question_answer

The ratio of potential difference between \[1\mu F\] and \[5\mu F\]capacitors -

A) \[1:2\]               

B) \[3:1\]

C) \[1:5\]               

D) \[10:1\]

Correct Answer: C

Solution :

\[2\,\mu F\,\And \,2\mu F\] are reduced to 1 capacitor of 1 \[\mu F\]and \[3\,\mu F\,\,\,\And \,\,\,1\,\mu F\] is reduced to \[4\mu F\]. \[\therefore \,\,{{V}_{{{4}_{\mu F}}}}=\frac{10\times 1}{5}=2\] (potential difference across \[4\,\mu F\] capacitor is calculated by voltage division Rule)             \[\therefore \,\,\,{{V}_{5}}_{_{\mu F}}=10\,V\] (potential, difference across \[5\,\mu F\] is same as Battery) \[\therefore \,\,\,\frac{{{V}_{{{1}_{\mu F}}}}}{{{V}_{{{5}_{\mu F}}}}}=\frac{2}{10}=\frac{1}{5}\]