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(A) The intensity of the bulb remains the same.In a $DC$ circuit,the inductor acts as a simple wire with some resistance (if not ideal) because the cu

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The intensity of the bulb remains the same

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(A) The intensity of the bulb remains the same.In a $DC$ circuit,the inductor acts as a simple wire with some resistance (if not ideal) because the current is steady $(dI/dt = 0)$.The inductive reactance $X_L = 2\pi fL$ is zero for $DC$ because the frequency $f = 0$.Therefore,the impedance of the circuit is determined only by the resistance of the bulb and the coil.Inserting a soft iron core increases the self-inductance $L$ of the coil,but since the current is steady $DC$,the inductance does not affect the steady-state current.The steady-state current is given by $I = \frac{V}{R}$,which remains unchanged. Thus,the intensity of the bulb remains the same.

$A$ coil and a bulb are connected in series with a $12 \,V$ direct current $(DC)$ source. $A$ soft iron core is now inserted into the coil. Then,

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