The diagram alongside shows three resistors connected across a cell of e.m.f. 1.8 V and internal resistance r. Calculate:

(i) The equivalent resistance R_p of resistances 3 Ω and 1.5 Ω joined in parallel is given by

$$\begin{aligned}\frac{1}{R_p} = \frac{1}{3} + \frac{2}{3} \space \space \big[\because 1.5 Ω = \frac{3}{2} Ω\big] \\[0.5em] \Rightarrow \frac{1}{R_p} = \frac{3}{3} \\[0.5em] \Rightarrow R_p = 1 Ω\end{aligned}$$

From, V = IR

the potential drop across this combination = 0.3 x 1 = 0.3 V

$$\begin{aligned}\therefore \text{Current through 3 Ω resistance} = \frac{\text{Voltage}}{\text{Resistance}} \\[0.5em] = \frac{0.3}{3} = 0.1 \text{ A}\end{aligned}$$

(ii) Total resistance of the circuit = R_p + 4 + r = (5 + r) Ω

Current in the circuit,

$$\begin{aligned}0.3 = \frac{1.8}{5 + r} \\[0.5em] 0.3 \times (5 + r) = 1.8 \\[0.5em] (0.3 \times 5) + (0.3 \times r) = 1.8 \\[0.5em] 1.5 + (0.3 \times r) = 1.8 \\[0.5em] 0.3 \times r = 1.8 - 1.5 \\[0.5em] r = \frac{0.3}{0.3} = 1 Ω\end{aligned}$$

∴ Internal resistance r = 1 Ω