question_answer

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

Correct Answer: C

Solution :

Net emf in the circuit here \[E={{E}_{2}}-{{E}_{1}}=16-6=10\,V\] 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 \[{{V}_{1}}=\frac{q}{{{C}_{1}}}=\frac{12}{2}=6V\]