy liberated is 8.5%.

Given,

• Rating of the heater, 1100 W-220 V
• Time, t = 4 hours

Electrical energy consumed :

$$\begin{aligned}\text W = \text{Pt} \\[1em] \text W = \left(\frac{1100}{1000}\ \text{kW}\right) \times 4\ \text h \\[1em] \text W = 1.1 \times 4 \\[1em] \text W = 4.4\ \text{kW-h}\end{aligned}$$

Resistance of the element :

$$\begin{aligned}\text R = \frac{\text V^2}{\text P} \\[1em] \text R = \frac{(220)^2}{1100} \\[1em] \text R = \frac{48400}{1100} \\[1em] \text R = 44\ \Omega\end{aligned}$$

Hence, the heater consumes 4.4 kW-h of electrical energy and the resistance of its element is 44 Ω.

Given,

• Rating of the bulb, 100 W–220 V
• Electronic charge, e = 1.6 × 10^-19 C

Current through the filament :

$$\begin{aligned}\text I = \frac{\text P}{\text V} \\[1em] \text I = \frac{100}{220} \\[1em] \text I = 0.4545\ \text A\end{aligned}$$

Number of electrons drifting per second : The current is the charge flowing per second, and if N electrons drift per second, then I = Ne. Hence

$$\begin{aligned}\text N = \frac{\text I}{\text e} \\[1em] \text N = \frac{0.4545}{1.6 \times 10^{-19}} \\[1em] \text N = 2.84 \times 10^{18}\end{aligned}$$

Hence, 2.84 × 10¹⁸ electrons drift per second through the filament.

Given,

• R₁ = 60 Ω, R₂ = 90 Ω, joined in parallel
• Power consumed by R₁, P₁ = 15 W

In a parallel combination the same potential difference V acts across both the resistors, and the power consumed is

$\text P = \frac{\text V^2}{\text R} \quad \Rightarrow \quad \text V^2 = \text{PR}$

For the resistor R₁,

$$\begin{aligned}\text V^2 = \text P_1\text R_1 \\[1em] \text V^2 = 15 \times 60 \\[1em] \text V^2 = 900\end{aligned}$$

For the resistor R₂, the same V acts, so

$$\begin{aligned}\text P_2 = \frac{\text V^2}{\text R_2} \\[1em] \text P_2 = \frac{900}{90} \\[1em] \text P_2 = 10\ \text W\end{aligned}$$

Hence, the power consumed by the resistor R₂ is 10 W.

Given,

• Power of the bulb = 60 W, power of the heater = 240 W
• Mains voltage, V = 120 V
• Resistance of the connecting leads, r = 6 Ω

Resistances of the bulb and the heater :

$$\begin{aligned}\text R_\text b = \frac{\text V^2}{\text P_\text b} \\[1em] \text R_\text b = \frac{(120)^2}{60} \\[1em] \text R_\text b = 240\ \Omega \\[1em] \text R_\text h = \frac{\text V^2}{\text P_\text h} \\[1em] \text R_\text h = \frac{(120)^2}{240} \\[1em] \text R_\text h = 60\ \Omega\end{aligned}$$

When only the bulb is switched on : The bulb is in series with the leads,

$$\begin{aligned}\text I = \frac{120}{240 + 6} \\[1em] \text I = \frac{120}{246}\ \text A\end{aligned}$$

Voltage at the bulb,

$$\begin{aligned}\text V_1 = \text{IR}_\text b \\[1em] \text V_1 = \frac{120}{246} \times 240 \\[1em] \text V_1 = 117.1\ \text V\end{aligned}$$

When the heater is also turned on : The bulb and the heater are in parallel,

$$\begin{aligned}\text R_\text p = \frac{240 \times 60}{240 + 60} \\[1em] \text R_\text p = \frac{14400}{300} \\[1em] \text R_\text p = 48\ \Omega \\[1em] \text I' = \frac{120}{48 + 6} \\[1em] \text I' = \frac{120}{54}\ \text A\end{aligned}$$

Voltage at the bulb,

$$\begin{aligned}\text V_2 = \text I'\text R_\text p \\[1em] = \frac{120}{54} \times 48 \\[1em] \text V_2 = 106.7\ \text V\end{aligned}$$

Change in the voltage at the bulb :

$$\begin{aligned}\text V_1 - \text V_2 = 117.1 - 106.7 \\[1em] \text V_1 - \text V_2 = 10.4\ \text V\end{aligned}$$

Hence, the voltage at the bulb will reduce by 10.4 V.

Given,

• Resistors of 2 Ω, 4 Ω and 5 Ω joined in parallel
• Emf of the battery, V = 20 V, internal resistance negligible

(a) Equivalent resistance :

$$\begin{aligned}\frac{1}{\text R} = \frac{1}{2} + \frac{1}{4} + \frac{1}{5} \\[1em] \frac{1}{\text R} = \frac{10 + 5 + 4}{20} \\[1em] \frac{1}{\text R} = \frac{19}{20} \\[1em] \text R = \frac{20}{19}\ \Omega\end{aligned}$$

(b) Current drawn from the battery :

$$\begin{aligned}\text I = \frac{\text V}{\text R} \\[1em] \text I = \frac{20}{\left(\frac{20}{19}\right)} \\[1em] \text I = 19\ \text A\end{aligned}$$

(c) Current through each resistor : In parallel the same potential difference of 20 V acts across each resistor,

$$\begin{aligned}\text I_1 = \frac{20}{2} \\[1em] \text I_1 = 10\ \text A \\[1em] \text I_2 = \frac{20}{4} \\[1em] \text I_2 = 5\ \text A \\[1em] \text I_3 = \frac{20}{5} \\[1em] \text I_3 = 4\ \text A\end{aligned}$$

Hence, the equivalent resistance is $\frac{20}{19}$ Ω, the current drawn from the battery is 19 A, and the currents through the 2 Ω, 4 Ω and 5 Ω resistors are 10 A, 5 A and 4 A respectively.

Given,

• Resistance of the heater, R_H = 150 Ω
• Maximum current it can take, I = 1.1 A
• Supply voltage, V = 220 V

The total resistance of the circuit needed to limit the current to 1.1 A is

$$\begin{aligned}\text R = \frac{\text V}{\text I} \\[1em] \text R = \frac{220}{1.1} \\[1em] \text R = 200\ \Omega\end{aligned}$$

If R' is the resistance joined in series with the heater, then

$$\begin{aligned}\text R_\text H + \text R' = \text R \\[1em] \text R' = 200 - 150 \\[1em] \text R' = 50\ \Omega\end{aligned}$$

Hence, a minimum resistance of 50 Ω should be joined in series with the heater.

Given,

• Resistance of the filament lamp, R_L = 200 Ω
• Rated supply, V = 150 V
• New supply, V' = 210 V

Current at which the lamp works properly :

$$\begin{aligned}\text I = \frac{\text V}{\text R_\text L} \\[1em] \text I = \frac{150}{200} \\[1em] \text I = 0.75\ \text A\end{aligned}$$

To use the lamp at 210 V the same current must flow through it, so the total resistance of the circuit must be

$$\begin{aligned}\text R = \frac{\text V'}{\text I} \\[1em] \text R = \frac{210}{0.75} \\[1em] \text R = 280\ \Omega\end{aligned}$$

If R' is the resistance joined in series, then

$$\begin{aligned}\text R' = \text R - \text R_\text L \\[1em] \text R' = 280 - 200 \\[1em] \text R' = 80\ \Omega\end{aligned}$$

Hence, a resistance of 80 Ω should be joined in series with the lamp.

Given,

• Each bulb takes a current of 2 A at 50 V
• The two bulbs are joined in series
• Supply voltage, V = 120 V, required current I = 2 A

Resistance of each bulb :

$$\begin{aligned}\text R_\text b = \frac{50}{2} \\[1em] \text R_\text b = 25\ \Omega\end{aligned}$$

Resistance of the two bulbs in series :

$$\begin{aligned}\text R_\text S = 25 + 25 \\[1em] \text R_\text S = 50\ \Omega\end{aligned}$$

Total resistance needed for a current of 2 A at 120 V :

$$\begin{aligned}\text R = \frac{\text V}{\text I} \\[1em] \text R = \frac{120}{2} \\[1em] \text R = 60\ \Omega\end{aligned}$$

Resistance to be added in series :

$$\begin{aligned}\text R' = \text R - \text R_\text S \\[1em] \text R' = 60 - 50 \\[1em] \text R' = 10\ \Omega\end{aligned}$$

Hence, a resistance of 10 Ω should be added in series with the lamps.

Carbon is a material whose resistivity decreases with a rise in temperature.

Carbon is a semiconductor, and in it the number of free electrons per unit volume n is very small at ordinary temperatures. On heating, a large number of additional free electrons are produced, so that n increases greatly. Since the resistivity is $\rho = \frac{m}{ne^2\tau}$, and the increase in n is much larger than the accompanying decrease in the relaxation time τ, the resistivity decreases.

By Ohm's law, $\text I = \frac{\text V}{\text R}$. If the potential difference V is kept constant, then

$\text I \propto \frac{1}{\text R}$

Hence the graph between the current and the resistance is a rectangular hyperbola.

The variation of resistance with temperature is given by

$\text R_t = \text R_0(1 + \alpha t)$

where R₀ and R_t are the resistances at 0°C and t°C.

Temperature coefficient of resistance : It is the change in resistance per unit original resistance per degree rise of temperature,

$\alpha = \frac{\text R_t - \text R_0}{\text R_0 t}$

Difference for carbon : For common conductors (metals) α is positive, so their resistance increases with a rise in temperature. Carbon is a semiconductor and its temperature coefficient of resistance is negative, so its resistance decreases with a rise in temperature.

Ohm's law is not valid for a conductor when its physical state, in particular its temperature, does not remain constant.

Explanation:
If the temperature of the conductor changes while the current flows through it, its resistance no longer remains constant and the ratio V/I ceases to be a constant.

Given,

• The two bulbs are marked 200 V-60 W and 200 V-100 W
• They are joined in series with a 200 V AC source

(i) Ratio of the resistances : From $\text R = \frac{\text V^2}{\text P}$ at the same rated voltage $\text R \propto \frac{1}{\text P}$

$$\begin{aligned}\frac{\text R_1}{\text R_2} = \frac{\text P_2}{\text P_1} \\[1em] \frac{\text R_1}{\text R_2} = \frac{100}{60} \\[1em] \frac{\text R_1}{\text R_2} = \frac{5}{3}\end{aligned}$$

(ii) Ratio of the powers consumed in series : In series the same current I flows through both, and P = I²R,

$$\begin{aligned}\frac{\text P_1}{\text P_2} = \frac{\text R_1}{\text R_2} \\[1em] \frac{\text P_1}{\text P_2} = \frac{5}{3}\end{aligned}$$

(iii) Ratio of the potential differences : Since V = I R,

$$\begin{aligned}\frac{\text V_1}{\text V_2} = \frac{\text R_1}{\text R_2} \\[1em] \frac{\text V_1}{\text V_2} = \frac{5}{3}\end{aligned}$$

Hence, all the three ratios are 5 : 3.

The drift velocity in terms of the potential difference applied is

$v_d = \frac{e\text V}{ml}\tau$

(i) When V is doubled : v_d is directly proportional to V, so the drift velocity is doubled.

(ii) When l is doubled : v_d is inversely proportional to l, so the drift velocity is halved.

(i) The terms in the equation I = n e A v_d are :

• n — the free-electron density, that is, the number of free electrons per unit volume of the conductor
• e — the magnitude of the charge on an electron
• A — the area of cross-section of the conductor
• v_d — the drift velocity of the free electrons

(ii) Expression for v_d : The electric field set up in the conductor of length l across which a potential difference V is maintained is

$\text E = \frac{\text V}{l}$

Each free electron is accelerated by this field, $a = \frac{e\text E}{m}$, and if τ is the relaxation time, the average velocity acquired is

$$\begin{aligned}v_d = a\tau \\[1em] v_d = \frac{e\text E}{m}\tau\end{aligned}$$

Substituting for E,

$v_d = \frac{e\text V}{ml}\tau$

Given,

• The wires A and B are of equal masses and of the same material
• $d_\text A = \frac{d_\text B}{2}$
• R_A = 24 Ω

Since the wires are of the same material and of equal masses, their volumes V are equal. For such a wire,

$$\begin{aligned}\text R = \rho\frac{l}{\text A} \\[1em] \text R = \frac{\rho\text V}{\text A^2} \\[1em] \text R \propto \frac{1}{\text A^2} \propto \frac{1}{d^4}\end{aligned}$$

Hence

$$\begin{aligned}\frac{\text R_\text A}{\text R_\text B} = \left(\frac{d_\text B}{d_\text A}\right)^4 \\[1em] \frac{\text R_\text A}{\text R_\text B} = (2)^4 \\[1em] \frac{\text R_\text A}{\text R_\text B} = 16 \\[1em] \text R_\text B = \frac{24}{16} \\[1em] \text R_\text B = 1.5\ \Omega\end{aligned}$$

Hence, the resistance of the wire B is 1.5 Ω.

Given,

• Emf of the battery, E = 3.5 V, internal resistance negligible
• Resistance of the wire, R = 2.0 Ω
• Area of cross-section, A = 0.70 × 10^-6 m²

Current in the wire :

$$\begin{aligned}\text I = \frac{\text E}{\text R} \\[1em] \text I = \frac{3.5}{2.0} \\[1em] \text I = 1.75\ \text A\end{aligned}$$

Current density :

$$\begin{aligned}j = \frac{\text I}{\text A} \\[1em] j = \frac{1.75}{0.70 \times 10^{-6}} \\[1em] j = 2.5 \times 10^{6}\ \text A/\text m^2\end{aligned}$$

Hence, the electric current density in the wire is 2.5 × 10⁶ A/m².

Silicon and germanium are semiconductors, for which the temperature coefficient of resistance is negative. With a rise in temperature the number of free electrons n increases greatly, so that the resistance $\text R = \frac{ml}{ne^2\tau\text A}$ decreases.

Hence the graph of resistance R against temperature T is as shown.

The new resistivity remains the same, that is, ρ.

Explanation:
The resistivity is a characteristic property of the material of the wire, $\rho = \frac{m}{ne^2\tau}$. On stretching, only the length and the area of cross-section of the wire change; the material remains the same and, as the temperature is unchanged, n and τ also remain unchanged. Hence the resistivity does not change, although the resistance becomes 16 times.

(i) (a) Drift velocity : When an electric field is applied across a conductor, its free electrons get drifted towards the positive end of the conductor. The average velocity with which they drift is called the drift velocity, $v_d = \frac{e\text E}{m}\tau$.

(b) Relaxation time : The average time interval that elapses between two successive collisions of a free electron with the metal ions is called the relaxation time. It equals the mean free path of the electron divided by its rms velocity, $\tau = \frac{\lambda}{v_\text{rms}}$, and its value is of the order of 10^-14 s.

(ii) The current I is the same at every cross-section of the plug.

The drift velocity is

$v_d = \frac{\text I}{ne\text A}$

so $v_d \propto \frac{1}{\text A}$. From the figure the area of cross-section goes on increasing from the end B to the end A. Hence the drift velocity of the free electrons decreases as they move from B to A.

From the figure the colour bands of the carbon resistor in order are Red, Yellow, Brown, and there is no fourth band.

Using the colour code, Red corresponds to the figure 2, Yellow to the figure 4 and Brown to the multiplier 10¹. As the fourth band is absent, the tolerance is 20%.

$$\begin{aligned}\text R = 24 \times 10^1\ \Omega \pm 20% \\[1em] \text R = 240\ \Omega \pm 20%\end{aligned}$$

Hence, the resistance of the carbon resistor is 240 Ω ± 20%.

The temperature coefficient of resistance is the change in resistance per unit original resistance per degree rise of temperature,

$\alpha = \frac{\text R_t - \text R_0}{\text R_0 t}$

Hence the statement means that when the temperature of the metal wire is raised by 1°C, the fractional change in its resistance is 4 × 10^-3, that is, its resistance increases by 0.4% of its original value for every 1°C rise of temperature.

Current density : Current density is the current flowing per unit area of cross-section held normal to the direction of flow of charge,

$\text J = \frac{\text I}{\text A}$

It is a vector quantity, its direction being the direction of flow of positive charge, and its SI unit is A m^-2.

Vector equation for an ohmic conductor :

$\vec{\text J} = \sigma\vec{\text E}$

where σ is the specific conductance of the material. This is Ohm's law in vector form.

metals

Explanation:
Ohm's law is not a universal law. It is true only for metallic conductors, and even for them only so long as the temperature and other physical states of the conductor do not change.

Given,

• Potential difference, V = 3 V
• Length of the wire, l = 60 cm = 0.6 m
• Current density, j = 1 × 10^-7 Am^-2

Electric field in the wire :

$$\begin{aligned}\text E = \frac{\text V}{l} \\[1em] \text E = \frac{3}{0.6} \\[1em] \text E = 5\ \text V\ \text m^{-1}\end{aligned}$$

Conductivity : By Ohm's law in vector form, j = σE, so

$$\begin{aligned}\sigma = \frac{j}{\text E} \\[1em] \sigma = \frac{1 \times 10^{-7}}{5} \\[1em] \sigma = 2 \times 10^{-8}\ \Omega^{-1}\ \text m^{-1}\end{aligned}$$

Hence, the conductivity of the material of the wire is 2 × 10^-8 Ω^-1 m^-1.

Given,

• Radius of the first wire = r, drift velocity = v_d
• Radius of the second wire = 2r, and the same current I flows through it

The current in terms of the drift velocity is

$$\begin{aligned}\text I = ne\text A v_d \\[1em] v_d = \frac{\text I}{ne\text A} \\[1em] v_d = \frac{\text I}{ne\pi r^2}\end{aligned}$$

For the same material and the same current, $v_d \propto \frac{1}{r^2}$. Hence for the wire of radius 2r,

$$\begin{aligned}v_d' = \frac{\text I}{ne\pi(2r)^2} \\[1em] v_d' = \frac{1}{4}\left(\frac{\text I}{ne\pi r^2}\right) \\[1em] v_d' = \frac{v_d}{4}\end{aligned}$$

Hence, the drift velocity through the thicker wire is $\frac{v_d}{4}$.

Temperature coefficient of resistance : The temperature coefficient of resistance of the material of a conductor is the change in resistance per unit original resistance per degree rise of temperature,

$\alpha = \frac{\text R_t - \text R_0}{\text R_0 t}$

where R₀ and R_t are the resistances of the conductor at 0°C and t°C respectively. Its unit is °C^-1.

Given,

• R = 470 Ω = 47 × 10¹ Ω
• Tolerance = 5%

The first significant figure is 4, for which the colour is Yellow.

The second significant figure is 7, for which the colour is Violet.

The multiplier is 10¹, for which the colour of the third band is Brown.

A tolerance of 5% corresponds to a Gold fourth band.

Hence, the colour code is Yellow, Violet, Brown, Gold.

140 Ω

Explanation:

Given,

• Arm ab = 10 Ω
• Arm bc = 20 Ω, where c is a free end
• Arm be = 50 Ω
• Arm de = 40 Ω, where d is a free end
• Arm ef = 80 Ω

The 20 Ω resistance ends at the free point c and the 40 Ω resistance starts from the free point d. Since no closed path can be completed through c or through d, no current can flow in these two resistances and they may be ignored.

Hence the current flows only along the path a → b → e → f, in which the 10 Ω, 50 Ω and 80 Ω resistances are in series,

$$\begin{aligned}\text R = 10 + 50 + 80 \\[1em] \text R = 140\ \Omega\end{aligned}$$

(i) For a metallic wire : The resistance of a metal increases with a rise in temperature, because the relaxation time τ of the free electrons decreases. The variation is nearly linear.

(ii) For a piece of carbon : Carbon is a semiconductor, and its temperature coefficient of resistance is negative. On heating, the number of free electrons n increases greatly, so its resistance decreases rapidly with a rise in temperature.

260 Ω ± 5%

Explanation:

From the figure the colour bands in order are Red, Blue, Brown and Gold.

Using the colour code, Red corresponds to the figure 2, Blue to the figure 6, Brown to the multiplier 10¹, and Gold to a tolerance of 5%. Hence

$$\begin{aligned}\text R = 26 \times 10^1\ \Omega \pm 5% \\[1em] \text R = 260\ \Omega \pm 5%\end{aligned}$$

Given,

• Resistance at 0°C, R₀ = 3.0 Ω
• Resistance at 150°C, R_t = 4.8 Ω

The variation of resistance with temperature is R_t = R₀(1 + αt). Substituting the values,

$$\begin{aligned}4.8 = 3.0(1 + \alpha \times 150) \\[1em] 1.6 = 1 + 150\alpha \\[1em] \alpha = \frac{0.6}{150} \\[1em] \alpha = 4 \times 10^{-3}\ ^{\circ}\text C^{-1}\end{aligned}$$

Hence, the temperature coefficient of resistance of the material is 4 × 10^-3 °C^-1.

$\frac{\text R}{n^2}$

Explanation:
When a wire of resistance R is cut into n equal parts, the length of each part becomes $\frac{l}{n}$ while the area of cross-section is unchanged. Hence the resistance of each part is

$\text R' = \frac{\text R}{n}$

These n equal resistances are then joined in parallel, so the equivalent resistance is

$$\begin{aligned}\text R_\text P = \frac{\text R'}{n} \\[1em] \text R_\text P = \frac{\text R}{n^2}\end{aligned}$$

Drift speed of free electrons : When an electric field is applied across a conductor, its free electrons get drifted towards the positive end of the conductor. The average speed of this drift is called the drift speed.

It is given by $v_d = \frac{e\text E}{m}\tau$. Its value is only of the order of 10^-5 m s^-1, far smaller than the thermal speeds of the electrons at room temperature, which are of the order of 10⁶ m s^-1.

Given,

• R₁ = 60 Ω and R₂ = 90 Ω joined in parallel
• Power consumed by R₁, P₁ = 15 W

In parallel the same potential difference V acts across both the resistors, and

$$\begin{aligned}\text P = \frac{\text V^2}{\text R} \\[1em] \text V^2 = \text{PR}\end{aligned}$$

For R₁,

$$\begin{aligned}\text V^2 = 15 \times 60 \\[1em] \text V^2 = 900\end{aligned}$$

For R₂, the same V acts, so

$$\begin{aligned}\text P_2 = \frac{\text V^2}{\text R_2} \\[1em] \text P_2 = \frac{900}{90} \\[1em] \text P_2 = 10\ \text W\end{aligned}$$

Hence, the power consumed by the resistor R₂ is 10 W.

24 × 10⁶ Ω ± 10%

Explanation:
Using the colour code, Red corresponds to the figure 2, Yellow to the figure 4, Blue to the multiplier 10⁶, and Silver to a tolerance of 10%. Hence

$\text R = 24 \times 10^6\ \Omega \pm 10%$

Current density : Current density is the current flowing per unit area of cross-section held normal to the direction of flow of charge,

$j = \frac{\text I}{\text A}$

Relation with drift speed : Since I = n e A v_d,

$$\begin{aligned}j = \frac{ne\text A v_d}{\text A} \\[1em] j = ne v_d\end{aligned}$$

where n is the free-electron density and e the electronic charge.

Given,

• Rating of the bulb, 200 V, 100 W
• Five such bulbs are joined in series to a 200 V supply

Resistance of the filament :

$$\begin{aligned}\text R = \frac{\text V^2}{\text P} \\[1em] \text R = \frac{(200)^2}{100} \\[1em] \text R = \frac{40000}{100} \\[1em] \text R = 400\ \Omega\end{aligned}$$

Current through the series combination : The equivalent resistance of five such bulbs in series is

$$\begin{aligned}\text R_\text S = 5 \times 400 \\[1em] \text R_\text S = 2000\ \Omega \\[1em] \text I = \frac{\text V}{\text R_\text S} \\[1em] \text I = \frac{200}{2000} \\[1em] \text I = 0.1\ \text A\end{aligned}$$

Hence, the resistance of the filament is 400 Ω and the current flowing through the bulbs is 0.1 A.

Given,

• Resistance of the wire, R = 20 Ω
• The wire is bent to form a complete circle

When the wire is bent into a circle, the two diametrically opposite points divide it into two equal halves. The resistance of each half is

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

These two halves are connected in parallel between the two diametrically opposite points,

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

Hence, the resistance between the two diametrically opposite points is 5 Ω.

Given,

• Area of cross-section, A = 1 mm² = 1 × 10^-6 m²
• Current, I = 5 A
• Free electron number density, n = 2 × 10²⁸ m^-3
• Electronic charge, e = 1.6 × 10^-19 C

From I = n e A v_d,

$v_d = \frac{\text I}{ne\text A}$

Substituting the values,

$$\begin{aligned}v_d = \frac{5}{(2 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (1 \times 10^{-6})} \\[1em] v_d = \frac{5}{3.2 \times 10^{3}} \\[1em] v_d = 1.56 \times 10^{-3}\ \text m/\text s\end{aligned}$$

Hence, the drift speed of the free electrons is 1.56 × 10^-3 m/s.

Given,

• Rating of the bulb, 250 V, 750 W
• Supply voltage, V = 250 V

(i) Electric current flowing through it :

$$\begin{aligned}\text I = \frac{\text P}{\text V} \\[1em] \text I = \frac{750}{250} \\[1em] \text I = 3\ \text A\end{aligned}$$

(ii) Resistance of its filament :

$$\begin{aligned}\text R = \frac{\text V}{\text I} \\[1em] \text R = \frac{250}{3} \\[1em] \text R = 83.3\ \Omega\end{aligned}$$

Hence, the current flowing is 3 A and the resistance of the filament is 83.3 Ω.

For a metallic conductor the current density is directly proportional to the electric field intensity established inside it. This is Ohm's law in vector form,

$\vec{\text J} = \sigma\vec{\text E}$

where σ is the conductivity of the material of the conductor. Since σ is a characteristic property of the material, the ratio J/E is constant and does not depend on the electric field applied.

R₂ = 2R₁

Explanation:
Both the bulbs are designed to operate on the same voltage V, and the resistance of a bulb from its rating is

$\text R = \frac{\text V^2}{\text P}$

so that $\text R \propto \frac{1}{\text P}$. Hence

$$\begin{aligned}\frac{\text R_2}{\text R_1} = \frac{\text P_1}{\text P_2} \\[1em] \frac{\text R_2}{\text R_1} = \frac{100}{50} \\[1em] \frac{\text R_2}{\text R_1} = 2 \\[1em] \text R_2 = 2\text R_1\end{aligned}$$

Ω^-1 m^-1

Explanation:
The electrical conductivity is the reciprocal of the resistivity, $\sigma = \frac{1}{\rho}$. Since the SI unit of resistivity is ohm-metre (Ω m), the SI unit of conductivity is (Ω m)^-1, that is, Ω^-1 m^-1.

16R

Explanation:
On stretching, the volume of the wire remains constant. When the radius becomes half, the area of cross-section becomes $\frac{\text A}{4}$, and hence the length becomes 4l. Therefore

$$\begin{aligned}\text R' = \rho\frac{4l}{\left(\frac{\text A}{4}\right)} \\[1em] \text R' = 16\left(\rho\frac{l}{\text A}\right) \\[1em] \text R' = 16\text R\end{aligned}$$

temperature of the wire

Explanation:
The specific resistance is a characteristic property of the material of the wire, $\rho = \frac{m}{ne^2\tau}$. It does not depend on the length, the area of cross-section or the volume of the wire. It depends only on the nature of the material and on its temperature, since a change of temperature changes the relaxation time τ and, in semiconductors, the free-electron density n as well.

2 : 1

Explanation:
In a parallel combination the same potential difference V acts across both the bulbs, and the power dissipated is

$\text P = \frac{\text V^2}{\text R} \quad \Rightarrow \quad \text P \propto \frac{1}{\text R}$

Since the resistances are in the ratio 1 : 2, the powers dissipated are in the inverse ratio, that is, 2 : 1.

5 × 10^-7 Ωm

Explanation:

Given, V = 2 V, l = 1 m and j = 4 × 10⁶ Am^-2.

The electric field in the wire is

$$\begin{aligned}\text E = \frac{\text V}{l} \\[1em] \text E = \frac{2}{1} \\[1em] \text E = 2\ \text V\ \text m^{-1}\end{aligned}$$

By Ohm's law in vector form, $j = \frac{\text E}{\rho}$, so

$$\begin{aligned}\rho = \frac{\text E}{j} \\[1em] \rho = \frac{2}{4 \times 10^{6}} \\[1em] \rho = 5 \times 10^{-7}\ \Omega\text m\end{aligned}$$

The heating power developed in a wire of resistance R carrying a current I is

$\text P = \text I^2\text R$

so that $\text P \propto \text I^2$. The graph of P against I is therefore a parabola passing through the origin, whose slope goes on increasing as I increases.

Hence, the graph shown in option (c) is the correct one.

In the relation I = n e A v_d, the symbol n stands for the number of free electrons per unit volume of the wire, that is, the free-electron density.

4 times

Explanation:
The heating power developed in a wire of resistance R carrying a current I is

$\text P = \text I^2\text R$

so that $\text P \propto \text I^2$ for a given wire. If the current is doubled, then

$$\begin{aligned}\frac{\text P'}{\text P} = \left(\frac{2\text I}{\text I}\right)^2 \\[1em] \frac{\text P'}{\text P} = 4\end{aligned}$$

Hence the heating power becomes 4 times.

increases

Explanation:
In a semiconductor the number of free electrons n per unit volume is very small at ordinary temperatures, but on heating a large number of additional free electrons are produced, so that n increases greatly. Although the relaxation time τ decreases, the increase in n is much larger. Hence the resistivity decreases, that is, the electrical conductivity of a semiconductor increases with a rise in temperature.

A semiconductor crystal has a negative temperature coefficient of resistance. With a rise in temperature the number of free electrons n increases greatly, so the resistance $\text R = \frac{ml}{ne^2\tau\text A}$ goes on decreasing, rapidly at first and then more slowly.

Hence, the graph shown in option (c) is the correct one.

becomes double

Explanation:
The drift speed of the free electrons in terms of the potential difference applied across the ends of the wire is

$v_d = \frac{e\text V}{ml}\tau$

Hence $v_d \propto \text V$ for a given wire, and on doubling the potential difference the drift speed becomes double.

length = x, area of cross-section = 2y

Explanation:
The resistance of a wire is

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

so R is minimum when the length is smallest and the area of cross-section is largest. Among the given wires, the smallest length is x and the largest area of cross-section is 2y.

Given,

• B₁ (230 V, 40 W), B₂ (230 V, 60 W) and B₃ (230 V, 100 W)
• The three bulbs are joined in series to a 230 V supply

The resistance of a bulb from its rating is $\text R = \frac{\text V^2}{\text P}$, so $\text R \propto \frac{1}{\text P}$. Hence the bulb of the lowest wattage has the highest resistance,

$\text R_1 \gt \text R_2 \gt \text R_3$

In series the same current I flows through all the three bulbs, and the power actually consumed by a bulb is

$\text P = \text I^2\text R$

so the bulb of the greatest resistance consumes the greatest power and glows the brightest.

Hence, the bulb B₁ glows the brightest.

Given,

• Area of cross-section, A = 6 × 10^-6 m²
• Current, I = 3 A
• Free electron number density, n = 5 × 10²⁸ m^-3
• Electronic charge, e = 1.6 × 10^-19 C

(i) Current density :

$$\begin{aligned}j = \frac{\text I}{\text A} \\[1em] j = \frac{3}{6 \times 10^{-6}} \\[1em] j = 5 \times 10^{5}\ \text{Am}^{-2}\end{aligned}$$

(ii) Drift speed : Since j = n e v_d,

$$\begin{aligned}v_d = \frac{j}{ne} \\[1em] v_d = \frac{5 \times 10^{5}}{(5 \times 10^{28}) \times (1.6 \times 10^{-19})} \\[1em] v_d = \frac{5 \times 10^{5}}{8 \times 10^{9}} \\[1em] v_d = 6.25 \times 10^{-5}\ \text{ms}^{-1}\end{aligned}$$

Hence, the current density is 5 × 10⁵ Am^-2 and the drift speed of the free electrons is 6.25 × 10^-5 ms^-1.

1.6 × 10^-5 Ω m

Reason —

Given, V = 5 V, l = 0.1 m, v_d = 2.5 × 10^-4 m s^-1 and n = 8 × 10²⁸ m^-3.

The electric field in the wire is

$$\begin{aligned}\text E = \frac{\text V}{l} \\[1em] \text E = \frac{5}{0.1} \\[1em] \text E = 50\ \text V\ \text m^{-1}\end{aligned}$$

The current density is

$$\begin{aligned}j = ne v_d \\[1em] j = (8 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (2.5 \times 10^{-4}) \\[1em] j = 3.2 \times 10^{6}\ \text A\ \text m^{-2}\end{aligned}$$

By Ohm's law in vector form, $j = \frac{\text E}{\rho}$, so

$$\begin{aligned}\rho = \frac{\text E}{j} \\[1em] \rho = \frac{50}{3.2 \times 10^{6}} \\[1em] \rho = 1.6 \times 10^{-5}\ \Omega\ \text m\end{aligned}$$

$\frac{1875}{64}\ \mu\Omega$

Reason —

Given, side of the Al bar = 7 mm, side of the Fe core = 2 mm, length l = 50 mm = 0.05 m, ρ_Al = 2.7 × 10^-8 Ω m and ρ_Fe = 1.0 × 10^-7 Ω m.

Areas of cross-section :

$$\begin{aligned}\text A_{\text{Fe}} = (2)^2 \\[1em] \text A_{\text{Fe}} = 4\ \text{mm}^2 \\[1em] \text A_{\text{Fe}} = 4 \times 10^{-6}\ \text m^2 \\[1em] \text A_{\text{Al}} = (7)^2 - (2)^2 \\[1em] \text A_{\text{Al}} = 45\ \text{mm}^2 \\[1em] \text A_{\text{Al}} = 45 \times 10^{-6}\ \text m^2\end{aligned}$$

Resistances of the two parts :

$$\begin{aligned}\text R_{\text{Al}} = \frac{(2.7 \times 10^{-8}) \times 0.05}{45 \times 10^{-6}} \\[1em] \text R_{\text{Al}} = 3 \times 10^{-5}\ \Omega \\[1em] \text R_{\text{Al}} = 30\ \mu\Omega \\[1em] \text R_{\text{Fe}} = \frac{(1.0 \times 10^{-7}) \times 0.05}{4 \times 10^{-6}} \\[1em] \text R_{\text{Fe}} = 1.25 \times 10^{-3}\ \Omega \\[1em] \text R_{\text{Fe}} = 1250\ \mu\Omega\end{aligned}$$

The two parts carry the current side by side between the faces P and Q, so they are in parallel,

$$\begin{aligned}\text R = \frac{\text R_{\text{Al}}\text R_{\text{Fe}}}{\text R_{\text{Al}} + \text R_{\text{Fe}}} \\[1em] \text R = \frac{30 \times 1250}{30 + 1250} \\[1em] \text R = \frac{37500}{1280} \\[1em] \text R = \frac{1875}{64}\ \mu\Omega\end{aligned}$$

current

Reason — For a steady current the same charge crosses every cross-section of the conductor in a given time, so the current I is the same at every section. The current density $\left(j = \frac{\text I}{\text A}\right)$, the drift velocity $\left(v_d = \frac{\text I}{ne\text A}\right)$ and the electric field $\left(\text E = \rho j\right)$ all depend on the area of cross-section, which is not uniform. Hence only the current remains constant.

26 Ω

Reason —

Given, the bulb is rated 500 W, 100 V and the main supply is 230 V.

For the bulb to work perfectly it must consume 500 W at 100 V, so the current through it is

$$\begin{aligned}\text I = \frac{\text P}{\text V} \\[1em] \text I = \frac{500}{100} \\[1em] \text I = 5\ \text A\end{aligned}$$

The remaining voltage is dropped across the series resistance R,

$$\begin{aligned}\text V_\text R = 230 - 100 \\[1em] \text V_\text R = 130\ \text V \\[1em] \text R = \frac{\text V_\text R}{\text I} \\[1em] \text R = \frac{130}{5} \\[1em] \text R = 26\ \Omega\end{aligned}$$

Linear increase for Cu, exponential decrease for Si.

Reason — Copper is a metal. On heating, the relaxation time τ of its free electrons decreases while the free-electron density n remains practically unchanged, so its resistance increases nearly linearly with temperature.

Undoped silicon is a semiconductor. On heating, the number of free electrons n increases greatly, and this increase far outweighs the decrease in τ. Hence the resistance of silicon decreases exponentially with a rise in temperature.

Reason — The line charge produces a radial electric field inside the conducting material, which drives a radial current. This current carries charge away from the axis, so the enclosed charge, and hence the field, goes on decreasing with time.

Let λ be the charge per unit length on the axis at any instant t, and λ₀ its value at t = 0. At a point distant r from the axis, the field in the material and, by Ohm's law, the current density are

$$\begin{aligned}\text E = \frac{\lambda}{2\pi\varepsilon r} \\[1em] j = \sigma\text E \\[1em] j = \frac{\sigma\lambda}{2\pi\varepsilon r}\end{aligned}$$

The current flowing out through a coaxial cylindrical surface of radius r and length l reduces the charge λl enclosed by it,

$$\begin{aligned}\frac{\text d(\lambda l)}{\text{dt}} = -j(2\pi r l) \\[1em] \frac{\text d\lambda}{\text{dt}} = -\frac{\sigma\lambda}{2\pi\varepsilon r}(2\pi r) \\[1em] \frac{\text d\lambda}{\text{dt}} = -\frac{\sigma}{\varepsilon}\lambda\end{aligned}$$

Separating the variables and integrating from t = 0 to t,

$$\begin{aligned}\int_{\lambda_0}^{\lambda}\frac{\text d\lambda}{\lambda} = -\frac{\sigma}{\varepsilon}\int_{0}^{t}\text{dt} \\[1em] \ln\frac{\lambda}{\lambda_0} = -\frac{\sigma}{\varepsilon}t \\[1em] \lambda = \lambda_0 e^{-\sigma t/\varepsilon}\end{aligned}$$

Since j ∝ λ, the magnitude of the current density falls off exponentially from an initial value and tends to zero,

$j(t) = j_0 e^{-\sigma t/\varepsilon}$

Hence, the graph shown in option (d) is the correct one.

$\frac{a^3 \text R}{6b}$

Reason —

Given, Q = at − bt².

The current is the rate of flow of charge,

$$\begin{aligned}i = \frac{\text{dQ}}{\text{dt}} \\[1em] i = \frac{\text d}{\text{dt}}(at - bt^2) \\[1em] i = a - 2bt\end{aligned}$$

The current becomes zero when

$$\begin{aligned}a - 2bt = 0 \\[1em] t = \frac{a}{2b}\end{aligned}$$

so the charge flows only up to this instant. The total heat produced is

$$\begin{aligned}\text H = \int_{0}^{a/2b} i^2\text R\thinspace\text{dt} \\[1em] \text H = \text R\int_{0}^{a/2b}(a - 2bt)^2\thinspace\text{dt}\end{aligned}$$

Expanding the square and integrating term by term,

$$\begin{aligned}\text H = \text R\int_{0}^{a/2b}\left(a^2 - 4abt + 4b^2t^2\right)\text{dt} \\[1em] \text H = \text R\left[a^2t - \frac{4abt^2}{2} + \frac{4b^2t^3}{3}\right]_{0}^{a/2b} \\[1em] \text H = \text R\left[a^2t - 2abt^2 + \frac{4b^2t^3}{3}\right]_{0}^{a/2b}\end{aligned}$$

Substituting the limits,

$$\begin{aligned}\text H = \text R\left[a^2\left(\frac{a}{2b}\right) - 2ab\left(\frac{a}{2b}\right)^2 + \frac{4b^2}{3}\left(\frac{a}{2b}\right)^3\right] \\[1em] \text H = \text R\left[\frac{a^3}{2b} - \frac{a^3}{2b} + \frac{a^3}{6b}\right] \\[1em] \text H = \frac{a^3\text R}{6b}\end{aligned}$$

n²R

Reason — When the wire is melted and stretched, its volume remains constant. If the length becomes n times, the area of cross-section becomes $\frac{1}{n}$ times. Hence

$$\begin{aligned}\text R' = \rho\frac{nl}{\left(\frac{\text A}{n}\right)} \\[1em] \text R' = n^2\left(\rho\frac{l}{\text A}\right) \\[1em] \text R' = n^2\text R\end{aligned}$$

1.0 m²/Vs

Reason —

Given, I = 5 A, ρ = 1.7 × 10^-8 Ωm, r = 5 mm = 5 × 10^-3 m and v_d = 1.1 × 10^-3 m/s.

Area of cross-section :

$$\begin{aligned}\text A = \pi r^2 \\[1em] \text A = 3.14 \times (5 \times 10^{-3})^2 \\[1em] \text A = 7.85 \times 10^{-5}\ \text m^2\end{aligned}$$

Current density :

$$\begin{aligned}j = \frac{\text I}{\text A} \\[1em] j = \frac{5}{7.85 \times 10^{-5}} \\[1em] j = 6.37 \times 10^{4}\ \text A\ \text m^{-2}\end{aligned}$$

Electric field :

$$\begin{aligned}\text E = \rho j \\[1em] \text E = (1.7 \times 10^{-8}) \times (6.37 \times 10^{4}) \\[1em] \text E = 1.08 \times 10^{-3}\ \text V\ \text m^{-1}\end{aligned}$$

Mobility :

$$\begin{aligned}\mu = \frac{v_d}{\text E} \\[1em] \mu = \frac{1.1 \times 10^{-3}}{1.08 \times 10^{-3}} \\[1em] \mu = 1.0\ \text m^2/\text{Vs}\end{aligned}$$

0.02

Reason —

Given, I = 1.5 A, A = 5 mm² = 5 × 10^-6 m², n = 9 × 10²⁸ m^-3 and e = 1.6 × 10^-19 C.

From I = n e A v_d,

$$\begin{aligned}v = \frac{\text I}{ne\text A} \\[1em] v = \frac{1.5}{(9 \times 10^{28}) \times (1.6 \times 10^{-19}) \times (5 \times 10^{-6})} \\[1em] v = \frac{1.5}{7.2 \times 10^{4}} \\[1em] v = 2.08 \times 10^{-5}\ \text m/\text s\end{aligned}$$

Converting to mm/s,

$$\begin{aligned}v = 2.08 \times 10^{-5} \times 10^{3} \\[1em] v = 0.0208 \approx 0.02\ \text{mm/s}\end{aligned}$$

$\frac{\rho}{4\pi}\left(\frac{1}{a} - \frac{1}{b}\right)$

Reason — Consider a thin spherical shell of the medium of radius x and thickness dx lying between the two spheres. The current flows radially, so for this shell the length is dx and the area of cross-section is 4πx². Its resistance is

$\text{dR} = \rho\frac{\text{dx}}{4\pi x^2}$

All such shells from x = a to x = b are in series, so the total resistance is

$$\begin{aligned}\text R = \int_{a}^{b}\frac{\rho\thinspace\text{dx}}{4\pi x^2} \\[1em] \text R = \frac{\rho}{4\pi}\left[-\frac{1}{x}\right]_{a}^{b} \\[1em] \text R = \frac{\rho}{4\pi}\left(\frac{1}{a} - \frac{1}{b}\right)\end{aligned}$$

$\frac{5}{3}\ \Omega$

Reason —

Given, the original resistance is 3 Ω and the wire is elongated to double its length.

On elongating, the volume remains constant, so $\text R \propto l^2$ and

$$\begin{aligned}\text R' = (2)^2 \times 3 \\[1em] \text R' = 12\ \Omega\end{aligned}$$

This wire is bent into a circle. Two points making an angle 60° at the centre divide the circle into an arc of 60° and an arc of 300°. The resistance of an arc is proportional to the angle it subtends, so

$$\begin{aligned}\text R_1 = \frac{60}{360} \times 12 \\[1em] \text R_1 = 2\ \Omega \\[1em] \text R_2 = \frac{300}{360} \times 12 \\[1em] \text R_2 = 10\ \Omega\end{aligned}$$

The two arcs are in parallel between the two points,

$$\begin{aligned}\text R_{\text{eq}} = \frac{2 \times 10}{2 + 10} \\[1em] \text R_{\text{eq}} = \frac{20}{12} \\[1em] \text R_{\text{eq}} = \frac{5}{3}\ \Omega\end{aligned}$$

$\frac{7}{64}\text R$

Reason —

Given, a wire of total resistance R is bent to form the square ABCD, and E is the mid-point of the arm CD.

The resistance of each side of the square is $\frac{\text R}{4}$, and since E is the mid-point of CD,

$\text R_{\text{EC}} = \text R_{\text{ED}} = \frac{\text R}{8}$

Between the points E and C there are two paths in parallel :

(i) the direct path E → C, of resistance $\frac{\text R}{8}$

(ii) the path E → D → A → B → C, of resistance

$\frac{\text R}{8} + \frac{\text R}{4} + \frac{\text R}{4} + \frac{\text R}{4} = \frac{7\text R}{8}$

Hence

$$\begin{aligned}\text R_{\text{eq}} = \frac{\left(\frac{\text R}{8}\right)\left(\frac{7\text R}{8}\right)}{\frac{\text R}{8} + \frac{7\text R}{8}} \\[1em] \text R_{\text{eq}} = \frac{\frac{7\text R^2}{64}}{\text R} \\[1em] \text R_{\text{eq}} = \frac{7}{64}\text R\end{aligned}$$

570 Ω

Reason —

Given, the actual value of the resistance is R = 30 Ω and the measured value is 5% less.

$$\begin{aligned}\text R_{\text{measured}} = 30 - (5% \text{ of } 30) \\[1em] \text R_{\text{measured}} = 30 - 1.5 \\[1em] \text R_{\text{measured}} = 28.5\ \Omega\end{aligned}$$

In the circuit shown the voltmeter is connected across R, so the ammeter measures the current through the parallel combination of R and the voltmeter resistance R_V. Hence the measured value is this parallel combination,

$$\begin{aligned}\frac{\text R\thinspace\text R_\text V}{\text R + \text R_\text V} = 28.5 \\[1em] \frac{30\thinspace\text R_\text V}{30 + \text R_\text V} = 28.5 \\[1em] 30\text R_\text V = 855 + 28.5\text R_\text V \\[1em] 1.5\text R_\text V = 855 \\[1em] \text R_\text V = 570\ \Omega\end{aligned}$$

$\text R(\text T) = \text R_0 e^{-\text T_0^2/\text T^2}$

Reason —

From the figure, the graph of ln R(T) against $\frac{1}{\text T^2}$ is a straight line with a negative slope, so

$\ln \text R(\text T) = \ln \text R_0 - \text T_0^2\left(\frac{1}{\text T^2}\right)$

where ln R₀ is the intercept and $\text T_0^2$ the magnitude of the slope. Taking antilogarithms,

$\text R(\text T) = \text R_0 e^{-\text T_0^2/\text T^2}$

0.5 Ω

Reason —

Given, two cells each of emf 1.5 V and internal resistance r are joined in series, the 15 Ω and 10 Ω resistances are in parallel, and the 2 Ω is in series with them. The voltmeter reads 2 V across the 10 Ω.

Equivalent resistance of the parallel part :

$$\begin{aligned}\text R_\text P = \frac{15 \times 10}{15 + 10} \\[1em] \text R_\text P = \frac{150}{25} \\[1em] \text R_\text P = 6\ \Omega\end{aligned}$$

Current in the circuit : The voltmeter is ideal, so the potential difference across the parallel combination is 2 V. Hence

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

Applying the circuit equation : The total emf is 1.5 + 1.5 = 3 V and the total resistance is (6 + 2 + 2r) Ω,

$$\begin{aligned}3 = \frac{1}{3}(6 + 2 + 2r) \\[1em] 9 = 8 + 2r \\[1em] r = 0.5\ \Omega\end{aligned}$$

500 Ω

Reason — A resistance of 200 Ω is written as 20 × 10¹ Ω, so its first three colour bands are Red (2), Black (0) and Brown (10¹).

If the red colour of the first band is replaced by green, the first significant figure becomes 5, and the code now reads Green (5), Black (0) and Brown (10¹). Hence the new resistance is

$$\begin{aligned}\text R = 50 \times 10^1 \\[1em] \text R = 500\ \Omega\end{aligned}$$

20 mA

Reason —

Given, the colour bands are green, black, red and brown, and the power rating of the resistor is 2 W.

Using the colour code, Green corresponds to the figure 5, Black to the figure 0 and Red to the multiplier 10²,

$$\begin{aligned}\text R = 50 \times 10^2 \\[1em] \text R = 5000\ \Omega\end{aligned}$$

The maximum current which can be passed is obtained from P = I²R,

$$\begin{aligned}\text I = \sqrt{\frac{\text P}{\text R}} \\[1em] \text I = \sqrt{\frac{2}{5000}} \\[1em] \text I = \sqrt{4 \times 10^{-4}} \\[1em] \text I = 2 \times 10^{-2}\ \text A \\[1em] \text I = 20\ \text{mA}\end{aligned}$$

V₁ = V₂ and I₁ = I₂

Reason —

In Circuit-1 the voltmeter V₁ is connected across the 10 Ω resistance, and in Circuit-2 the voltmeter V₂ is connected in series with an additional 10 Ω and this arrangement is joined across the first 10 Ω.

An ideal voltmeter has infinite resistance, so it draws no current. Hence in Circuit-2 no current flows through the additional 10 Ω and there is no potential drop across it. Both the voltmeters therefore read the potential difference across the 10 Ω resistance of the main circuit, that is,

$\text V_1 = \text V_2$

Since the voltmeters draw no current, the main circuit is the same in both cases and the ammeters also read the same current,

$\text I_1 = \text I_2$

11 × 10^-5 W

Reason —

Given, I = 2 mA = 2 × 10^-3 A and the power dissipated P = 4.4 W.

Resistance of the unknown resistor :

$$\begin{aligned}\text R = \frac{\text P}{\text I^2} \\[1em] \text R = \frac{4.4}{(2 \times 10^{-3})^2} \\[1em] \text R = \frac{4.4}{4 \times 10^{-6}} \\[1em] \text R = 1.1 \times 10^{6}\ \Omega\end{aligned}$$

Power dissipated at 11 V :

$$\begin{aligned}\text P' = \frac{\text V^2}{\text R} \\[1em] \text P' = \frac{(11)^2}{1.1 \times 10^{6}} \\[1em] \text P' = \frac{121}{1.1 \times 10^{6}} \\[1em] \text P' = 1.1 \times 10^{-4} \\[1em] \text P' = 11 \times 10^{-5}\ \text W\end{aligned}$$

P₁ = 16 W, P₂ = 4 W

Reason —

Given, the bulbs are rated 25 W, 220 V and 100 W, 220 V, joined in series across 220 V.

Resistances of the bulbs :

$$\begin{aligned}\text R_1 = \frac{(220)^2}{25} \\[1em] \text R_1 = 1936\ \Omega \\[1em] \text R_2 = \frac{(220)^2}{100} \\[1em] \text R_2 = 484\ \Omega\end{aligned}$$

Current in the series circuit :

$$\begin{aligned}\text I = \frac{220}{1936 + 484} \\[1em] \text I = \frac{220}{2420} \\[1em] \text I = \frac{1}{11}\ \text A\end{aligned}$$

Powers drawn :

$$\begin{aligned}\text P_1 = \text I^2\text R_1 \\[1em] \text P_1 = \frac{1}{121} \times 1936 \\[1em] \text P_1 = 16\ \text W \\[1em] \text P_2 = \text I^2\text R_2 \\[1em] \text P_2 = \frac{1}{121} \times 484 \\[1em] \text P_2 = 4\ \text W\end{aligned}$$

9 : 4

Reason —

Let the resistance of each bulb be R. Section A has three bulbs in parallel and section B has three bulbs in parallel, and the two sections are joined in series across the source of emf E.

(i) When all the six are glowing :

$$\begin{aligned}\text R_\text A = \frac{\text R}{3}, \qquad \text R_\text B = \frac{\text R}{3} \\[1em] \text R_1 = \frac{\text R}{3} + \frac{\text R}{3} \\[1em] \text R_1 = \frac{2\text R}{3} \\[1em] \text P_1 = \frac{\text E^2}{\text R_1} \\[1em] \text P_1 = \frac{3\text E^2}{2\text R}\end{aligned}$$

(ii) When two from A and one from B are glowing :

$$\begin{aligned}\text R_\text A' = \frac{\text R}{2}, \qquad \text R_\text B' = \text R \\[1em] \text R_2 = \frac{\text R}{2} + \text R \\[1em] \text R_2 = \frac{3\text R}{2} \\[1em] \text P_2 = \frac{\text E^2}{\text R_2} \\[1em] \text P_2 = \frac{2\text E^2}{3\text R}\end{aligned}$$

Ratio :

$$\begin{aligned}\frac{\text P_1}{\text P_2} = \frac{\frac{3\text E^2}{2\text R}}{\frac{2\text E^2}{3\text R}} \\[1em] \frac{\text P_1}{\text P_2} = \frac{3}{2} \times \frac{3}{2} \\[1em] \frac{\text P_1}{\text P_2} = \frac{9}{4}\end{aligned}$$

1

Reason — The two conductors are of the same material, so they have the same resistivity ρ, and they have the same length l and the same resistance R. From

$$\begin{aligned}\text R = \rho\frac{l}{\text A} \\[1em] \text A = \frac{\rho l}{\text R}\end{aligned}$$

Since ρ, l and R are the same for both, their areas of cross-section must also be equal, whatever be the shape of the cross-section. Hence

$\frac{\text A_1}{\text A_2} = 1$

ρ_M > ρ_A > ρ_C

Reason — Of the four materials, mercury has by far the largest resistivity (about 98 × 10^-8 Ω m), followed by tungsten, aluminium and copper. Among aluminium and copper, copper has the lower resistivity, which is why copper wires are used as connecting wires. Hence the correct order among these three materials is

$\rho_\text M \gt \rho_\text A \gt \rho_\text C$

96%

Reason —

Given, the power delivered P = 1 kW = 1000 W at V = 220 V, and the total resistance of the line R = 2 Ω.

Current in the line :

$$\begin{aligned}\text I = \frac{\text P}{\text V} \\[1em] \text I = \frac{1000}{220} \\[1em] \text I = 4.545\ \text A\end{aligned}$$

Power lost in the line :

$$\begin{aligned}\text P_{\text{loss}} = \text I^2\text R \\[1em] \text P_{\text{loss}} = (4.545)^2 \times 2 \\[1em] \text P_{\text{loss}} = 41.3\ \text W\end{aligned}$$

Efficiency of transmission :

$$\begin{aligned}\eta = \frac{\text{power delivered}}{\text{power delivered} + \text{power lost}} \times 100 \\[1em] \eta = \frac{1000}{1041.3} \times 100 \\[1em] \eta = 96%\end{aligned}$$

1.95 V

Reason —

Given, R₁ = 400 Ω and R₂ = 800 Ω in series across a 6 V battery, and a voltmeter of resistance R_V = 10 kΩ = 10000 Ω is connected across the 400 Ω.

Resistance of the 400 Ω with the voltmeter across it :

$$\begin{aligned}\text R' = \frac{400 \times 10000}{400 + 10000} \\[1em] \text R' = \frac{4 \times 10^{6}}{10400} \\[1em] \text R' = 384.6\ \Omega\end{aligned}$$

Current drawn from the battery :

$$\begin{aligned}\text I = \frac{6}{384.6 + 800} \\[1em] \text I = \frac{6}{1184.6} \\[1em] \text I = 5.06 \times 10^{-3}\ \text A\end{aligned}$$

Potential difference measured :

$$\begin{aligned}\text V = \text I\text R' \\[1em] \text V = (5.06 \times 10^{-3}) \times 384.6 \\[1em] \text V = 1.95\ \text V\end{aligned}$$

Ammeter is always connected in series and voltmeter in parallel

Reason — An ammeter measures the current through the resistor, so the whole current must pass through it and it has to be joined in series. It is given a very low resistance so that it does not appreciably alter the current in the circuit.

A voltmeter measures the potential difference across the resistor, so it has to be joined in parallel with it. It is given a very high resistance so that it draws practically no current from the circuit.

0.10 W

Reason —

Given, emf of the battery E = 3.0 V, power dissipated in the external resistor P = 0.5 W and terminal voltage V = 2.5 V.

Current in the circuit : The power dissipated in the resistor is P = VI,

$$\begin{aligned}\text I = \frac{\text P}{\text V} \\[1em] \text I = \frac{0.5}{2.5} \\[1em] \text I = 0.2\ \text A\end{aligned}$$

Potential drop across the internal resistance :

$$\begin{aligned}\text V_r = \text E - \text V \\[1em] \text V_r = 3.0 - 2.5 \\[1em] \text V_r = 0.5\ \text V\end{aligned}$$

Power dissipated within the internal resistance :

$$\begin{aligned}\text P_r = \text V_r\text I \\[1em] \text P_r = 0.5 \times 0.2 \\[1em] \text P_r = 0.10\ \text W\end{aligned}$$

(A) - (R), (B) - (S), (C) - (P), (D) - (Q)

Reason — Taking the terms of Column-I one by one :

(A) Drift Velocity : The drift velocity acquired by a free electron in a field E is $v_d = \frac{e\text E}{m}\tau$, which is (R).

(B) Electrical Resistivity : From Ohm's law in vector form, $\text J = \sigma\text E = \frac{\text E}{\rho}$, so $\rho = \frac{\text E}{\text J}$, which is (S).

(C) Relaxation Period : Since $\rho = \frac{m}{ne^2\tau}$, the relaxation time is $\tau = \frac{m}{ne^2\rho}$, which is (P).

(D) Current Density : The current density is $\text J = ne v_d$, which is (Q).

4 Ω

Reason — Let the resistance of each wire be R'. The four wires are of equal length, equal area of cross-section and the same material, so they are of equal resistance.

In parallel :

$$\begin{aligned}\text R_\text P = \frac{\text R'}{4} \\[1em] \text R_\text P = 0.25 \\[1em] \text R' = 1\ \Omega\end{aligned}$$

In series :

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

$\frac{r_2}{r_2 + r_3}$

Reason —

From the figure, the current I₁ enters at A through r₁ and then divides into I₂ through r₂ and I₃ through r₃, since r₂ and r₃ are in parallel. Hence

$\text I_1 = \text I_2 + \text I_3$

The potential difference across the parallel combination is the same for both branches,

$$\begin{aligned}\text I_2 r_2 = \text I_3 r_3 \\[1em] \text I_2 = \frac{\text I_3 r_3}{r_2}\end{aligned}$$

Substituting in the first equation,

$$\begin{aligned}\text I_1 = \frac{\text I_3 r_3}{r_2} + \text I_3 \\[1em] \text I_1 = \text I_3\left(\frac{r_3 + r_2}{r_2}\right) \\[1em] \frac{\text I_3}{\text I_1} = \frac{r_2}{r_2 + r_3}\end{aligned}$$

97 m

Reason —

Given, the two wires are of equal length l and each of diameter 2 mm, so the radius is 1 mm = 10^-3 m. Their resistivities are

$$\begin{aligned}\rho_1 = 12\ \mu\Omega\ \text{cm} \\[1em] \rho_1 = 12 \times 10^{-8}\ \Omega\ \text m \\[1em] \rho_2 = 51\ \mu\Omega\ \text{cm} \\[1em] \rho_2 = 51 \times 10^{-8}\ \Omega\ \text m\end{aligned}$$

Area of cross-section :

$$\begin{aligned}\text A = \pi r^2 \\[1em] \text A = 3.14 \times (10^{-3})^2 \\[1em] \text A = 3.14 \times 10^{-6}\ \text m^2\end{aligned}$$

Resistances of the two wires :

$$\begin{aligned}\text R_1 = \frac{(12 \times 10^{-8})l}{3.14 \times 10^{-6}} \\[1em] \text R_1 = 3.82 \times 10^{-2}\thinspace l \\[1em] \text R_2 = \frac{(51 \times 10^{-8})l}{3.14 \times 10^{-6}} \\[1em] \text R_2 = 16.24 \times 10^{-2}\thinspace l\end{aligned}$$

In parallel, the equivalent resistance is 3 Ω :

$$\begin{aligned}\frac{\text R_1\text R_2}{\text R_1 + \text R_2} = 3 \\[1em] \frac{(3.82 \times 10^{-2})(16.24 \times 10^{-2})l^2}{(3.82 + 16.24) \times 10^{-2}\thinspace l} = 3 \\[1em] \frac{(62.0 \times 10^{-4})l}{20.06 \times 10^{-2}} = 3 \\[1em] (3.09 \times 10^{-2})l = 3 \\[1em] l = 97\ \text m\end{aligned}$$

245 Ω

Reason —

Given, the shunt resistance S = 5 Ω and 2% of the total current passes through the galvanometer.

Let the total current be I. Then the current through the galvanometer is 0.02 I and that through the shunt is 0.98 I.

The galvanometer and the shunt are in parallel, so the potential difference across them is the same,

$$\begin{aligned}\text I_\text g\text R_\text g = \text I_\text s\text S \\[1em] (0.02\thinspace\text I)\text R_\text g = (0.98\thinspace\text I) \times 5 \\[1em] \text R_\text g = \frac{0.98 \times 5}{0.02} \\[1em] \text R_\text g = 245\ \Omega\end{aligned}$$

(7500 ± 375) Ω

Reason —

From the figure the colour bands in order are Violet, Green, Red and Gold.

Using the colour code, Violet corresponds to the figure 7, Green to the figure 5, Red to the multiplier 10², and Gold to a tolerance of 5%. Hence

$$\begin{aligned}\text R = 75 \times 10^2 \\[1em] \text R = 7500\ \Omega \\[1em] \text{Tolerance} = 5% \text{ of } 7500 \\[1em] \text{Tolerance} = 375\ \Omega\end{aligned}$$

Hence the resistance is (7500 ± 375) Ω.

2Ω are 4Ω are in parallel with 6Ω and 8Ω in series.

Reason — Taking the 2 Ω and the 4 Ω in parallel,

$$\begin{aligned}\text R_\text P = \frac{2 \times 4}{2 + 4} \\[1em] \text R_\text P = \frac{8}{6} \\[1em] \text R_\text P = \frac{4}{3}\ \Omega\end{aligned}$$

Joining the 6 Ω and the 8 Ω in series with this combination,

$$\begin{aligned}\text R = \frac{4}{3} + 6 + 8 \\[1em] \text R = \frac{4 + 18 + 24}{3} \\[1em] \text R = \frac{46}{3}\ \Omega\end{aligned}$$

which is the required equivalent resistance.

1Ω

Reason —

Given,

• Upper wire, from the left-hand corner towards A : 2 Ω up to the middle junction M, then 2 Ω from M
• Vertical arms between the upper and the lower wires, from left to right : 5 Ω, 2 Ω (at M), 3 Ω and 3 Ω
• A plain connecting wire joins the upper and the lower wires at the left-hand end, across the 5 Ω
• A is at the right-hand end of the upper wire and B at the right-hand end of the lower wire

The left-hand vertical connecting wire joins the two ends of the 5 Ω resistance, so the 5 Ω is short-circuited and carries no current. Hence the left-hand upper corner is at the same potential as B.

The two 3 Ω arms are joined directly between A and B, so

$$\begin{aligned}\text R_1 = \frac{3 \times 3}{3 + 3} \\[1em] \text R_1 = 1.5\ \Omega\end{aligned}$$

The 2 Ω vertical arm at the middle junction M and the 2 Ω which reaches B through the left-hand corner are both connected between M and B, that is, they are in parallel,

$$\begin{aligned}\text R_2 = \frac{2 \times 2}{2 + 2} \\[1em] \text R_2 = 1\ \Omega\end{aligned}$$

Adding the 2 Ω between A and M in series,

$$\begin{aligned}\text R_3 = 2 + 1 \\[1em] \text R_3 = 3\ \Omega\end{aligned}$$

Finally R₃ and R₁ are in parallel between A and B,

$$\begin{aligned}\text R = \frac{3 \times 1.5}{3 + 1.5} \\[1em] \text R = \frac{4.5}{4.5} \\[1em] \text R = 1\ \Omega\end{aligned}$$

11 × 10^-5 V m^-1

Reason —

Given, I = 5 A, A = 0.04 m², θ = 60° and ρ = 44 × 10^-8 Ωm.

The current is the flux of the current density through the area of cross-section,

$$\begin{aligned}\text I = \vec{j}\cdot\vec{\text A} \\[1em] \text I = j\text A\cos\theta \\[1em] j = \frac{\text I}{\text A\cos\theta} \\[1em] j = \frac{5}{0.04 \times \cos 60^{\circ}} \\[1em] j = \frac{5}{0.04 \times 0.5} \\[1em] j = \frac{5}{0.02} \\[1em] j = 250\ \text A\ \text m^{-2}\end{aligned}$$

By Ohm's law in vector form, E = ρ j,

$$\begin{aligned}\text E = (44 \times 10^{-8}) \times 250 \\[1em] \text E = 11000 \times 10^{-8} \\[1em] \text E = 1.1 \times 10^{-4} \\[1em] \text E = 11 \times 10^{-5}\ \text V\ \text m^{-1}\end{aligned}$$

0.858 mΩ

Reason —

Given, l = 25 cm = 0.25 m, A = 3 mm² = 3 × 10^-6 m², ρ_Cu = 1.7 × 10^-8 Ωm and ρ_Al = 2.6 × 10^-8 Ω m.

Resistance of the copper rod :

$$\begin{aligned}\text R_{\text{Cu}} = \frac{(1.7 \times 10^{-8}) \times 0.25}{3 \times 10^{-6}} \\[1em] \text R_{\text{Cu}} = 1.417 \times 10^{-3}\ \Omega\end{aligned}$$

Resistance of the aluminium rod :

$$\begin{aligned}\text R_{\text{Al}} = \frac{(2.6 \times 10^{-8}) \times 0.25}{3 \times 10^{-6}} \\[1em] \text R_{\text{Al}} = 2.167 \times 10^{-3}\ \Omega\end{aligned}$$

From the figure the two rods are joined in parallel between A and B,

$$\begin{aligned}\text R = \frac{\text R_{\text{Cu}}\text R_{\text{Al}}}{\text R_{\text{Cu}} + \text R_{\text{Al}}} \\[1em] \text R = \frac{(1.417 \times 2.167) \times 10^{-6}}{(1.417 + 2.167) \times 10^{-3}} \\[1em] \text R = \frac{3.070 \times 10^{-6}}{3.584 \times 10^{-3}} \\[1em] \text R = 0.857 \times 10^{-3}\ \Omega \approx 0.858\ \text m\Omega\end{aligned}$$

Drift velocity of electron increases.

Reason —

For a steady current the same current I flows through every cross-section of the conductor PQ, so the current does not decrease. Since r₂ < r₁, the area of cross-section goes on decreasing as one moves from P to Q.

The drift velocity is

$v_d = \frac{\text I}{ne\text A}$

so $v_d \propto \frac{1}{\text A}$, and the drift velocity increases from P to Q.

The electric field is E = ρ j, and the current density $j = \frac{\text I}{\text A}$ also increases, so the electric field increases as well. Hence only the second statement is correct.

0.003°C^-1

Reason —

Given, R₁ = 16 Ω at t₁ = 15°C and R₂ = 20 Ω at t₂ = 100°C.

Using R_t = R₀(1 + αt) for the two temperatures,

$\text R_1 = \text R_0(1 + \alpha t_1), \qquad \text R_2 = \text R_0(1 + \alpha t_2)$

Eliminating R₀,

$\alpha = \frac{\text R_2 - \text R_1}{\text R_1 t_2 - \text R_2 t_1}$

Substituting the values,

$$\begin{aligned}\alpha = \frac{20 - 16}{(16 \times 100) - (20 \times 15)} \\[1em] \alpha = \frac{4}{1600 - 300} \\[1em] \alpha = \frac{4}{1300} \\[1em] \alpha = 3.08 \times 10^{-3} \approx 0.003\ ^{\circ}\text C^{-1}\end{aligned}$$

56%

Reason — On stretching, the volume of the wire remains constant, so

$\text R \propto l^2$

The length increases by 25%, so l' = 1.25 l. Hence

$$\begin{aligned}\frac{\text R'}{\text R} = (1.25)^2 \\[1em] \frac{\text R'}{\text R} = 1.5625 \\[1em] \frac{\text R' - \text R}{\text R} \times 100 = (1.5625 - 1) \times 100 \\[1em] \frac{\text R' - \text R}{\text R} \times 100 = 56.25%\end{aligned}$$

To the nearest integer, the percentage change in resistance is 56%.

$\frac{1}{4}\frac{\text{VA}}{\rho l}$

Reason — The resistance of the wire is $\text R = \frac{\rho l}{\text A}$, so the current is

$$\begin{aligned}\text I = \frac{\text V}{\text R} \\[1em] \text I = \frac{\text{VA}}{\rho l}\end{aligned}$$

When the length is doubled and the area of cross-section is halved,

$$\begin{aligned}\text R' = \frac{\rho(2l)}{\left(\frac{\text A}{2}\right)} \\[1em] \text R' = \frac{4\rho l}{\text A} \\[1em] \text R' = 4\text R\end{aligned}$$

Hence the resultant current is

$$\begin{aligned}\text I' = \frac{\text V}{4\text R} \\[1em] \text I' = \frac{1}{4}\frac{\text{VA}}{\rho l}\end{aligned}$$

11250 C

Reason —

Given, i = α₀t + βt² with α₀ = 20 As^-1 and β = 8 As^-2, and t = 15 s.

The current is the rate of flow of charge, so

$$\begin{aligned}\text q = \int_{0}^{15} i\thinspace\text{dt} \\[1em] \text q = \int_{0}^{15}(20t + 8t^2)\thinspace\text{dt} \\[1em] \text q = \left[\frac{20t^2}{2} + \frac{8t^3}{3}\right]_{0}^{15} \\[1em] \text q = \left[10t^2 + \frac{8t^3}{3}\right]_{0}^{15} \\[1em] \text q = \left[10 \times (15)^2\right] + \left[\frac{8 \times (15)^3}{3}\right] \\[1em] \text q = 2250 + \frac{8 \times 3375}{3} \\[1em] \text q = 2250 + 9000 \\[1em] \text q = 11250\ \text C\end{aligned}$$

Note: The current in the question is printed as i = α₀t + βt^2t. The exponent 2t is a misprint; the units given for β (As^-2) show that the term must be βt², and only then does the integration give the answer in the options.

625 × 10²⁵

Reason —

Given, I = 10 A, A = 5 mm² = 5 × 10^-6 m², v_d = 2 × 10^-3 ms^-1 and e = 1.6 × 10^-19 C.

From I = n e A v_d,

$$\begin{aligned}n = \frac{\text I}{e\text A v_d} \\[1em] n = \frac{10}{(1.6 \times 10^{-19}) \times (5 \times 10^{-6}) \times (2 \times 10^{-3})} \\[1em] n = \frac{10}{1.6 \times 10^{-27}} \\[1em] n = 6.25 \times 10^{27} \\[1em] n = 625 \times 10^{25}\ \text m^{-3}\end{aligned}$$

(A) and (D)

Reason — The drift velocity in terms of the potential difference applied across a wire of length l is

$v_d = \frac{e\text V}{ml}\tau$

(A) is correct : With a rise in temperature the metal ions vibrate more vigorously, so the relaxation time τ decreases and hence the drift velocity decreases.

(D) is correct : From the above relation $v_d \propto \frac{1}{l}$, so the drift velocity is inversely proportional to the length of the conductor.

(B) is wrong, because the expression contains no area of cross-section. (C) is wrong, because v_d is directly proportional to V. (E) is wrong, being the opposite of (A).