question_answer

In steady-state, the charge on \[6\mu F\] capacitor is

A) \[10\mu \operatorname{C}\]

B) \[20\mu \operatorname{C}\]

C) \[30\mu \operatorname{C}\]

D) \[40\mu \operatorname{C}\]

Correct Answer: B

Solution :

If \[{{V}_{1}}\] and\[{{V}_{2}}\] are the P.d. across \[3\mu F\] and \[6\mu F\] capacitor respectively

then\[{{V}_{1}}+{{V}_{2}}=(20-10)\] and \[3{{V}_{1}}=6{{V}_{2}}\]

On solving \[{{V}_{1}}=\frac{20}{3}V\]and\[{{V}_{2}}=\frac{10}{3}V\]

Charge on \[6\mu F,q=C{{V}_{2}}=6\times \frac{10}{3}=20\mu F.\]