The resistance R is the same as that of the c | vedclass

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(A) When the switch $S$ is closed, the current in the branch containing the inductor $L$ grows slowly due to the back $EMF$ induced by the inductor, w

Vedclass · Accepted Answer

The bulb $B_2$ lights up earlier than $B_1$ and finally both the bulbs shine equally bright.

Vedclass · Answer

(A) When the switch $S$ is closed, the current in the branch containing the inductor $L$ grows slowly due to the back $EMF$ induced by the inductor, which opposes the change in current $(e = -L \frac{di}{dt})$. In the branch containing the resistor $R$, the current reaches its steady state value almost instantaneously. Therefore, the bulb $B_2$ (in the resistor branch) lights up earlier than $B_1$ (in the inductor branch). However, the problem states that the resistance of the coil $L$ is equal to the resistance $R$. In the steady state, the inductor acts as a simple wire with resistance. Since the total resistance of both branches is the same (assuming the bulbs have identical resistance), the steady-state current in both branches will be equal. Thus, both bulbs will eventually shine with equal brightness. Wait, re-evaluating the standard problem: If the coil $L$ has internal resistance $r_L$ and we are given $r_L = R$, then the total resistance of the top branch is $R_{total} = r_L + R_{bulb1}$ and the bottom branch is $R_{total} = R + R_{bulb2}$. If $R_{bulb1} = R_{bulb2}$, then the final brightness is equal. Looking at the provided options, option $(A)$ is the standard physical outcome for this circuit configuration.

The resistance $R$ is the same as that of the coil that makes $L$. Which of the following statements gives the correct description of the happenings when the switch $S$ is closed?

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