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 :
Show Hint
- 0.64 A
- 0.45 A
- 2.2 A
- 0.32 A
The Correct Option is A
Approach Solution - 1
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 (Vrms) = 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 (Vrms) 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 (Irms) 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 (Ipeak) is related to the RMS current by the formula:
\(I_{\text{peak}} = I_{\text{rms}} \times \sqrt{2}\)
Substitute the value of Irms:
\(I_{\text{peak}} = 0.4545 \times \sqrt{2} \approx 0.4545 \times 1.414 \approx 0.643 A\)
Thus, the peak value of current through the bulb is approximately 0.64 A.
Therefore, the correct answer is 0.64 A.
Approach Solution -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.
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