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 $L_1$.

List-$I$	List-$II$
$A$. Purely capacitive circuit	$I$. $I$ leads $V$ by $90^{\circ}$
$B$. Purely inductive circuit	$II$. $I$ and $V$ are in phase
$C$. $LCR$ series at resonance	$III$. $V$ leads $I$ by angle $\theta$
$D$. $LCR$ series circuit	$IV$. $V$ leads $I$ by $90^{\circ}$

Similar Questions

The self-inductance of a choke coil is $10\, mH$. When it is connected to a $10\, V$ $dc$ source,the power loss is $20\, W$. When it is connected to a $10\, V$ $ac$ source,the power loss is $10\, W$. The frequency of the $ac$ source is......$Hz$.

If $C, R, L$ and $I$ denote capacity, resistance, inductance and electric current respectively, the quantities having the same dimensions of time are :
$(1)$ $C R$
$(2)$ $\frac{L}{R}$
$(3)$ $\sqrt{L C}$
$(4)$ $L I^2$

Statement-$I$: $A$ capacitor can be used in an $a.c.$ circuit in place of a choke coil.
Statement-$II$: $A$ capacitor blocks $d.c.$ and allows $a.c.$ to pass.

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