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

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.