Consider the following circuit given below. T | vedclass

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(C) In the given circuit,the bulb is connected in series with switch $S_3$. For the bulb to light up,the circuit must be complete,meaning current must

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$S_2$ and $S_3$ are closed but $S_1$ is open

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(C) In the given circuit,the bulb is connected in series with switch $S_3$. For the bulb to light up,the circuit must be complete,meaning current must flow through the bulb.$1$. If $S_2$ and $S_3$ are closed and $S_1$ is open,the current flows from the source through $S_2$ and $S_3$ to the bulb,completing the circuit. Thus,the bulb lights up.$2$. If $S_1$ and $S_3$ are closed,the circuit is short-circuited through the path containing $S_1$ and $S_3$,bypassing the bulb or causing a direct path from the source,which may not light the bulb depending on the exact configuration,but $S_2$ and $S_3$ closed is the standard condition for the bulb to be in the main current path.$3$. Therefore,the bulb will light up when $S_2$ and $S_3$ are closed while $S_1$ is open. Option $(c)$ is correct.

Consider the following circuit given below. The bulb will light up,if

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