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=12\,\,V $ ... (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=8\,\,V $