Question

Four cells, each of e.m.f. 1.5 V and internal resistance 2 Ω are connected in parallel, to form a battery. The battery is connected to the external resistance of 1.5 Ω and a set of two resistances of 3 Ω and 1.5 Ω in parallel. Draw a circuit diagram and answer the following:

1. What is the current in the main circuit?
2. What is the current in 3 Ω wire?
3. What is the drop in potential?

Answer 1

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

Total internal resistance of 4 cells in parallel

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

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

∴ Equivalent resistance of \(1.5 \Omega\) and \(3 \Omega\) in parallel

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

\(\therefore \quad \mathrm{R}_{\mathrm{p}}=1 \Omega\)

∴ Total external resistance \(=1+1.5=2.5 \Omega\)

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

(b) P.D. across the parallel circuit, \(\mathrm{V}=\mathrm{I} . \mathrm{R}_{\mathrm{p}}=0.5 \times 1=0.5 \mathrm{~V}\)

\[ \therefore \text { Current in } 3 \Omega \text { wire, } I=\frac{V}{R}=\frac{0.5}{3}=0.167 \text{A}\]

(c) Drop in potential across the terminals of the battery, \(\mathrm{E}-\mathrm{V}=\mathrm{Ir}=0.5 \times 0.5=0.25 \text{V}\)