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Q.Show that the total energy of LC circuit is V=VE=12qm2CV = V_E = \dfrac{1}{2}\dfrac{q_m^2}{C}.

Nagaland NbseNagaland Board of School Education 2020Subjective· 3mImportance★★★★★
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Adding the instantaneous capacitor energy and inductor energy for an oscillating LC circuit, the time-dependence cancels via cos⁡2+sin⁡2=1\cos^2+\sin^2=1, leaving a constant total energy qm2/2Cq_m^2/2C.

In an ideal (resistanceless) LC circuit, once a capacitor of capacitance CC (initially charged to qmq_m) is connected to an inductor LL, the charge oscillates sinusoidally:

q(t)=qmcos⁡(ωt),ω=1LCq(t) = q_m\cos(\omega t), \qquad \omega = \frac{1}{\sqrt{LC}}

Energy stored in the capacitor at any instant:

UE=q22C=qm2cos⁡2ωt2CU_E = \frac{q^2}{2C} = \frac{q_m^2\cos^2\omega t}{2C}

Current in the circuit (rate of flow of charge):

i=dqdt=−qmωsin⁡(ωt)i = \frac{dq}{dt} = -q_m\omega\sin(\omega t)

Energy stored in the inductor at the same instant:

UB=12Li2=12L qm2ω2sin⁡2ωtU_B = \frac{1}{2}Li^2 = \frac{1}{2}L\,q_m^2\omega^2\sin^2\omega t

Since ω2=1LC\omega^2 = \dfrac{1}{LC}:

UB=12L qm2(1LC)sin⁡2ωt=qm2sin⁡2ωt2CU_B = \frac{1}{2}L\,q_m^2\left(\frac{1}{LC}\right)\sin^2\omega t = \frac{q_m^2\sin^2\omega t}{2C}

Total energy at any instant tt:

U=UE+UB=qm2cos⁡2ωt2C+qm2sin⁡2ωt2C=qm22C(cos⁡2ωt+sin⁡2ωt)U = U_E+U_B = \frac{q_m^2\cos^2\omega t}{2C}+\frac{q_m^2\sin^2\omega t}{2C} = \frac{q_m^2}{2C}\big(\cos^2\omega t+\sin^2\omega t\big)

Using the identity cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta = 1: …

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