In an ac circuit, a coil of self-inductance L is connected in series with an electric bulb of resistance R. If the frequency of ac supply increases, then
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → Brightness of the bulb decreases
The coil's inductive reactance is \( X_L = \omega L = 2\pi f L \), so it grows with the frequency. The circuit is a series LR circuit:
\( \displaystyle Z = \sqrt{R^{2} + X_L^{2}} \qquad I_{\mathrm{rms}} = \frac{V_{\mathrm{rms}}}{Z} \qquad \tan\phi = \frac{X_L}{R} \)
Option (1) Impedance of the circuit decreases — Incorrect. A larger \( X_L \) makes \( Z \) increase.
Option (2) Effective current in the circuit increases — Incorrect. With \( Z \) larger, \( I_{\mathrm{rms}} = V_{\mathrm{rms}}/Z \) decreases.
Option (3) Brightness of the bulb decreases — Correct. The bulb's brightness depends on the power \( I_{\mathrm{rms}}^{2}R \); the current falls, so the bulb becomes dimmer.
Option (4) Phase difference between the current and the voltage decreases — Incorrect. \( \tan\phi = X_L/R \) increases, so the phase difference increases.