Q144 / 180 · COMEDK 2025 (10 May Shift 3) · Physics
A series LCR circuit having $\mathrm{R}=44 \Omega, \mathrm{~L}=2 \mathrm{H}$ and $\mathrm{C}=25 \mu \mathrm{~F}$ is connected to a variable frequency of 220 V . What is the average power transferred to the circuit in one complete cycle if the frequency of supply equals the natural frequency of the circuit?
- Correct answerA1.1 kW
- B4.2 kW
- C1.2 kW
- D2.1 kW
Solution
The natural frequency of a series LCR circuit is given by the resonant frequency $f_0 = \dfrac{1}{2\pi \sqrt{LC}}$.
At resonance, the inductive reactance $X_L$ equals the capacitive reactance $X_C$, which implies the net impedance of the circuit is equal to the resistance $R$.
Given $R = 44 \Omega$, $L = 2 \text{ H}$, and $C = 25 \times 10^{-6} \text{ F}$.
The voltage $V = 220 \text{ V}$ is the RMS voltage. At resonance, the current in the circuit is $I = \dfrac{V}{Z} = \dfrac{V}{R}$.
$I = \dfrac{220}{44} = 5 \text{ A}$.
The average power $P$ transferred to the circuit in an LCR circuit is given by $P = V_{rms} I_{rms} \cos \phi$.
At resonance, the phase angle $\phi = 0$, so $\cos \phi = 1$.
$P = V_{rms} I_{rms} = 220 \text{ V} \times 5 \text{ A} = 1100 \text{ W}$.
Converting to kilowatts, $P = 1.1 \text{ kW}$.
Answer: 1.1 kW
At resonance, the inductive reactance $X_L$ equals the capacitive reactance $X_C$, which implies the net impedance of the circuit is equal to the resistance $R$.
Given $R = 44 \Omega$, $L = 2 \text{ H}$, and $C = 25 \times 10^{-6} \text{ F}$.
The voltage $V = 220 \text{ V}$ is the RMS voltage. At resonance, the current in the circuit is $I = \dfrac{V}{Z} = \dfrac{V}{R}$.
$I = \dfrac{220}{44} = 5 \text{ A}$.
The average power $P$ transferred to the circuit in an LCR circuit is given by $P = V_{rms} I_{rms} \cos \phi$.
At resonance, the phase angle $\phi = 0$, so $\cos \phi = 1$.
$P = V_{rms} I_{rms} = 220 \text{ V} \times 5 \text{ A} = 1100 \text{ W}$.
Converting to kilowatts, $P = 1.1 \text{ kW}$.
Answer: 1.1 kW
Hence, the correct answer is option: 'A'.