In the circuit shown,X is joined to Y for a l | vedclass

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Question

(A) When $X$ is connected to $Y$ for a long time,the inductor $L$ acts as a short circuit (steady state). The current flowing through the inductor is

Vedclass · Accepted Answer

$\frac{L E^2}{2 R_1^2}$

Vedclass · Answer

(A) When $X$ is connected to $Y$ for a long time,the inductor $L$ acts as a short circuit (steady state). The current flowing through the inductor is $I_0 = \frac{E}{R_1}$.The energy stored in the inductor is $U = \frac{1}{2} L I_0^2 = \frac{1}{2} L \left(\frac{E}{R_1}\right)^2 = \frac{L E^2}{2 R_1^2}$.When $X$ is switched to $Z$,the inductor $L$ and resistor $R_2$ form a closed series circuit. The energy stored in the inductor is dissipated as heat in the resistor $R_2$ as the current decays to zero.Since all the energy stored in the inductor is dissipated as heat in the circuit,the total heat produced in $R_2$ is equal to the initial energy stored in the inductor.Therefore,the total heat $H = U = \frac{L E^2}{2 R_1^2}$.

In the circuit shown,$X$ is joined to $Y$ for a long time,and then $X$ is joined to $Z$. The total heat produced in $R_2$ is:

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