A) \[120\mu C\]
B) \[80\mu C\]
C) \[40\mu C\]
D) \[20\mu C\]
Correct Answer: C
Solution :
Common potential \[V=\frac{6\times 20+3\times 0}{(6+3)}=\frac{120}{9}Volt\] So, charge on \[3\,\mu F\] capacitor \[{{Q}_{2}}=3\times {{10}^{-6}}\times \frac{120}{9}=40\,\mu C\]You need to login to perform this action.
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