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

(i) Current through 3 Ω resistor.

(ii) The internal resistance r.

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

$\dfrac{1}{R$p} = \dfrac{1}{3} + \dfrac{2}{3} \space \space \big[\because 1.5 Ω = \dfrac{3}{2} Ω\big] \\[0.5em] \Rightarrow \dfrac{1}{Rp} = \dfrac{3}{3} \\[0.5em] \Rightarrow R_p = 1 Ω

From, V = IR

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

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

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

Current in the circuit,

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

∴ Internal resistance r = 1 Ω