question_answer

Two capacitors of \[2\,\mu F\]and \[4\,\mu F\]are connected in parallel. A third capacitor of \[6\,\mu F\]is connected in series. The combination is connected across a 12 V battery. The voltage across \[2\,\mu F\]capacitor is:

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 \[=\text{ }12\text{ }V\text{ }-\text{ }6\text{ }V\] \[\Rightarrow \,\,V=6\,V\]