Question

Three resistors of 2 Ω, 4 Ω and 6 Ω are connected in series to a battery of e.m.f. 12 V and negligible internal resistance. Draw the circuit diagram and calculate (a) current drawn from the cell (b) p.d. at the ends of each resistor.

Answer 1

Resistance of 2 Ω, 4 Ω and 6 Ω in series

R = 2 + 4 + 6 = 12 Ω

(a) \[ \therefore \text{Current drawn from cell (I)} = \frac{\mathrm{V}}{\mathrm{R}} = \frac{12}{12} = 1 \ \text{A}\]

(b) P.D. across \(2 \Omega\) resistor \(\mathrm{V}_{1}=\mathrm{I} . \mathrm{R}=1 \times 2= \mathrm{2V}\)

P.D. across \(4 \Omega\) resistor \(\mathrm{V}_{2}=\mathrm{I} . \mathrm{R}=1 \times 4=\mathrm{4 V}\)

P.D. across \(6 \Omega\) resistor \(\mathrm{V}_{3}=\mathrm{I} . \mathrm{R}=1 \times 6=\mathrm{6V}\)