A light bulb and an open coil inductor are connected to an ac source through a key in series. The key is closed and after some time, an iron rod is inserted into the interior of the inductor. The glow of the light bulb will be
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → decreased
In an ac circuit the inductor opposes the current through its inductive reactance \( X_L = \omega L \). Inserting an iron rod raises the permeability of the core, so the self-inductance L increases greatly.
\( X_L = \omega L \)
\( \displaystyle I = \frac{V}{\sqrt{R^{2} + X_L^{2}}} \)
A larger \( X_L \) increases the impedance and reduces the current, so the bulb glows less brightly.
Why the other options are incorrect
Option (2) increased — Incorrect. The current can only fall, because inserting iron increases the reactance.
Option (3) unaffected — Incorrect. The glow would be unaffected with a dc source, where a steady current is not opposed by the inductance; with ac the reactance matters.
Option (4) zero — Incorrect. The current is reduced, not stopped: the impedance becomes larger but stays finite.
NCERT (2026-27):
AC Voltage Applied to a Capacitor – “… Example 7.5 A light bulb and an open coil inductor are connected to an ac source through a key as shown in Fig. 7.9. 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.”
AC Voltage Applied to a Capacitor – “… Solution As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.”