An electric bulb,an open coil inductor,an AC | vedclass

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(C) When an iron rod is inserted into the inductor,the self-inductance $L$ of the coil increases because the permeability of the core increases. The i

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The glow of the bulb decreases

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(C) When an iron rod is inserted into the inductor,the self-inductance $L$ of the coil increases because the permeability of the core increases. The inductive reactance of the circuit is given by $X_L = \omega L$. As $L$ increases,$X_L$ increases. The total impedance of the series circuit is $Z = \sqrt{R^2 + X_L^2}$,where $R$ is the resistance of the bulb. Since $X_L$ increases,the total impedance $Z$ of the circuit increases. The current in the circuit is given by $I = V/Z$. As $Z$ increases,the current $I$ flowing through the bulb decreases. Since the power dissipated in the bulb is $P = I^2 R$,a decrease in current leads to a decrease in the power,and thus the glow of the bulb decreases.

An electric bulb,an open coil inductor,an $AC$ source,and a key are all connected in series to form a closed circuit. The key is closed and after some time,an iron rod is inserted into the interior of the inductor. Then:

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