Question

Four cells, each with an e.m.f. 1.5 V and internal resistance of 0.25 Ω are connected in series. This battery of cells is connected to an external resistance of 5 Ω.

1. Draw the figure for the circuit diagram.
2. Calculate total resistance of circuit.
3. Calculate total current flowing through external circuit.
4. Drop in potential across the terminals of the cell.

Answer 1

(i)

(ii) Total resistance of circuit

Total e.m.f. of the battery

\[ \mathrm{E} = 4 \times 1.5 \]

\[ = 6 \ \text{V} \]

Total internal resistance of the battery

\[ r = 4 \times 0.25 \]

\[ = 1 \ \Omega \]

External resistance

\[ \mathrm{R} = 5 \ \Omega \]

\[ \text{Total resistance} = \mathrm{R} + r \]

\[ = 5 + 1 \]

\[ = 6 \ \Omega \]

(iii) Total current flowing through external circuit

\[ \mathrm{I} = \frac{\mathrm{E}}{\mathrm{R} + r} \]

\[ = \frac{6}{6} \]

\[ = 1 \ \text{A} \]

(iv) Drop in potential across the terminals of cell

Terminal voltage

\[ \mathrm{V} = \mathrm{IR} \]

\[ = 1 \times 5 \]

\[ = 5 \ \text{V} \]

\[ \therefore \text{Drop in potential} = \mathrm{E} - \mathrm{V} \]

\[ = 6 - 5 \]

\[ = 1 \ \text{V} \]