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 \[{{C}_{1}}\] when only \[{{S}_{1}}\] is closed is

A)  \[3\,\mu C\]                 

B)  \[9\,\mu C\]

C)  \[6\,\mu C\]                 

D)  \[12\,\mu C\]

Correct Answer: B

Solution :

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