Question

If an electric bulb of resistance \(400 \Omega\) is connected to an ac source of peak voltage 282.8 V, then the electric power of the bulb is

• A. \(200 \text{ W}\)
• B. \(150 \text{ W}\)
• C. \(300 \text{ W}\)
• D. \(100 \text{ W}\)
• E. \(250 \text{ W}\)

Hint:

\(V_{\text{rms}} = \frac{V_0}{\sqrt{2}}\) for sinusoidal AC.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
For AC, power \(P = \frac{V_{\text{rms}}^2}{R}\). \(V_{\text{rms}} = \frac{V_0}{\sqrt{2}}\).

Step 2: Detailed Explanation:
\(V_{\text{rms}} = \frac{282.8}{\sqrt{2}} \approx \frac{282.8}{1.414} = 200 \text{ V}\)
\(P = \frac{(200)^2}{400} = \frac{40000}{400} = 100 \text{ W}\)

Step 3: Final Answer:
Power = \(100 \text{ W}\).