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

Brightness of the bulb decreases

The correct answer is Option (3) → Brightness of the bulb decreases

$\text{Impedance of the circuit: } Z = \sqrt{R^2 + (\omega L)^2}, \ \ \omega = 2\pi f.$

$\text{As frequency } f \uparrow \ \Rightarrow \ \omega L \uparrow \ \Rightarrow \ Z \uparrow.$

$I = \frac{V}{Z} \ \Rightarrow \ \text{effective current decreases.}$

$\text{Power in bulb } P = I^2 R \ \Rightarrow \ \text{brightness decreases.}$

$\text{Phase difference } \phi = \tan^{-1}\!\left(\frac{\omega L}{R}\right) \ \uparrow \ \Rightarrow \ \text{it increases, not decreases.}$

${\ \text{Brightness of the bulb decreases}\ }$