Question

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

• A. 0.79W
• B. 2.74W
• C. 0.43W
• D. 1.12W

Answer 1

The Correct Option is A

To solve this problem, we analyze the given RLC series circuit and calculate the power loss.

Given:

• \( L = 20\,\text{mH} = 20 \times 10^{-3}\,\text{H} \)
• \( C = 100\,\mu\text{F} = 100 \times 10^{-6}\,\text{F} \)
• \( R = 50\,\Omega \)
• \( V(t) = 10 \sin(314t)\,\text{V} \)

Angular frequency: \( \omega = 314\,\text{rad/s} \)

Step 1: Reactances

\[ X_L = \omega L = 314 \times 20 \times 10^{-3} = 6.28\,\Omega \]

\[ X_C = \frac{1}{\omega C} = \frac{1}{314 \times 100 \times 10^{-6}} = 31.85\,\Omega \]

Step 2: Impedance

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

\[ Z = \sqrt{50^2 + (6.28 - 31.85)^2} \]

\[ Z = \sqrt{2500 + (-25.57)^2} = \sqrt{3153.1} \approx 56.15\,\Omega \]

Step 3: RMS Voltage

\[ V_{\text{rms}} = \frac{10}{\sqrt{2}} \approx 7.07\,\text{V} \]

Step 4: Current

\[ I = \frac{V_{\text{rms}}}{Z} = \frac{7.07}{56.15} \approx 0.126\,\text{A} \]

Step 5: Power Loss

\[ P = I^2 R = (0.126)^2 \times 50 \approx 0.79\,\text{W} \]

Final Answer: \( 0.79\,\text{W} \)