Question

What is the potential difference across $2\mu F$ capacitor in the circuit shown?

• A. 12 V
• B. 4 V
• C. 6V
• D. 18 V

Answer 1

Answer: (C)

Net emf in the circuit here $E=E_2-E_1=16-6=10V$ While the equivalent capacitance $C=\frac{C_1{C}_2}{C_1+C_2}=\frac{2\times 3}{2+3}=\frac{6}{5}\mu F$ Charge on each capacitor $q=CV=\frac{6}{5}\times 10=12\mu C$ $\therefore$ Potential difference across $2\mu F$ capacitor $V_1=\frac{q}{C_1}=\frac{12}{2}=6V$