Question

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

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

Answer 1

The Correct Option is D

Impedance and Power Calculation 

The following formulas and calculations are used to determine the impedance and power in an RLC circuit.

1. Reactance of Capacitor \( X_C \):

The reactance of the capacitor is given by:

\(X_C = \frac{1}{\omega C} = \frac{1}{340 \times 50 \times 10^{-6}} = 58.8 \, \Omega\)

2. Reactance of Inductor \( X_L \):

The reactance of the inductor is given by:

\(X_L = \omega L = 340 \times 20 \times 10^{-3} = 6.8 \, \Omega\)

3. Total Impedance \( Z \):

The total impedance \( Z \) of the RLC circuit is given by:

\(Z = \sqrt{R^2 + (X_C - X_L)^2}\)

Substitute the known values:

\(Z = \sqrt{40^2 + (58.8 - 6.8)^2}\)

\(Z = \sqrt{4304} \, \Omega\)

4. Power \( P \):

To calculate the power, we use the formula:

\(P = i_{\text{rms}}^2 R = \left( \frac{V_{\text{rms}}}{Z} \right)^2 R\)

Substitute the given values:

\(P = \left( \frac{10 / \sqrt{2}}{\sqrt{4304}} \right)^2 \times 40\)

After simplifying:

\(P = \frac{50 \times 40}{4304} \approx 0.51 \, W\)

Conclusion:

The power dissipated in the circuit is approximately \( 0.51 \, W \).