In a series circuit C=2μF,L=1mHandR=10Ω, when the current in the circuit is maximum, at that time the ratio of the energies stored in the capacitor and the inductor will be
A
1 : 1
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B
1 : 2
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C
2 : 1
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D
1 : 5
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Solution
The correct option is D 1 : 5 Current will be maximum in the condition of resonance so imax=VR=V10A Energy stored in the coil WL=12Li2max=12L(V10)2 12×10−3(V2100)=12×10−5V2joule ∴Energy stored in the capacitor WC=12CV2=12×2×10−6V2=10−6V2joule∴WCWL=15