Question

Study the circuit diagram carefully and calculate:

1. current in the main circuit.
2. current in each of the resistors in the parallel circuit.

Answer 1

Equivalent resistance of 1.5 Ω and 3 Ω in parallel

\[\frac{1}{R_{p_1}} = \frac{1}{1.5} + \frac{1}{3}\]

\[\frac{1}{R_{p_1}}= \frac{2}{3} + \frac{1}{3}\]

\[\frac{1}{R_{p_1}}= \frac{3}{3}\]

\[\frac{1}{R_{p_1}}= 1\]

\[\therefore R_{p_1} = 1\Omega\]

Equivalent resistance of 6 Ω and 12 Ω in parallel

\[\frac{1}{R_{p_2}} = \frac{1}{6} + \frac{1}{12}\]

\[\frac{1}{R_{p_2}} = \frac{2 + 1}{12}\]

\[\frac{1}{R_{p_2}} = \frac{3}{12}\]

\[\frac{1}{R_{p_2}} = \frac{1}{4}\]

\[\therefore R_{p_2} = 4\Omega\]

\[\therefore \text{Total equivalent resistance in series} = 1 + 4 = 5\Omega\]

\[\text{(a) }\therefore \text{Current in main circuit, }I = \frac{V}{R} = \frac{9}{5} = 1.8\text{A}\]

\[\text{(b) P.D. across }1.5\Omega\text{ and }3\Omega\text{ combination, }V_1 = I.R_{p_1} = 1.8 \times 1 = 1.8\text{V}\]

\[\therefore \text{Current in }1.5\Omega\text{ resistor} = \frac{V_1}{R} = \frac{1.8}{1.5} = 1.2\text{A}\]

\[\text{Current in }3\Omega\text{ resistor} = \frac{V_1}{R} = \frac{1.8}{3} = 0.6\text{A}\]

\[\text{P.D. across }6\Omega\text{ and }12\Omega\text{ combination, }V_2 = I.R_{p_2} = 1.8 \times 4 = 7.2\text{V}\]

\[\therefore \text{Current in }6\Omega\text{ resistor} = \frac{V_2}{R} = \frac{7.2}{6} = 1.2\text{A}\]

\[\text{Current in }12\Omega\text{ resistor} = \frac{V_2}{R} = \frac{7.2}{12} = 0.6\text{A}\]