A bulb and an iron core inductor are connected to an AC source through key as shown in figure:
The bulb glows with certain brightness. Now iron rod is taken out of the inductor. Then the brightness of bulb.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → increases
The bulb and the inductor are in series, so the current through the bulb is set by the impedance of the circuit. The iron core has a large relative permeability \( \mu_r \), which makes the self-inductance large.
\( \displaystyle L = \mu_r \mu_0 n^{2} A l \)
Taking out the iron rod leaves only air inside the coil (\( \mu_r \approx 1 \)), so \( L \) falls a lot. Then
\( \displaystyle X_L = \omega L \;\text{decreases}, \qquad Z = \sqrt{R^{2} + X_L^{2}} \;\text{decreases} \)
\( \displaystyle I = \frac{V}{Z} \;\text{increases} \)
A larger current gives more power \( I^{2}R \) in the bulb, so its brightness increases.