Question

A series LCR circuit having \(R = 44\Omega\), \(L = 2\,\text{H}\) and \(C = 25\mu F\) is connected to a variable frequency of \(220\,\text{V}\). 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?

• A. \(2.1\,\text{kW}\)
• B. \(4.2\,\text{kW}\)
• C. \(1.1\,\text{kW}\)
• D. \(1.2\,\text{kW}\)

Hint:

At resonance in LCR circuit, impedance becomes minimum and equals resistance, so use \(P = \frac{V^2}{R}\).

Answer 1

The Correct Option is C

Step 1: Understand resonance condition.
At resonance in a series LCR circuit, the inductive reactance and capacitive reactance cancel each other:
\[ X_L = X_C \]
Thus, total impedance becomes purely resistive:
\[ Z = R \]

Step 2: Use power formula.
Average power in AC circuit is:
\[ P = \frac{V^2}{R} \]
At resonance, power factor = 1.

Step 3: Substitute given values.
\[ V = 220\,\text{V}, \quad R = 44\Omega \]
\[ P = \frac{220^2}{44} \]

Step 4: Simplify.
\[ P = \frac{48400}{44} \]
\[ P = 1100\,\text{W} \]

Step 5: Convert to kW.
\[ P = 1.1\,\text{kW} \]

Step 6: Physical interpretation.
At resonance, circuit draws maximum power because impedance is minimum and purely resistive.

Step 7: Final answer.
\[ \boxed{1.1\,\text{kW}} \]