Question

Two cells, each of e.m.f. 1.5 V and internal resistance 1 Ω are connected in parallel, to form a battery. The battery is connected to an external resistance of 0.5 Ω and two resistances of 3 Ω and 1.5 Ω in parallel.

1. Draw the circuit diagram.
2. Calculate the current in the main circuit.
3. Calculate the current in the 1.5 Ω resistor.
4. Calculate the drop in potential across the terminals of the battery.

Answer 1

(a)

Total e.m.f. of 2 cells in parallel = 1.5 V.

Total int. resistance of 2 cells in parallel

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

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

∴ Equivalent resistance of 3 Ω and 1.5 Ω in parallel

\[ \frac{1}{R_{p}}=\frac{1}{3}+\frac{1}{1.5}=\frac{1+2}{3}=\frac{3}{3}=1\]

\[ \therefore R_{p}=1 \Omega\]

Total external resistance \[ = 1 + 0.5 = 1.5 \Omega\]

\[ \text{(b) } \therefore \text{Current in main circuit, } I = \frac{E}{R+r_{p}} = \frac{1.5}{1.5+0.5} = \frac{1.5}{2} = 0.75 \text{ A}\]

(c) P.D. across the parallel circuit \[ V = I R_{p} = 0.75 \times 1 = 0.75 \text{ V}\]

\[ \therefore \text{Current in 1.5 Ω resistor, } I = \frac{V}{R} = \frac{0.75}{1.5} = 0.50 \text{ A}\]

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

\[ E - V = Ir = 0.75 \times 0.5 = 0.375 \text{ V}\]