A series LCR circuit with `R=22Omega, L=1.5 H` and `C=40 mu F` is connected to a variable frequency 220 V ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power transferred to the circuit in one complete cycle?

Options

A. 2000 W
B. 2200 W
C. 2400 W
D. 2500 W

Explanation

To solve the problem, we need to determine the average power transferred to a series LCR circuit when the frequency of the supply matches the natural frequency of the circuit. Here are the steps:

### Step 1: Identify the given values
- Resistance, \( R = 22 \, \Omega \)
- Inductance, \( L = 1.5 \, H \)
- Capacitance, \( C = 40 \, \mu F = 40 \times 10^{-6} \, F \)
- Voltage, \( V_{rms} = 220 \, V \)

### Step 2: Understand the condition of resonance
When the frequency of the supply equals the natural frequency of the circuit, the circuit is said to be in resonance. At resonance, the impedance \( Z \) of the circuit is equal to the resistance \( R \). Therefore, we have:
\[ Z = R \]

### Step 3: Calculate the average power
The average power \( P \) in an LCR circuit at resonance can be calculated using the formula:
\[ P = \frac{V_{rms}^2}{R} \]

### Step 4: Substitute the values into the power formula
Now, substituting the known values into the power formula:
\[ P = \frac{(220 \, V)^2}{22 \, \Omega} \]

### Step 5: Perform the calculations
Calculating \( (220)^2 \):
\[ (220)^2 = 48400 \]
Now, divide by \( R \):
\[ P = \frac{48400}{22} \]
\[ P = 2200 \, W \]

### Conclusion
The average power transferred to the circuit in one complete cycle is:
\[ P = 2200 \, W \]