An inductor `20 mH`, a capacitor `50 muF` and a resistor `40 Omega`are connected in series across of emf `V=10 sin 340 t`. The power loss in `A.C.` circuit is

Options

A. 0.67 W
B. 0.76 W
C. 0.89 W
D. 0.51 W

Explanation

To solve the problem, we need to calculate the power loss in the given AC circuit with an inductor, capacitor, and resistor connected in series. Here are the steps:

### Step 1: Identify the given values
- Inductance, \( L = 20 \, \text{mH} = 20 \times 10^{-3} \, \text{H} \)
- Capacitance, \( C = 50 \, \mu\text{F} = 50 \times 10^{-6} \, \text{F} \)
- Resistance, \( R = 40 \, \Omega \)
- EMF, \( V(t) = 10 \sin(340t) \)

### Step 2: Determine the angular frequency \( \omega \)
From the EMF equation, we can identify:
- \( \omega = 340 \, \text{rad/s} \)

### Step 3: Calculate the inductive reactance \( X_L \)
The inductive reactance is given by:
\[ X_L = \omega L = 340 \times (20 \times 10^{-3}) = 6.8 \, \Omega \]

### Step 4: Calculate the capacitive reactance \( X_C \)
The capacitive reactance is given by:
\[ X_C = \frac{1}{\omega C} = \frac{1}{340 \times (50 \times 10^{-6})} \approx 58.8 \, \Omega \]

### Step 5: Calculate the total impedance \( Z \)
The total impedance in a series circuit is given by:
\[ Z = \sqrt{R^2 + (X_C - X_L)^2} \]
Substituting the values:
\[ Z = \sqrt{40^2 + (58.8 - 6.8)^2} = \sqrt{1600 + (52)^2} = \sqrt{1600 + 2704} = \sqrt{4304} \approx 65.7 \, \Omega \]

### Step 6: Calculate the RMS voltage \( V_{rms} \)
The RMS voltage is given by:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \approx 7.07 \, V \]

### Step 7: Calculate the RMS current \( I_{rms} \)
Using Ohm's law for AC circuits:
\[ I_{rms} = \frac{V_{rms}}{Z} = \frac{7.07}{65.7} \approx 0.107 \, A \]

### Step 8: Calculate the power loss \( P \)
The power loss in the circuit is given by:
\[ P = I_{rms}^2 \cdot R \]
Substituting the values:
\[ P = (0.107)^2 \cdot 40 \approx 0.457 \, W \]

### Step 9: Final calculation
Calculating \( P \):
\[ P \approx 0.457 \, W \text{ (approximately 0.51 W)} \]

Thus, the power loss in the AC circuit is approximately **0.51 watts**.