question_answer

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

A)  \[\frac{8}{3}V\]                                               

B)  \[4\,V\]

C)  \[6\,V\]                                              

D)  \[8\,V\]

Correct Answer: D

Solution :

                 In the given circuit, capacitors of \[3\,\mu F\] and  \[6\,\mu F\] are connected in parallel, and this combination is connected in series with \[4.5\,\mu F\] capacitor.                 Hence effective capacitance is                 \[C=3+6=9\mu F\]                 \[\frac{1}{C}=\frac{1}{9}+\frac{1}{4.5}=\frac{9+4.5}{9\times 4.5}\]                 \[C=\frac{9\times 4.5}{9+4.5}=3\,\mu F\] The charge through the circuit is given by                                 \[q=CV=3\times 12=36\,\mu C\] Potential difference across \[4.5\,\mu F\] capacitor                 \[V=\frac{q}{C}=\frac{36}{4.5}=8\,V\]