question_answer

In the circuit shown in the given figure the potential difference across \[3\mu F\]capacitor is:

A)  16V          

B)  10V

C)  4V             

D)  6V

Correct Answer: B

Solution :

The resultant emf E which is responsible for flow of current is \[E=20-4=16\,V\] Given capacitors are connected in series, therefore \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}\] \[{{C}_{1}}=3\,\mu F,\,{{C}_{2}}=5\mu F\] \[\therefore \] \[\frac{1}{C}=\frac{1}{3}+\frac{1}{5}=\frac{8}{15}\] \[\Rightarrow \] \[C=\frac{15}{8}\mu F\] Total charge stored in capacitor is \[q=CV=\frac{15}{8}\times 16=30\,\mu C\] Potential difference across \[3\mu F\] capacitor is \[q=\frac{q}{C}=\frac{30}{3}=10\,V\]