In series LCR circuit `R = 18 Omega` and impedence is `33 Omega`. An Vrms voltage `220 V` is applied across the circuit . The true power consumed in AC circuit is

Options

A. 220 W
B. 400 W
C. 600 W
D. 800 W

Answer

To solve the problem, we need to find the true power consumed in an AC circuit with the given parameters. Let's break it down step by step.

### Step 1: Write down the given values
- Resistance, \( R = 18 \, \Omega \)
- Impedance, \( Z = 33 \, \Omega \)
- RMS Voltage, \( V_{rms} = 220 \, V \)

### Step 2: Calculate the power factor \( \cos \phi \)
The power factor \( \cos \phi \) can be calculated using the formula:
\[ \cos \phi = \frac{R}{Z} \]
Substituting the values:
\[ \cos \phi = \frac{18}{33} \]

### Step 3: Calculate the RMS current \( I_{rms} \)
The RMS current can be calculated using the formula:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Substituting the values:
\[ I_{rms} = \frac{220}{33} \]

### Step 4: Calculate the true power \( P \)
The true power consumed in the circuit is given by the formula:
\[ P = V_{rms} \times I_{rms} \times \cos \phi \]
Substituting the values we calculated:
\[ P = 220 \times \left(\frac{220}{33}\right) \times \left(\frac{18}{33}\right) \]

### Step 5: Simplify the expression
1. Calculate \( I_{rms} \):
\[ I_{rms} = \frac{220}{33} = \frac{220 \div 11}{33 \div 11} = \frac{20}{3} \, A \]

2. Calculate \( \cos \phi \):
\[ \cos \phi = \frac{18}{33} = \frac{18 \div 9}{33 \div 9} = \frac{2}{3} \]

3. Substitute these back into the power formula:
\[ P = 220 \times \left(\frac{20}{3}\right) \times \left(\frac{2}{3}\right) \]
\[ P = 220 \times \frac{40}{9} \]
\[ P = \frac{8800}{9} \approx 977.78 \, W \]

### Step 6: Final calculation
To find the exact value:
\[ P = \frac{8800}{9} \approx 977.78 \, W \]

### Conclusion
The true power consumed in the AC circuit is approximately **977.78 W**.

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