A series `LCR` circuit with `R = 20 Omega, L = 1.5 H` and `C = 35 mu F` is connected to a variable frequency `200 V` ac supply. When the frequency of the supply equals the natural frequency of the circuit, what is the average power in `Kw` transferred to the circuit in one complete cycle?

Explanation

To solve the problem step-by-step, we will follow the principles of alternating current (AC) circuits, particularly focusing on a series LCR circuit at resonance.

### Step 1: Identify the given values
We have the following values:
- Resistance, \( R = 20 \, \Omega \)
- Inductance, \( L = 1.5 \, H \)
- Capacitance, \( C = 35 \, \mu F = 35 \times 10^{-6} \, F \)
- Voltage of the AC supply, \( V_{rms} = 200 \, V \)

### Step 2: Determine the resonant frequency
The resonant frequency \( f_0 \) of a series LCR circuit is given by the formula:
\[ f_0 = \frac{1}{2\pi\sqrt{LC}} \]
Substituting the values of \( L \) and \( C \):
\[ f_0 = \frac{1}{2\pi\sqrt{1.5 \times 35 \times 10^{-6}}} \]

### Step 3: Calculate the impedance at resonance
At resonance, the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \), and the impedance \( Z \) is equal to the resistance \( R \):
\[ Z = R = 20 \, \Omega \]

### Step 4: Calculate the current \( I_{rms} \)
The current \( I_{rms} \) can be calculated using Ohm's law:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Substituting the known values:
\[ I_{rms} = \frac{200 \, V}{20 \, \Omega} = 10 \, A \]

### Step 5: Calculate the average power \( P_{average} \)
The average power in an AC circuit is given by:
\[ P_{average} = V_{rms} \cdot I_{rms} \cdot \cos \phi \]
At resonance, \( \cos \phi = 1 \):
\[ P_{average} = V_{rms} \cdot I_{rms} \]
Substituting the values:
\[ P_{average} = 200 \, V \cdot 10 \, A = 2000 \, W \]

### Step 6: Convert power to kilowatts
To convert watts to kilowatts:
\[ P_{average} = \frac{2000 \, W}{1000} = 2 \, kW \]

### Final Answer
The average power transferred to the circuit in one complete cycle is \( 2 \, kW \).

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