Q144 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
A light bulb connected in series with a capacitor and an a.c.source is glowing with certain brightness. On reducing the value of capacitance and frequency respectively, the brightness of the bulb
- Correct answerAis reduced, is reduced
- Bis more, is more
- Cis more, is reduced
- Dis reduced, is more
Solution
The brightness of the bulb depends on the current flowing through it, which is given by $I = \dfrac{V}{Z}$.
The impedance of the series RC circuit is $Z = \sqrt{R^2 + X_C^2}$, where $X_C = \dfrac{1}{2\pi f C}$.
When the capacitance $C$ is reduced, the capacitive reactance $X_C$ increases. This increases the total impedance $Z$ of the circuit, which in turn reduces the current $I$. Consequently, the brightness of the bulb is reduced.
Similarly, when the frequency $f$ of the AC source is reduced, the capacitive reactance $X_C$ increases. This again increases the impedance $Z$ and reduces the current $I$. Thus, the brightness of the bulb is reduced.
Therefore, in both cases, the brightness of the bulb is reduced.
Answer: is reduced, is reduced
The impedance of the series RC circuit is $Z = \sqrt{R^2 + X_C^2}$, where $X_C = \dfrac{1}{2\pi f C}$.
When the capacitance $C$ is reduced, the capacitive reactance $X_C$ increases. This increases the total impedance $Z$ of the circuit, which in turn reduces the current $I$. Consequently, the brightness of the bulb is reduced.
Similarly, when the frequency $f$ of the AC source is reduced, the capacitive reactance $X_C$ increases. This again increases the impedance $Z$ and reduces the current $I$. Thus, the brightness of the bulb is reduced.
Therefore, in both cases, the brightness of the bulb is reduced.
Answer: is reduced, is reduced
Hence, the correct answer is option: 'A'.