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A light bulb and an open coil inductor are connected to an ac source through a key as shown.The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb \((a)\) increases; \((b)\)decreases; \((c)\) is unchanged, as the iron rod is inserted. Give your answer with reasons.
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A light bulb and an open coil inductor are connected to an ac source through a key as shown.The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The glow of the light bulb \((a)\) increases; \((b)\)decreases; \((c)\) is unchanged, as the iron rod is inserted. Give your answer with reasons.

Answer

When an iron rod is inserted inside the inductor which is connected to a light bulb than the magnetic field increases inside the inductor and therefore inductance increases. We know that the inductive reactance is directly proportional to the inductance so it increases. So the voltage drop increases across the inductor as
\(V=\)current\(×\) inductive reactance. Voltage drop across the bulb decreases and it will have lesser brightness.
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