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

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(A) The brightness of the bulb depends on the current $I$ flowing through the circuit,where $I = \frac{V}{Z}$. The impedance $Z$ of the $LR$ circuit i

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an iron rod is inserted in the coil.

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(A) The brightness of the bulb depends on the current $I$ flowing through the circuit,where $I = \frac{V}{Z}$. The impedance $Z$ of the $LR$ circuit is given by $Z = \sqrt{R^2 + X_L^2}$,where $X_L = \omega L = 2\pi f L$. For the brightness to decrease,the current $I$ must decrease,which means the impedance $Z$ must increase.$1$. When an iron rod is inserted into the coil,the self-inductance $L$ increases due to the increase in permeability. Consequently,$X_L$ increases,$Z$ increases,and the current $I$ decreases,leading to a decrease in brightness.$2$. If the frequency $f$ is decreased,$X_L$ decreases,$Z$ decreases,and the current $I$ increases,leading to an increase in brightness.$3$. If the number of turns $N$ is reduced,$L$ decreases (since $L \propto N^2$),$X_L$ decreases,$Z$ decreases,and the current $I$ increases,leading to an increase in brightness.$4$. Adding a capacitor such that the net reactance becomes zero (resonance) would increase the current,not decrease it.Therefore,the correct option is $A$.

$A$ coil of self-inductance $L$ is connected in series with a bulb and an a.c. source. Brightness of the bulb decreases when

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