A resistance of 40 `Omega` is connected in series with inductor of self-inductance 5 H and a capacitor of capacitance 80 `mu`F. This combination is connected to an AC source of rms voltage 220 V. frequency of AC source can changed continuously.
What is the average power consumed by circuit ?

Options

A. 605 W
B. 1210 `sqrt(2)` W
C. 1210 W
D. 1210/`sqrt(2)` W

Explanation

To solve the problem, we need to find the average power consumed by the LCR circuit when connected to an AC source. The circuit consists of a resistor (R), an inductor (L), and a capacitor (C) connected in series.

### Step-by-step Solution:

1. **Identify Given Values:**
- Resistance, \( R = 40 \, \Omega \)
- Inductance, \( L = 5 \, H \)
- Capacitance, \( C = 80 \, \mu F = 80 \times 10^{-6} \, F \)
- RMS Voltage, \( V_{rms} = 220 \, V \)

2. **Calculate the Angular Frequency (\( \omega \)):**
- The angular frequency \( \omega \) can vary, but at resonance, \( \omega \) is such that the inductive reactance \( X_L \) equals the capacitive reactance \( X_C \).
- The formulas for reactance are:
\[ X_L = \omega L \]
\[ X_C = \frac{1}{\omega C} \]
- At resonance, \( X_L = X_C \), so:
\[ \omega L = \frac{1}{\omega C} \]
- Rearranging gives:
\[ \omega^2 = \frac{1}{LC} \]
- Therefore, \( \omega = \frac{1}{\sqrt{LC}} \).

3. **Calculate \( X_L \) and \( X_C \) at Resonance:**
- Since \( X_L = X_C \) at resonance, we can calculate either:
\[ X_L = \omega L = \frac{1}{\sqrt{LC}} \cdot L \]
- Substituting \( L = 5 \, H \) and \( C = 80 \times 10^{-6} \, F \):
\[ X_L = \frac{1}{\sqrt{5 \times 80 \times 10^{-6}}} \cdot 5 \]

4. **Calculate the Impedance \( Z \) at Resonance:**
- At resonance, the impedance \( Z \) is equal to the resistance \( R \):
\[ Z = R = 40 \, \Omega \]

5. **Calculate the Power Factor \( \cos \phi \):**
- At resonance, the phase angle \( \phi = 0 \), hence:
\[ \cos \phi = 1 \]

6. **Calculate the Average Power \( P \):**
- The formula for average power in an AC circuit is:
\[ P = \frac{V_{rms}^2}{Z} \cdot \cos \phi \]
- Substituting the values:
\[ P = \frac{(220)^2}{40} \cdot 1 \]
- Calculate \( P \):
\[ P = \frac{48400}{40} = 1210 \, W \]

### Final Answer:
The average power consumed by the circuit is \( 1210 \, W \).

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