An inductor L,a resistance R and two identica | vedclass

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Question

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

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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 gradually due to the back $EMF$ induced by the inductor,which opposes the change in current. The current in the branch containing only the resistor $R$ reaches its steady-state value almost instantaneously.Therefore,the bulb $B_2$ (in the resistor branch) lights up earlier than the bulb $B_1$ (in the inductor branch).As time passes,the current in the inductor branch increases and eventually reaches its steady-state value. Since the total resistance in both branches is the same (the inductor coil also has resistance $R$),the steady-state current in both branches will be equal.Thus,finally,both bulbs $B_1$ and $B_2$ will shine with equal brightness.

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