Observe the given circuit diagram and answer the questions that follow:

(a) 5Ω and 3Ω are in series so their equivalent resistance (R_s) = 5 + 3 = 8Ω

Now 2Ω and R_s are in parallel.

$$\begin{aligned}\frac{\text{1}}{\text{R}_\text{p}} = \frac{\text{1}}{\text{R}_\text{s}} + \frac{\text{1}}{\text{2}} \\[1em] \Rightarrow \frac{\text{1}}{\text{R}_\text{p}} = \frac{\text{1}}{\text{2}} + \frac{\text{1}}{\text{8}} \\[1em] \Rightarrow \frac{\text{1}}{\text{R}_\text{p}} = \frac{4+1}{8} \\[1em] \Rightarrow \frac{\text{1}}{\text{R}_\text{p}} = \frac{5}{8} \\[1em] \therefore \text{R}_\text{p} = \frac{8}{5} = 1.6 \space Ω\\[1em]\end{aligned}$$

From circuit diagram,

Internal Resistance = 0.4 Ω

Total resistance of circuit = 1.6 + 0.4 = 2 Ω

(b) Current drawn from the battery

$$\begin{aligned}\text{I} = \frac{\text{e.m.f}}{\text{Total Resistance}} \\[1em] = \frac{4}{2} = 2 \text{ A}\end{aligned}$$

Now, the current I divides in 2 parts.

Let current across 2Ω be I₁ and across the series combination of 5Ω and 3Ω be I₂

I = I₁ + I₂ = 2 A .....(1)

Terminal Voltage of cell V = IR_p = 2 x 1.6 = 3.2 V

∴ P.d. across 2 Ω resistor = 3.2 V

I₁ = $\frac{3.2 \text{ V}}{2 \text{ Ω}}$

= 1.6 A

Substituting value of I₁ in Eq 1,

2 = 1.6 + I₂

⇒ I₂ = 2 - 1.6 = 0.4 A

Hence, current across 3Ω = 0.4 A