question_answer

The charge deposited on 4\[\mu \]F capacitor in the circuit is

A)  \[6\times {{10}^{-6}}C\]              

B)  \[12\times {{10}^{-6}}C\]

C)  \[24\times {{10}^{-6}}C\]            

D)  \[36\times {{10}^{-6}}C\]

Correct Answer: C

Solution :

As the capacitors \[4\mu F\] and \[2\mu F\] are connected in parallel and are in series with \[6\mu F\]capacitor, their equivalent capacitance is                 \[\frac{(2+4)\times 6}{2+4+6}=3\mu F\] Charge in the circuit,                 \[Q=3\mu F\times 12\,V=36\,\mu C\] Since, the capacitors \[4\mu F\] and \[2\mu F\] are connected in parallel, therefore potential difference across them is same. \[\Rightarrow \]               \[\frac{{{Q}_{1}}}{{{Q}_{2}}}=\frac{{{C}_{1}}}{{{C}_{2}}}=\frac{4}{2}\]or \[{{Q}_{1}}=2{{Q}_{2}}\] Also,                      \[{{Q}_{1}}=2{{Q}_{2}}\] \[\therefore \] \[36\mu C=2{{Q}_{2}}+{{Q}_{2}}\] or \[{{Q}_{2}}=\frac{36\mu C}{3}=12\mu C\]                 \[{{Q}_{1}}=Q-{{Q}_{2}}=36\mu C-12\mu C\]                 \[=24\mu C=24\times {{10}^{-6}}C\]