The adjoining figure shows two bulbs B_1 and | vedclass

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(C) When the switch $S$ is opened,the circuit containing the battery is broken. For the branch containing bulb $B_1$ and resistor $R$,there is no sour

Vedclass · Accepted Answer

$B_1$ dies out promptly but $B_2$ dies out with some delay.

Vedclass · Answer

(C) When the switch $S$ is opened,the circuit containing the battery is broken. For the branch containing bulb $B_1$ and resistor $R$,there is no source of electromotive force $(EMF)$ to maintain the current,so the current through $B_1$ drops to zero almost instantaneously. However,the inductor $L$ in the branch with bulb $B_2$ opposes any change in current due to self-induction. When the switch is opened,the inductor generates an induced $EMF$ that maintains the current in the loop formed by the inductor $L$ and the bulb $B_2$. Consequently,the current through $B_2$ decays gradually according to the time constant of the $L-R$ circuit,causing $B_2$ to dim slowly before extinguishing. Therefore,$B_1$ dies out promptly,while $B_2$ dies out with some delay.

The adjoining figure shows two bulbs $B_1$ and $B_2$,a resistor $R$,and an inductor $L$. When the switch $S$ is turned off,what happens to the bulbs?

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