Question

Four cells, each of e.m.f. 1.5 V and internal resistance 2 Ω each are connected in parallel to form a battery. The battery is connected to an external resistance of 0.5 Ω and three resistances of 12 Ω, 6 Ω and 4 Ω in parallel.

1. Draw the circuit diagram.
2. Calculate the current in the main circuit.
3. Calculate the current in the 4 Ω resistor.
4. Calculate the drop in potential across the terminals of the battery.

Answer 1

(a)

Total e.m.f. of 4 cells in parallel = 1.5 V.

Total int. resistance of 4 cells in parallel

\[ \frac{1}{r_{\mathrm{p}}}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{4}{2}=\frac{2}{1} \]

\[ \therefore r_{p}=\frac{1}{2}=0.5 \Omega\]

∴ Equivalent resistance of 12 Ω, 6 Ω and 4 Ω in parallel

\[ \frac{1}{R_{p}}=\frac{1}{12}+\frac{1}{6}+\frac{1}{4}=\frac{1+2+3}{12}=\frac{6}{12}=\frac{1}{2}\]

\[ \therefore R_{p}=2 \Omega\]

Total external resistance \[ = 2 + 0.5 = 2.5 \Omega\]

\[ \text{(b) } \therefore \text{Current in main circuit, } I = \frac{E}{R+r_{p}} = \frac{1.5}{2.5+0.5} = \frac{1.5}{3} = 0.5 \text{ A}\]

(c) P.D. across the parallel circuit \[ V = I R_{p} = 0.5 \times 2 = 1 \text{ V}\]

\[ \therefore \text{Current in 4 Ω resistor, } I = \frac{V}{R} = \frac{1}{4} = 0.25 \text{ A}\]

(d) Drop in potential across the terminals of the battery

\[ E - V = Ir = 0.5 \times 0.5 = 0.25 \text{ V}\]