A coil of self-inductance L is connected in s | vedclass

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(C) The circuit consists of an inductor $L$ and a bulb $B$ in series with an $AC$ source. The impedance of the circuit is $Z = \sqrt{R^2 + X_L^2}$,whe

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an iron rod is inserted into the coil.

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(C) The circuit consists of an inductor $L$ and a bulb $B$ in series with an $AC$ source. The impedance of the circuit is $Z = \sqrt{R^2 + X_L^2}$,where $R$ is the resistance of the bulb and $X_L = \omega L$ is the inductive reactance.When an iron rod is inserted into the coil,the permeability of the core increases,which significantly increases the self-inductance $L$ of the coil.Since $X_L = \omega L$,an increase in $L$ leads to an increase in the inductive reactance $X_L$.As the total impedance $Z = \sqrt{R^2 + X_L^2}$ increases,the total 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 thus the brightness of the bulb decreases.

$A$ coil of self-inductance $L$ is connected in series with a bulb $B$ and an $AC$ source. The brightness of the bulb decreases when:

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