In an LC circuit,the capacitor has a maximum | vedclass

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Question

(A) In an $LC$ oscillation circuit,the energy oscillates between the capacitor and the inductor.At any instant,the voltage across the capacitor is $V_

Vedclass · Accepted Answer

$\frac{q_0}{LC}$

Vedclass · Answer

(A) In an $LC$ oscillation circuit,the energy oscillates between the capacitor and the inductor.At any instant,the voltage across the capacitor is $V_C = \frac{q}{C}$ and the induced emf across the inductor is $V_L = L \frac{di}{dt}$.From Kirchhoff's voltage law for the $LC$ loop,$V_C = V_L$,which implies $\frac{q}{C} = L \frac{di}{dt}$.To find the maximum value of $\frac{di}{dt}$,we must consider the maximum voltage across the capacitor,which occurs when the charge on the capacitor is at its maximum value $q_0$.Thus,$\left( \frac{di}{dt} \right)_{\max} = \frac{1}{L} \left( \frac{q}{C} \right)_{\max} = \frac{q_0}{LC}$.

In an $LC$ circuit,the capacitor has a maximum charge $q_0$. The value of $\left. \frac{di}{dt} \right|_{\max }$ is:

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