An LCR circuit is equivalent to a damped pend | vedclass

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(A) Applying Kirchhoff's Voltage Law $(KVL)$ to the $LCR$ circuit at any time $t$:$\frac{q}{C} - iR - L \frac{di}{dt} = 0$Since $i = -\frac{dq}{dt}$,w

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(A) Applying Kirchhoff's Voltage Law $(KVL)$ to the $LCR$ circuit at any time $t$:$\frac{q}{C} - iR - L \frac{di}{dt} = 0$Since $i = -\frac{dq}{dt}$,we have:$\frac{q}{C} + R \frac{dq}{dt} + L \frac{d^2q}{dt^2} = 0$Rearranging gives the differential equation for a damped harmonic oscillator:$\frac{d^2q}{dt^2} + \frac{R}{L} \frac{dq}{dt} + \frac{q}{LC} = 0$The charge on the capacitor decays as $q(t) = Q_0 e^{-\frac{Rt}{2L}} \cos(\omega' t + \phi)$.The maximum charge at any cycle is given by the envelope $Q_{Max} = Q_0 e^{-\frac{Rt}{2L}}$.Squaring this,we get $Q_{Max}^2 = Q_0^2 e^{-\frac{Rt}{L}}$.Comparing the decay constant $\lambda = \frac{R}{L}$,we see that for a fixed $R$,a smaller $L$ results in a larger decay constant,meaning the charge decays faster.Since $L_1 > L_2$,the decay constant for $L_2$ is larger than for $L_1$.Therefore,the curve for $L_2$ decays faster than the curve for $L_1$.

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