Q146 / 150 · MHT CET 2026 (15 April Shift 2) · Physics
An electric lamp connected in series with a capacitor and an a.c source is glowing with certain brightness. On reducing the frequency of source the brightness of the lamp
- Ais increased
- Correct answerBis reduced
- Cremain the same
- Dbecomes zero
Solution
The impedance of the series RC circuit is given by $Z = \sqrt{R^2 + X_C^2}$, where $X_C = \dfrac{1}{2\pi f C}$.
When the frequency $f$ of the AC source is reduced, the capacitive reactance $X_C$ increases.
As a result, the total impedance $Z$ of the circuit increases.
The current in the circuit is $I = \dfrac{V}{Z}$. Since $Z$ increases, the current $I$ decreases.
The brightness of the lamp is proportional to the power dissipated, $P = I^2 R$. With a decrease in current, the power dissipated decreases, and hence the brightness of the lamp is reduced.
Answer: is reduced
When the frequency $f$ of the AC source is reduced, the capacitive reactance $X_C$ increases.
As a result, the total impedance $Z$ of the circuit increases.
The current in the circuit is $I = \dfrac{V}{Z}$. Since $Z$ increases, the current $I$ decreases.
The brightness of the lamp is proportional to the power dissipated, $P = I^2 R$. With a decrease in current, the power dissipated decreases, and hence the brightness of the lamp is reduced.
Answer: is reduced
Hence, the correct answer is option: 'B'.