A) \[\frac{1}{2}\]
B) 2
C) \[\frac{1}{4}\]
D) 4
Correct Answer: B
Solution :
In series voltage is divided in inverse ratio of capacitance \[\frac{{{V}_{1}}}{{{V}_{2}}}=\frac{{{C}_{2}}}{{{C}_{1}}}=\frac{0.6}{0.3}=\frac{2}{1}\] \[{{\operatorname{E}}_{1}}=\frac{1}{2}\times {{C}_{1}}{{V}_{1}}^{2}=\frac{1}{2}\times 0.3\mu F\times {{V}_{1}}^{2}\] \[{{\operatorname{E}}_{2}}=\frac{1}{2}\times {{C}_{2}}{{V}_{2}}^{2}=\frac{1}{2}\times 0.6\mu F\times {{V}_{2}}^{2}\] \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{2}{1}\]You need to login to perform this action.
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