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.

$$\begin{aligned}\frac{1}{\text R_\text P} = \frac{1}{\text R_\text s} + \frac{1}{2} \\[1 em] ⇒ \frac{1}{\text R_\text P} = \frac{1}{8} + \frac{1}{2} = \frac{1+4}{8}\\[1 em] ⇒ \text R_\text P = \frac{8}{5} = 1.6\ \text Ω\end{aligned}$$

From circuit diagram,

Internal Resistance = 0.4 Ω

Total resistance of circuit = 1.6 + 0.4 = 2 Ω

(b) Current drawn from the battery :

$$\begin{aligned}\text I = \frac{\text {emf}}{\text {Total resistance}} \\[1em] ⇒ \text I = \frac{4}{2} = 2\ \text A\end{aligned}$$

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 = $\frac{3.2}{2}$ = 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