NEET2026PhysicsChapterActual
An electric bulb and a capacitor are connected in series across an AC source of constant rms voltage but variable frequency. If the frequency of the AC source is increased, how will the brightness of the bulb and the potential difference across the capacitor change?
Options
- ABrightness decreases, and the potential difference across the capacitor increases.
- BBrightness increases, and the potential difference across the capacitor decreases.
- CBoth the brightness and the potential difference across the capacitor increase.
- DBrightness remains unchanged, and the potential difference across the capacitor decreases.
Correct answer
B. Brightness increases, and the potential difference across the capacitor decreases.
Step-by-step solution
The impedance of the series RC circuit is given by Z = R^2 + X_C^2 , where X_C = 1 2 f C . When the frequency f of the AC source is increased, the capacitive reactance X_C decreases. As a result, the total impedance Z of the circuit decreases, which leads to an increase in the rms current I = V Z . The brightness of the bulb depends on the power dissipated, P = I^2 R . Since the current I increases, the power dissipated increases, causing the brightness of the bulb to increase. The potential difference across the b