A) 12 V
B) 4 V
C) 6V
D) 18 V
Correct Answer: C
Solution :
Net emf in the circuit here \[E={{E}_{2}}-{{E}_{1}}=16-6=10\,V\] While the equivalent capacitance \[C=\frac{{{C}_{1}}{{C}_{2}}}{{{C}_{1}}+{{C}_{2}}}=\frac{2\times 3}{2+3}=\frac{6}{5}\mu F\] Charge on each capacitor \[q=CV=\frac{6}{5}\times 10=12\mu C\] \[\therefore \] Potential difference across \[2\mu F\] capacitor \[{{V}_{1}}=\frac{q}{{{C}_{1}}}=\frac{12}{2}=6V\]You need to login to perform this action.
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