Question

Three resistors of 4 Ω, 6 Ω and 12 Ω are connected in parallel. The combination of these resistors is connected in series to a resistance of 2 Ω and then to a battery of e.m.f. 6 V and negligible internal resistance

1. Draw the circuit diagram.
2. Calculate the current in the main circuit.
3. Calculate the current in each of the resistors in parallel.

Answer 1

(a)

(b) Equivalent resistance of 4 Ω, 6 Ω and 12 Ω in parallel

\[\frac{1}{R_p} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12}\]

\[\frac{1}{R_p} = \frac{3 + 2 + 1}{12}\]

\[\frac{1}{R_p} = \frac{6}{12}\]

\[\frac{1}{R_p} = \frac{1}{2}\]

\[\therefore R_p = 2\Omega\]

\[\therefore \text{Total equivalent resistance in series} = 2 + 2 = 4\Omega\]

\[\therefore \text{Current in main circuit, }I = \frac{V}{R} = \frac{6}{4} = 1.5\text{A}\]

\[\text{(c) P.D. across parallel circuit, }V = I.R = 1.5 \times 2 = 3\text{V}\]

\[\therefore \text{Current in }4\Omega\text{ resistor, }I_1 = \frac{V}{R} = \frac{3}{4} = 0.75\text{A}\]

\[\text{Similarly, current in }6\Omega\text{ resistor, }I_2 = \frac{V}{R} = \frac{3}{6} = 0.5\text{A}\]

\[\text{Current in }12\Omega\text{ resistor, }I_3 = \frac{V}{R} = \frac{3}{12} = 0.25\text{A}\]