A solenoid with an iron core and a bulb are connected to an a.c. source. The bulb is glowing with certain brightness. When the iron core is removed from the solenoid, the brightness of the bulb
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → increases
The solenoid in an a.c. circuit offers inductive reactance \( X_L = \omega L \), and its inductance depends on the core: \( L = \mu_r \mu_0 n^{2} A l \). An iron core has a large relative permeability \( \mu_r \).
Removing the iron core makes \( \mu_r \) fall to about 1, so \( L \) and \( X_L \) decrease. The impedance of the bulb-solenoid circuit therefore falls:
\( \displaystyle I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{\sqrt{R^{2} + X_L^{2}}} \)
With the same source voltage, a smaller \( X_L \) gives a larger current, so the bulb glows brighter.
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. … 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.”