Question

An electric bulb rated as 100 W-220 V is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is :

• A. 0.64 A
• B. 0.45 A
• C. 2.2 A
• D. 0.32 A

Hint:

For a resistive load like an electric bulb connected to an AC source, the power \( P = V_{rms} I_{rms} \). Use the given power and rms voltage to find the rms current. The peak value of current in an AC circuit is related to the rms current by \( I_0 = \sqrt{2} I_{rms} \).

Answer 1

The Correct Option is A

Determine the peak current through a 100 W bulb connected to a 220 V AC source.

Given:

• Power (P) = 100 W
• RMS Voltage (V_rms) = 220 V

Calculate the bulb's resistance (R) using the power formula:

\(P = \frac{(V_{\text{rms}})^2}{R}\)

Rearrange for R:

\(R = \frac{(V_{\text{rms}})^2}{P}\)

Substitute values:

\(R = \frac{(220)^2}{100} = \frac{48400}{100} = 484 \, \Omega\)

Calculate the RMS current (I_rms) using Ohm's Law:

\(I_{\text{rms}} = \frac{V_{\text{rms}}}{R}\)

Substitute values:

\(I_{\text{rms}} = \frac{220}{484} \approx 0.4545 \, \text{A}\)

Calculate the peak current (I_peak):

\(I_{\text{peak}} = I_{\text{rms}} \times \sqrt{2}\)

Substitute I_rms:

\(I_{\text{peak}} = 0.4545 \times \sqrt{2} \approx 0.4545 \times 1.414 \approx 0.643 \, \text{A}\)

The peak current is approximately 0.64 A.

Therefore, the correct answer is 0.64 A.