Target Exam

CUET

Subject

Physics

Chapter

Alternating Current

Question:

An a.c. voltage source of fixed amplitude $ε_0$ and variable angular frequency $ω$ is connected with an electric bulb of resistance R (inductance zero) and a capacitor of capacitance C in series. On increasing the angular frequency of the source, the total impedance of the circuit

Options:

increases and the bulb glows brighter

decreases and the bulb glows dimmer

remains the same and the bulb glows brighter

decreases and the bulb glows brighter

Correct Answer:

decreases and the bulb glows brighter

Explanation:

The correct answer is Option (4) → decreases and the bulb glows brighter

Given: Series AC circuit with bulb (resistance $R$) and capacitor ($C$)

Impedance of the circuit: $Z = \sqrt{R^2 + \frac{1}{(\omega C)^2}}$

As angular frequency $\omega$ increases, $\frac{1}{\omega C}$ decreases, so $Z$ decreases.

Current through the circuit: $I = \frac{\varepsilon_0}{Z}$. As $Z$ decreases, $I$ increases.

Power dissipated in bulb: $P = I^2 R$. As $I$ increases, $P$ increases → bulb glows brighter.

Total impedance decreases and the bulb glows brighter