question_answer

In the circuit shown in the figure, the potential difference across the\[4.5\,\mu F\]capacitor is:        

A)  8 V                             

B)  6 V

C)  4 V                             

D) \[\frac{8}{3}\,V\]

Correct Answer: A

Solution :

The given circuit is simplified as \[\therefore \]    \[C=\frac{4.5\times 9}{4.5+9}=3\mu F\] \[\therefore \]    \[Q=CV=3\times 12=36\,\mu C\] Now, the Potential diff. across\[4.5\,\mu C\]capacitor is             \[V=\frac{36\,\mu C}{4.5\,\mu F}=8V\]  Hence, the correction option is [a].