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(B) The circuit consists of an $AC$ source,a resistor (bulb),and an inductor $L$ in series.When an iron rod is inserted into the inductor,its self-ind

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gets dimmer

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(B) The circuit consists of an $AC$ source,a resistor (bulb),and an inductor $L$ in series.When an iron rod is inserted into the inductor,its self-inductance $L$ increases because the permeability of the core increases.The inductive reactance is given by $X_L = \omega L$.As $L$ increases,$X_L$ increases.The total impedance of the circuit is $Z = \sqrt{R^2 + X_L^2}$. As $X_L$ increases,the total impedance $Z$ of the circuit increases.Since the source voltage $V$ is constant,the current in the circuit $I = \frac{V}{Z}$ decreases.The brightness of the bulb depends on the power dissipated,$P = I^2 R$.Since the current $I$ decreases,the power dissipated by the bulb decreases,and therefore,the bulb gets dimmer.

Switch $S$ is closed at $t = 0$. After a sufficiently long time,an iron rod is inserted into the inductor $L$. Then,the light bulb:

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