A light bulb and an hollow coil inductor are connected to an ac source through a key as shown in figure. The switch is closed and after sometime an iron rod is inserted into the interior of the inductor. What will be the effect on the glow of light bulbs?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → decreases
The bulb and the inductor are in series across an ac source. The glow of the bulb depends on the current, and the current is \( I = \frac{V}{Z} \) with the impedance \( Z = \sqrt{R^{2}+X_L^{2}} \), where \( X_L = \omega L \).
Inserting the iron rod into the coil greatly increases the magnetic flux for the same current, because the permeability of iron is large. The self-inductance L increases, so the inductive reactance \( X_L = \omega L \) increases. Then Z increases, the current falls, and the bulb glows dimmer.
NCERT (2026-27):
AC Voltage Applied to a Capacitor – …
