Given: Capacitance, C=0.2μF=0.2×10−6 F Inductance L=0.5mH=0.5×10−3H Current I = ? Using energy conservation 21CV2=21CV12+21LI2 21×0.2×10−6×102+0 =21×0.2×10−6×52+21×0.5×10−3I2 ∴I=3×10−1 A=0.17 A
JEE Main 2018 — Physics Electromagnetism
An ideal capacitor of capacitance 0.2μF is charged to a potential difference of 10 V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5mH. The current at a time when the potential difference across the capacitor is 5 V, is:
Held on 15 Apr 2018 · Verified 6 Jul 2026.
0.17 A
0.15 A
0.34 A
0.25 A
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