A) 2 V
B) 4 V
C) 1 V
D) 6 V
Correct Answer: D
Solution :
(d) 6V In series combination \[\frac{1}{C}=\frac{1}{{{C}_{1}}}+\frac{1}{{{C}_{2}}}+\frac{1}{{{C}_{3}}}\] \[\Rightarrow \,\,\,\,\,\frac{1}{C}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}\,\Rightarrow \frac{1}{C}=\frac{6+3+2}{6}\] \[C=\frac{6}{11}\mu F\] Now, \[C=\frac{Q}{V}\] So, \[Q=CV=\frac{6}{11}\times 11=6C\] So, potential difference across \[1\,\mu F\]capacitor is \[V=\frac{Q}{C}=\frac{6}{1}=6V\]You need to login to perform this action.
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