Question

The figure shows a circuit diagram containing 8 cells of em.f. 1.5 V and an internal resistance of 0.25 Ω. Calculate (a) total e.m.f. of battery (b) total internal resistance, (c) total external resistance (d) current recorded by ammeter (e) current flowing through (i) 12 Ω resistor (ii) 2 Ω resistor (f) drop in potential across the terminals of the cell.

Answer 1

(a) e.m.f. of 8 cells in series = 8 × 1.5 V = 12 V

(b) Internal resistance of 8 cells series = 8 × 0.25 Ω = 2 Ω

(c) Equivalent resistance 12 Ω and (4 + 2 = 6 Ω) in parallel.

\[ \frac{1}{\mathrm{R}_{p}} = \frac{1}{12} + \frac{1}{6}\]

\[ \frac{1}{\mathrm{R}_{p}} = \frac{3}{12}\]

\[ \frac{1}{\mathrm{R}_{p}} = \frac{1}{4} \]

\[ \therefore \mathrm{R}_{p} = 4 \ \Omega\]

∴ Total external resistance = 0.5 Ω + 1.5 Ω + 4 Ω = 6 Ω

(d) \[ \text{Current recorded by ammeter (I)} = \frac{\mathrm{E}}{\mathrm{R} + \mathrm{r}} = \frac{12}{6 + 2} = \frac{12}{8} = 1.5 \ \text{A}\]

(e) p.d. across parallel circuit, V = IR = 1.5 × 4 = 6 V

\[ \text{(i) Current in } 12 \ \Omega \text{ resistor, } \mathrm{I} = \frac{\mathrm{V}}{\mathrm{R}} = \frac{6}{12} = 0.5 \ \text{A}\]

\[ \text{(ii) Current in } (4 + 2 = 6 \ \Omega) \text{ resistor, } \mathrm{I} = \frac{\mathrm{V}}{\mathrm{R}} = \frac{6}{6} = 1.0 \ \text{A}\]

(f) Drop in potential across the terminal of the battery

E − V = Ir = 1.5 × 2 = 3 V