An LCR circuit behaves like a damped harmonic | vedclass

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Question

(A) For an $LCR$ circuit,applying Kirchhoff's Voltage Law $(KVL)$ to a closed loop with no external source:$-L \frac{di}{dt} - \frac{q}{C} - iR = 0$Si

Vedclass · Accepted Answer

$L \leftrightarrow m, C \leftrightarrow \frac{1}{k}, R \leftrightarrow b$

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(A) For an $LCR$ circuit,applying Kirchhoff's Voltage Law $(KVL)$ to a closed loop with no external source:$-L \frac{di}{dt} - \frac{q}{C} - iR = 0$Since $i = \frac{dq}{dt}$,we have:$L \frac{d^2q}{dt^2} + R \frac{dq}{dt} + \frac{1}{C}q = 0$For a mechanical damped harmonic oscillator,the equation of motion is given by:$m \frac{d^2x}{dt^2} + b \frac{dx}{dt} + kx = 0$Comparing the two differential equations term by term:$1$. The coefficient of the second derivative: $L$ corresponds to $m$.$2$. The coefficient of the first derivative: $R$ corresponds to $b$.$3$. The coefficient of the displacement/charge: $\frac{1}{C}$ corresponds to $k$,which implies $C$ corresponds to $\frac{1}{k}$.Thus,the correct equivalence is $L \leftrightarrow m, C \leftrightarrow \frac{1}{k}, R \leftrightarrow b$.

An $LCR$ circuit behaves like a damped harmonic oscillator. Comparing it with a physical spring-mass damped oscillator having a damping constant $b$,the correct equivalence would be:

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