A series RCL circuit contains a 222 `Omega` resistor, a 1.40 `muF` capacitor, and a 0.125 H inductor. The 444 Hz ac generator in the circuit has an rms voltage of 208 V. What is the average electric power dissipated by the circuit?

Options

A. 135 W
B. 166 W
C. 81 W
D. 191 W

Answer

To solve the problem of finding the average electric power dissipated by the series RCL circuit, we will follow these steps:

### Step 1: Identify the given values
- Resistor (R) = 222 Ω
- Capacitor (C) = 1.40 μF = 1.40 × 10^(-6) F
- Inductor (L) = 0.125 H
- Frequency (f) = 444 Hz
- RMS Voltage (V_rms) = 208 V

### Step 2: Calculate the capacitive reactance (X_C)
The formula for capacitive reactance is:
\[ X_C = \frac{1}{2 \pi f C} \]
Substituting the values:
\[ X_C = \frac{1}{2 \pi (444) (1.40 \times 10^{-6})} \]
Calculating this gives:
\[ X_C \approx 256.17 \, \Omega \]

### Step 3: Calculate the inductive reactance (X_L)
The formula for inductive reactance is:
\[ X_L = 2 \pi f L \]
Substituting the values:
\[ X_L = 2 \pi (444) (0.125) \]
Calculating this gives:
\[ X_L \approx 348.54 \, \Omega \]

### Step 4: Calculate the impedance (Z)
The formula for impedance in a series RCL circuit is:
\[ Z = \sqrt{R^2 + (X_C - X_L)^2} \]
Substituting the values:
\[ Z = \sqrt{(222)^2 + (256.17 - 348.54)^2} \]
Calculating this gives:
\[ Z \approx 950.03 \, \Omega \]

### Step 5: Calculate the average power (P_avg)
The formula for average power in an RCL circuit is:
\[ P_{avg} = \frac{V_{rms}^2}{Z^2} \cdot R \]
Substituting the values:
\[ P_{avg} = \frac{(208)^2}{(950.03)^2} \cdot 222 \]
Calculating this gives:
\[ P_{avg} \approx 166 \, \text{W} \]

### Conclusion
The average electric power dissipated by the circuit is approximately **166 Watts**.

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