A light bulb and an air core inductor are connected in series to an a.c. source through a switch. The switch is closed and after sometime, an iron rod is inserted into the interior of the inductor. The light bulb
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → glows with less brightness
The self inductance of a solenoid rises when it is filled with a material of high permeability, so inserting the iron rod makes L much larger. The inductive reactance \( X_L = 2\pi fL \) then rises, and so does the impedance of the series circuit.
\( Z = \sqrt{R^{2}+X_L^{2}} \)
The source voltage is fixed, so the current \( I = V/Z \) through the bulb falls and the bulb glows with less brightness. The bulb is not switched off, because the resistance R of its filament still allows a current.
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. … 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.”