An R-L-C circuit containing a `52Omega` resistor , a 230 mH inductor, and a 8.8`mu`F capacitor is driven by an AC voltage source that has an amplitude of 150 V and frequency f = 80 Hz . How much average power is dissipated by this circuit?

Options

A. 78.6 W
B. Zero
C. 19.6 W
D. 24.8 W

Explanation

To solve the problem of finding the average power dissipated in an R-L-C circuit, we will follow these steps:

### Step 1: Calculate the Inductive Reactance (X_L)
The inductive reactance \(X_L\) is given by the formula:
\[ X_L = \omega L \]
where \(\omega = 2\pi f\) and \(L\) is the inductance in henries.

Given:
- \(f = 80 \, \text{Hz}\)
- \(L = 230 \, \text{mH} = 230 \times 10^{-3} \, \text{H}\)

Calculating \(\omega\):
\[ \omega = 2\pi \times 80 \approx 502.65 \, \text{rad/s} \]

Now, substituting the values to find \(X_L\):
\[ X_L = 502.65 \times (230 \times 10^{-3}) \approx 115.6 \, \Omega \]

### Step 2: Calculate the Capacitive Reactance (X_C)
The capacitive reactance \(X_C\) is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
where \(C\) is the capacitance in farads.

Given:
- \(C = 8.8 \, \mu\text{F} = 8.8 \times 10^{-6} \, \text{F}\)

Now substituting the values to find \(X_C\):
\[ X_C = \frac{1}{502.65 \times (8.8 \times 10^{-6})} \approx 226 \, \Omega \]

### Step 3: Calculate the Impedance (Z)
The total impedance \(Z\) in the circuit can be calculated using:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Given:
- \(R = 52 \, \Omega\)

Substituting the values:
\[ Z = \sqrt{52^2 + (115.6 - 226)^2} \]
Calculating:
\[ Z = \sqrt{52^2 + (-110.4)^2} = \sqrt{2704 + 12188.16} \approx \sqrt{14892.16} \approx 122 \, \Omega \]

### Step 4: Calculate the RMS Current (I_rms)
The RMS current \(I_{rms}\) can be calculated using:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Given:
- \(V_{rms} = 150 \, V\)

Substituting the values:
\[ I_{rms} = \frac{150}{122} \approx 1.23 \, A \]

### Step 5: Calculate the Power Factor (cos φ)
The power factor \(cos \phi\) is given by:
\[ cos \phi = \frac{R}{Z} \]
Substituting the values:
\[ cos \phi = \frac{52}{122} \approx 0.426 \]

### Step 6: Calculate the Average Power (P_avg)
The average power \(P_{avg}\) is given by:
\[ P_{avg} = V_{rms} \cdot I_{rms} \cdot cos \phi \]
Substituting the values:
\[ P_{avg} = 150 \cdot 1.23 \cdot 0.426 \approx 78.6 \, W \]

### Final Answer
The average power dissipated by the circuit is approximately **78.6 watts**.

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