A) \[2400erg\]
B) \[1800erg\]
C) \[3600erg\]
D) \[5400erg\]
Correct Answer: B
Solution :
Total capacitance of given system \[{{C}_{eq}}=\frac{8}{5}\mu f\] \[U=\frac{1}{2}{{C}_{eq}}{{V}^{2}}\] \[=\frac{1}{2}\times \frac{8}{5}\times {{10}^{-6}}\times 225\] \[=180\times {{10}^{-6}}J\] \[-180\times {{10}^{-6}}\times {{10}^{-7}}erg=1800erg\]You need to login to perform this action.
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