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

We need to find the peak value of current through a bulb rated at 100 W when connected to an AC source of 220 V.

The given details are as follows:

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

First, we need to find the resistance (R) of the bulb. For an electric bulb, the power (P) is related to the voltage (V_rms) and resistance (R) by the formula:

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

Rearranging this formula to find R, we have:

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

Substitute the values:

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

Next, we find the RMS current (I_rms) using Ohm's Law:

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

Substitute the values:

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

The peak current (I_peak) is related to the RMS current by the formula:

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

Substitute the value of I_rms:

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

Thus, the peak value of current through the bulb is approximately 0.64 A.

Therefore, the correct answer is 0.64 A.

Answer 2

The power rating of the electric bulb is \( P = 100 \, \text{W} \) at an rms voltage \( V_{rms} = 220 \, \text{V} \). 
When the bulb is connected to an ac source of rms voltage 220 V, it will operate at its rated power. 
The relationship between power, rms voltage, and rms current \( I_{rms} \) is: \[ P = V_{rms} I_{rms} \] We can find the rms current through the bulb: \[ I_{rms} = \frac{P}{V_{rms}} = \frac{100 \, \text{W}}{220 \, \text{V}} = \frac{10}{22} \, \text{A} = \frac{5}{11} \, \text{A} \] The peak value of the current \( I_0 \) in an ac circuit is related to the rms current by: \[ I_0 = \sqrt{2} I_{rms} \] Substituting the value of \( I_{rms} \): \[ I_0 = \sqrt{2} \times \frac{5}{11} \, \text{A} \] We know that \( \sqrt{2} \approx 1.414 \). \[ I_0 \approx 1.414 \times \frac{5}{11} = \frac{7.07}{11} \approx 0.6427 \, \text{A} \] Rounding to two decimal places, the peak value of the current through the bulb is approximately 0.64 A.