A cell of e.m.f 1.5 V and internal resistance 1 Ω is connected to two resistors of resistances 6 Ω and 3 Ω in parallel and a resistor of resistance 2 Ω in…

Given,

• Emf of the cell (ε) = 1.5 V
• Internal resistance (r) = 1 Ω
• Resistances in parallel = 6 Ω and 3 Ω
• Series resistance = 2 Ω

As, 6 Ω and 3 Ω resistances are in parallel then let their net resistance be R_P

$$\begin{aligned}\frac{1}{\text{R}_\text{P}} = \frac{1}{\text{R}_1} + \frac{1}{\text{R}_2} \\[1em] \Rightarrow \frac{1}{\text{R}_\text{P}} = \frac{1}{6} + \frac{1}{3} \\[1em] \Rightarrow \frac{1}{\text{R}_\text{P}} = \frac{1 + 2}{6} \\[1em] \Rightarrow \frac{1}{\text{R}_\text{P}} = \frac{3}{6} \\[1em] \Rightarrow \frac{1}{\text{R}_\text{P}} = \frac{1}{2} \\[1em] \Rightarrow \text{R}_\text{P} = 2 Ω\end{aligned}$$

Now, this R_P is in series with 2 Ω and internal resistance of the cell.

Then,

Net series resistance (R_S) = R_P + 2 + r = 2 + 2 + 1 = 5 Ω

(a) Current through 2 Ω resistor = $\frac {\text ε}{\text R_\text S}= \frac{1.5}{5}$ = 0.3 A

So, current through 2 Ω resistor is 0.3 A.

(b) Now, 0.3 A current will split between 3 Ω and 6 Ω resistors.

Let I₁ is the current flowing through 6 Ω resistor and 0.3 - I₁ is flowing through 3 Ω resistor.

As, 6 Ω and 3 Ω are in parallel then the potential difference across them will be equal i.e.,

Potential difference across 6 Ω = Potential difference across 3 Ω

So, by using Ohm's law (V = IR)

6 x I₁ = 3 x (0.3 - I₁)

⇒ 6 x I₁ = 0.9 - 3 x I₁

⇒ 6 x I₁ + 3 x I₁ = 0.9

⇒ 9 x I₁ = 0.9

⇒ I₁ = $\frac {0.9}{9}$ = 0.1 A

So, current through 6 Ω resistance is 0.1 A.