A) 12 V
B) 4 V
C) 6 V
D) 18 V
Correct Answer: C
Solution :
(c) 6 V \[{{E}_{net}}=16-6=10V\] \[{{C}_{eq}}=\frac{2\times 3}{2+2}=\frac{6}{5}\mu F\] Potential difference across \[2\mu C\] capacitor. \[{{V}_{1}}=\frac{Q}{{{C}_{1}}}=\frac{12\mu C}{2\mu F}=6V\]You need to login to perform this action.
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