A) 2V
B) 8V
C) 6V
D) 1V
Correct Answer: C
Solution :
(c) 6V \[{{C}_{p}}=2+4=6\mu F\] \[\frac{1}{C}=\frac{1}{6}+\frac{1}{6}=\frac{2}{6}=\frac{1}{3}\] or \[C=3\mu F\] Total charge, \[Q=CV=3\times 12=36\mu C\] Voltage across \[6\,\mu F\]capacitor \[=\frac{36\,\mu C}{6\,\mu F}=6V\] \[\therefore \]Voltage across each \[2\mu F\]and \[4\,\mu F\]capacitors \[=\text{ }12\text{ }V\text{ }-\text{ }6\text{ }V\] \[\Rightarrow \,\,V=6\,V\]You need to login to perform this action.
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