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) \[\therefore \,\,\,\frac{{{V}_{{{1}_{\mu F}}}}}{{{V}_{{{5}_{\mu F}}}}}=\frac{2}{10}=\frac{1}{5}\]You need to login to perform this action.
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