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