Question

In the circuit shown in figure, switch S; is initially closed and $S_2$ is open. Find $V_a-V_{b.}$ .

• A. 4 V
• B. 8 V
• C. 12 V
• D. 16 V

Answer 1

Answer: (B)

Switch $S_2$ is open so capacitor is not in circuit.
Current through $3\Omega$ resistor $=\frac{24}{3+3}=4A$ Let potential of point $O$ shown in figure is $V_0$ then using Ohms law $V_0-V_a=3\times 4=12V$ ... (i) Now, current through $5\Omega$ resistor $=\frac{24}{5+1}$ $=4A$ So $V_0-V_b=4\times 1=4V$ ... (ii) From Eqs. (i) and (ii), we get $V_b-V_a=12-4=8V$