A coil of self inductance L is connected in series with a bulb B and an a.c. source. Brightness of the bulb decreases when (a) frequency of a.c. source is decreased (b) number of turns in the coil is reduced (c) an iron rod is inserted in the coil (d) a capacitance of reactance Xc is included in the same circuit

Answer: (c). Current I = (V)/(√(R² + XL²)) with XL = ω L; larger XL means dimmer bulb. Lower frequency or fewer turns reduces XL: bulb gets brighter. An iron rod increases L (higher permeability), so XL rises and current falls. So the answer is (c) an iron rod is inserted in the coil.

A coil of self inductance L is connected in series with a bulb B and… | MHT CET 2026 (14 May, Shift 1)

Q4MHT CET 2026 · 14 MAY · SHIFT 1Easy

RL and RC Series Circuits

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

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Answer: (c)

Current $I = \frac{V}{\sqrt{R^2 + X_L^2}}$ with $X_L = \omega L$ ; larger $X_L$ means dimmer bulb.
Lower frequency or fewer turns reduces $X_L$ : bulb gets brighter.
An iron rod increases $L$ (higher permeability), so $X_L$ rises and current falls.
So the answer is (c) an iron rod is inserted in the coil.

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