An inductor `20 mH`, a capacitor `100 muF` and a resistor `50 Omega` are connected in series across a source of emf, `V=10 sin 314 t`. The power loss in the circuit is

Options

A. `2.74 W`
B. `1.13 W`
C. `0.43 W`
D. `0.79 W`

Explanation

To solve the problem, we need to find the power loss in a series RLC circuit connected to an AC voltage source. The components given are:

- Inductor (L) = 20 mH = 20 × 10^-3 H
- Capacitor (C) = 100 µF = 100 × 10^-6 F
- Resistor (R) = 50 Ω
- Voltage source: V(t) = 10 sin(314t)

### Step 1: Calculate the angular frequency (ω)
The angular frequency (ω) is given in the voltage source:
\[ \omega = 314 \, \text{rad/s} \]

### Step 2: Calculate the inductive reactance (XL)
The inductive reactance (XL) can be calculated using the formula:
\[ X_L = \omega L \]
Substituting the values:
\[ X_L = 314 \times (20 \times 10^{-3}) = 6.28 \, \Omega \]

### Step 3: Calculate the capacitive reactance (XC)
The capacitive reactance (XC) can be calculated using the formula:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{314 \times (100 \times 10^{-6})} = \frac{1}{0.0314} \approx 31.85 \, \Omega \]

### Step 4: Calculate the impedance (Z)
The total impedance (Z) in the circuit can be calculated using the formula:
\[ Z = \sqrt{R^2 + (X_L - X_C)^2} \]
First, calculate \(X_L - X_C\):
\[ X_L - X_C = 6.28 - 31.85 = -25.57 \]
Now, substitute the values into the impedance formula:
\[ Z = \sqrt{50^2 + (-25.57)^2} = \sqrt{2500 + 653.0649} = \sqrt{3153.0649} \approx 56.14 \, \Omega \]

### Step 5: Calculate the power factor (cos φ)
The power factor (cos φ) can be calculated using:
\[ \cos \phi = \frac{R}{Z} \]
Substituting the values:
\[ \cos \phi = \frac{50}{56.14} \approx 0.890 \]

### Step 6: Calculate the maximum current (I0)
The maximum current (I0) can be calculated using:
\[ I_0 = \frac{V_0}{Z} \]
Where \(V_0\) is the maximum voltage, which is 10 V:
\[ I_0 = \frac{10}{56.14} \approx 0.178 \, A \]

### Step 7: Calculate the average power (P)
The average power (P) in the circuit can be calculated using:
\[ P = \frac{V_0 I_0}{2} \cos \phi \]
Substituting the values:
\[ P = \frac{10 \times 0.178}{2} \times 0.890 \approx 0.79 \, W \]

### Final Answer
The power loss in the circuit is approximately:
\[ \boxed{0.79 \, W} \]