An inductor `20 mH`, a capacitor `100muF` and a resistor `50Omega` are connected in series across a source of emf `V=10sin314t`. The power loss in the circuit is

Options

A. `1.13W`
B. `0.81W`
C. `2.74W`
D. `0.43W`

Explanation

To solve the problem of finding the power loss in a series LCR circuit consisting of an inductor (20 mH), a capacitor (100 µF), and a resistor (50 Ω) connected across an AC source with an emf of \( V = 10 \sin(314t) \), we can follow these steps:

### Step 1: Identify the Parameters
- Inductance \( L = 20 \, \text{mH} = 20 \times 10^{-3} \, \text{H} \)
- Capacitance \( C = 100 \, \mu\text{F} = 100 \times 10^{-6} \, \text{F} \)
- Resistance \( R = 50 \, \Omega \)
- Peak Voltage \( V_0 = 10 \, \text{V} \)
- Angular Frequency \( \omega = 314 \, \text{rad/s} \)

### Step 2: Calculate the Inductive Reactance \( X_L \)
The inductive reactance is given by 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 \( X_C \)
The capacitive reactance is given by the formula:
\[ X_C = \frac{1}{\omega C} \]
Substituting the values:
\[ X_C = \frac{1}{314 \times (100 \times 10^{-6})} = 31.84 \, \Omega \]

### Step 4: Calculate the Impedance \( Z \)
The impedance in a series LCR circuit is given by:
\[ Z = \sqrt{(R^2 + (X_L - X_C)^2)} \]
Calculating \( X_L - X_C \):
\[ X_L - X_C = 6.28 - 31.84 = -25.56 \, \Omega \]
Now substituting into the impedance formula:
\[ Z = \sqrt{(50^2 + (-25.56)^2)} = \sqrt{2500 + 653.6336} = \sqrt{3153.6336} \approx 56 \, \Omega \]

### Step 5: Calculate the RMS Voltage
The RMS voltage \( V_{RMS} \) is given by:
\[ V_{RMS} = \frac{V_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} \approx 7.07 \, \text{V} \]

### Step 6: Calculate the RMS Current \( I_{RMS} \)
Using Ohm's law for AC circuits:
\[ I_{RMS} = \frac{V_{RMS}}{Z} = \frac{7.07}{56} \approx 0.126 \, \text{A} \]

### Step 7: Calculate the Power Loss
The power loss in the circuit is given by:
\[ P = I_{RMS}^2 R \]
Substituting the values:
\[ P = (0.126)^2 \times 50 \approx 0.81 \, \text{W} \]

Thus, the power loss in the circuit is approximately **0.81 W**.

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