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III. Long Answer Questions · Q25

Q.Prove that the total energy is conserved during LC oscillations.

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Step 1. Case (i): q=Qmq=Q_m, i=0i=0: U=Qm2/2C+0=Qm2/2CU=Q_m^2/2C+0=Q_m^2/2C -- wholly electrical.

Step 2. Case (ii): q=0q=0, i=Imi=I_m: U=0+12LIm2U=0+\tfrac12LI_m^2 -- wholly magnetic; equating this to case (i)'s value gives Im=Qm/LCI_m=Q_m/\sqrt{LC}.

Step 3. Case (iii): at a general instant, q=Qmcos⁡ωtq=Q_m\cos\omega t, i=dq/dt=−Qmωsin⁡ωti=dq/dt=-Q_m\omega\sin\omega t; then U=Qm2cos⁡2ωt2C+12L(Qmω)2sin⁡2ωtU=\dfrac{Q_m^2\cos^2\omega t}{2C}+\dfrac12L(Q_m\omega)^2\sin^2\omega t.

Step 4. Substituting ω2=1/LC\omega^2=1/LC into the second term turns it into Qm22Csin⁡2ωt\dfrac{Q_m^2}{2C}\sin^2\omega t, so U=Qm22C(cos⁡2ωt+sin⁡2ωt)=Qm22CU=\dfrac{Q_m^2}{2C}(\cos^2\omega t+\sin^2\omega t)=\dfrac{Q_m^2}{2C}, using cos⁡2θ+sin⁡2θ=1\cos^2\theta+\sin^2\theta=1. …

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