Alternating Current

JEE Main level, single correct, medium

137 of 190

Question 137 of 190

An ideal capacitor of capacitance 0.2μF0.2\mu F is charged to a potential difference of 10V10V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5mH0.5mH. The current at a time when the potential difference across the capacitor is 5V5V is:

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Answer: (B) 0.17A0.17A

The energy stored in the capacitor is given by, Uc=12CV2{U}_{c}=\dfrac{1}{2}C{V}^{2}. The energy stored in the inductor is given by. UL=12LI2{U}_{L}=\dfrac{1}{2}L{I}^{2}. In LCLC circuit there is no loss of energy in the form of heat so, using energy conservation, Uci+ULi=Ucf+ULf{U}_{ci}+{U}_{Li}={U}_{cf}+{U}_{Lf}
⇒12×0.2×10−6×102+0=12×0.2×10−6×52+12×0.5×10−3I2\Rightarrow \dfrac{1}{2}\times 0.2\times {10}^{-6}\times {10}^{2}+0 =\dfrac{1}{2}\times 0.2\times {10}^{-6}\times {5}^{2}+\dfrac{1}{2}\times 0.5\times {10}^{-3}{I}^{2}
I=3×10−1A=0.17AI=\sqrt{3}\times {10}^{-1}A=0.17A

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