In given circuit, when $S_1$ and $S_3$ both are closed, then source is short circuited. Closing either $S_1$ or $S_3$ does not completes any of path. Hence, they produce no change in circuit.
On closing $S_2$ alone the circuit containing bulb is complete. Hence, bulb will light up when $S_2$ is closed alongwith $S_3$ but $S_1$ is open. Option $(c)$ is correct.
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$e=100$ $sin$ $20t$
$i=20sin$ $\left( {30t - \frac{\pi }{4}} \right)$
In one cycle of $A.C.$, the average power consumed by the circuit and the wattless current are, respectively
