question_answer

In the figure battery B supplies 12 V. Take \[{{C}_{1}}=1\,\mu F,\,\,{{C}_{2}}=2\,\mu F,\,\,{{C}_{3}}=3\,\mu F,\,{{C}_{4}}=\,4\mu F.\] Charge on capacitor g) when only\[{{C}_{1}}\]is closed is

A)  \[3\,\mu C\]                     

B)  \[9\,\mu C\]

C)  \[6\,\mu C\]                     

D)  \[12\,\mu C\]

Correct Answer: B

Solution :

 As \[{{\text{S}}_{\text{2}}}\] is open, hence \[{{\text{C}}_{1}}\And {{\text{C}}_{\text{3}}}\]are in series, also \[{{\text{C}}_{2}}\And \] \[{{\text{C}}_{4}}\]are in series combination. \[Q=\frac{3}{4}\times V=\frac{3}{4}\times 12=9\mu C;\] Charge on \[C_{1}^{'}=9\mu C;\] Charge on\[C_{1}^{'}=\frac{4}{3}\times 12=16\mu C\] Hence, charge on \[{{C}_{1}}\] and \[{{C}_{3}}\]is \[9\mu C,\] as both are in series combination.