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

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Frequency of the AC source is decreased

Number of turns in the coil is reduced

An iron rod is inserted in the coil

A capacitance of reactiance X_C = X_L is included in the same circuit

answer is D.

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Detailed Solution

$L = \frac{μ_{0} μ_{r} N^{2} A}{l}$

$\Rightarrow L \propto μ_{r}$

$i = \frac{E}{{(ω^{2} L^{2} + R^{2})}^{1 / 2}}$

When an iron rod is inserted in the coil, self-inductance increases which decreases the current .

Hence, the brightness of the bulb decreases.

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