In the circuit shown in the figure,if both th | vedclass

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(C) The bulb $B_1$ is in series with a capacitor $C$,and the bulb $B_2$ is in series with an inductor $L$.The capacitive reactance is given by $X_C =

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As the frequency is increased,the brightness of $B_1$ will increase and that of $B_2$ will decrease.

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(C) The bulb $B_1$ is in series with a capacitor $C$,and the bulb $B_2$ is in series with an inductor $L$.The capacitive reactance is given by $X_C = \frac{1}{2\pi f C}$,which decreases as frequency $f$ increases.The inductive reactance is given by $X_L = 2\pi f L$,which increases as frequency $f$ increases.As the frequency $f$ increases,$X_C$ decreases,allowing more current to flow through the branch containing $B_1$,thus increasing its brightness.Conversely,as $f$ increases,$X_L$ increases,reducing the current through the branch containing $B_2$,thus decreasing its brightness.Therefore,option $(c)$ is correct.

In the circuit shown in the figure,if both the bulbs $B_1$ and $B_2$ are identical,then:

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