Q176 / 180 · COMEDK 2024 (Afternoon Shift) · Physics
An AC voltage source of variable angular frequency $\omega$ and fixed amplitude $\mathrm{V}_0$ is connected in series with a capacitance $\mathrm{C}$ and an electric bulb of resistance $\mathrm{R}$ (inductance zero). When $\omega$ is decreased
- Atotal impedance of the circuit is unchanged
- Bthe bulb glows brighter
- Cthe bulb switches off
- Correct answerDthe bulb glows dimmer
Solution
The impedance $Z$ of an RC series circuit is given by $Z = \sqrt{R^2 + X_C^2}$, where $X_C = \dfrac{1}{\omega C}$ is the capacitive reactance.
When the angular frequency $\omega$ is decreased, the capacitive reactance $X_C = \dfrac{1}{\omega C}$ increases.
Since $Z = \sqrt{R^2 + \left(\dfrac{1}{\omega C}\right)^2}$, an increase in $X_C$ leads to an increase in the total impedance $Z$ of the circuit.
The current $I$ in the circuit is given by $I = \dfrac{V_0}{Z}$. As $Z$ increases, the current $I$ decreases.
The power dissipated by the bulb is $P = I^2 R$. Since the current $I$ decreases, the power dissipated by the bulb decreases, causing the bulb to glow dimmer.
Answer: the bulb glows dimmer
When the angular frequency $\omega$ is decreased, the capacitive reactance $X_C = \dfrac{1}{\omega C}$ increases.
Since $Z = \sqrt{R^2 + \left(\dfrac{1}{\omega C}\right)^2}$, an increase in $X_C$ leads to an increase in the total impedance $Z$ of the circuit.
The current $I$ in the circuit is given by $I = \dfrac{V_0}{Z}$. As $Z$ increases, the current $I$ decreases.
The power dissipated by the bulb is $P = I^2 R$. Since the current $I$ decreases, the power dissipated by the bulb decreases, causing the bulb to glow dimmer.
Answer: the bulb glows dimmer
Hence, the correct answer is option: 'D'.