A capacitor,an inductor,and an electric bulb | vedclass

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(C) The given circuit is a series $LCR$ circuit where the bulb acts as a resistor $R$.The current in the circuit is given by $I = \frac{V}{Z} = \frac{

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increase,reach a maximum,and then decrease in brightness.

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(C) The given circuit is a series $LCR$ circuit where the bulb acts as a resistor $R$.The current in the circuit is given by $I = \frac{V}{Z} = \frac{V}{\sqrt{R^2 + (\omega L - \frac{1}{\omega C})^2}}$.At resonance,the angular frequency is $\omega_0 = \frac{1}{\sqrt{LC}}$. At this frequency,the impedance $Z$ is minimum $(Z = R)$,and the current $I$ is maximum $(I = \frac{V}{R})$.As the frequency increases from a low value,the impedance $Z$ decreases until it reaches the resonance frequency,causing the current $I$ and the brightness of the bulb to increase to a maximum.As the frequency increases further beyond the resonance frequency,the impedance $Z$ increases again,causing the current $I$ and the brightness of the bulb to decrease.Therefore,the brightness increases,reaches a maximum at resonance,and then decreases.

$A$ capacitor,an inductor,and an electric bulb are connected in series to an a.c. supply of variable frequency. As the frequency of the supply is increased gradually,the electric bulb is found to

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