A 50 `Omega` resistor a `120 muF ` capacitor and an inductor of inductance 0.2 H are connected in series across an AC source of 10 V , 50 Hz . Calculate the average power and energy dissipated in 500 s.

Answer

To solve the problem step by step, we will calculate the average power and energy dissipated in 500 seconds for the given LCR circuit.

### Given Data:
- Resistance, \( R = 50 \, \Omega \)
- Capacitance, \( C = 120 \, \mu F = 120 \times 10^{-6} \, F \)
- Inductance, \( L = 0.2 \, H \)
- RMS Voltage, \( V_{rms} = 10 \, V \)
- Frequency, \( f = 50 \, Hz \)
- Time, \( t = 500 \, s \)

### Step 1: Calculate the Inductive Reactance (\( X_L \))
The inductive reactance is given by the formula:
\[ X_L = 2 \pi f L \]
Substituting the values:
\[ X_L = 2 \pi (50) (0.2) = 62.83 \, \Omega \]

### Step 2: Calculate the Capacitive Reactance (\( X_C \))
The capacitive reactance is given by the formula:
\[ X_C = \frac{1}{2 \pi f C} \]
Substituting the values:
\[ X_C = \frac{1}{2 \pi (50) (120 \times 10^{-6})} \approx 26.54 \, \Omega \]

### Step 3: Calculate the Total Impedance (\( Z \))
The total impedance in an LCR circuit is given by:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
Calculating \( X_L - X_C \):
\[ X_L - X_C = 62.83 - 26.54 = 36.29 \, \Omega \]
Now substituting into the impedance formula:
\[ Z = \sqrt{50^2 + (36.29)^2} \approx \sqrt{2500 + 1313.78} \approx \sqrt{3813.78} \approx 61.76 \, \Omega \]

### Step 4: Calculate the Current (\( I_{rms} \))
Using Ohm's law for AC circuits:
\[ I_{rms} = \frac{V_{rms}}{Z} \]
Substituting the values:
\[ I_{rms} = \frac{10}{61.76} \approx 0.162 \, A \]

### Step 5: Calculate the Power Factor (\( \cos \phi \))
The power factor is given by:
\[ \cos \phi = \frac{R}{Z} \]
Substituting the values:
\[ \cos \phi = \frac{50}{61.76} \approx 0.81 \]

### Step 6: Calculate the Average Power (\( P \))
The average power is given by:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos \phi \]
Substituting the values:
\[ P = 10 \cdot 0.162 \cdot 0.81 \approx 1.31 \, W \]

### Step 7: Calculate the Energy Dissipated (\( E \))
The energy dissipated over a time period \( t \) is given by:
\[ E = P \cdot t \]
Substituting the values:
\[ E = 1.31 \cdot 500 \approx 655 \, J \]

### Final Results:
- Average Power, \( P \approx 1.31 \, W \)
- Energy Dissipated in 500 seconds, \( E \approx 655 \, J \)

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