Question

If an electric bulb is rated at 50 W for a 220 V ac supply, then the resistance of the bulb and the peak voltage of the ac source are respectively:

• A. 968 $\Omega$, 220$\sqrt{2}$ V
• B. 484 $\Omega$, 220 V
• C. 968 $\Omega$, 220 V
• D. 484 $\Omega$, 220$\sqrt{2}$ V

Hint:

For AC circuits, the given voltage is typically the RMS value. Use $P = \frac{V_{\text{rms}}^2}{R}$ to find resistance, and $V_{\text{peak}} = V_{\text{rms}} \times \sqrt{2}$ to find the peak voltage. The resistance of a bulb is assumed constant for such calculations.

Answer 1

The Correct Option is A

Step 1: Determine the Resistance of the Bulb
The bulb is rated at 50 W for a 220 V AC supply. For AC circuits, the voltage given (220 V) is the RMS (root mean square) voltage. The power consumed by the bulb is given by: \[ P = \frac{V_{\text{rms}}^2}{R} \] Where $P = 50 \, \text{W}$, $V_{\text{rms}} = 220 \, \text{V}$, and $R$ is the resistance of the bulb. Rearrange to solve for $R$: \[ R = \frac{V_{\text{rms}}^2}{P} \] \[ V_{\text{rms}}^2 = (220)^2 = 48400 \] \[ R = \frac{48400}{50} = 968 \, \Omega \] So, the resistance of the bulb is 968 $\Omega$. Step 2: Calculate the Peak Voltage of the AC Source
For an AC source, the peak voltage $V_{\text{peak}}$ is related to the RMS voltage by: \[ V_{\text{peak}} = V_{\text{rms}} \times \sqrt{2} \] \[ V_{\text{peak}} = 220 \times \sqrt{2} = 220\sqrt{2} \, \text{V} \] The value of $\sqrt{2} \approx 1.414$, so: \[ 220 \times 1.414 \approx 311 \, \text{V} \] However, the standard form in such problems is to leave it as $220\sqrt{2}$ V, which matches the options. Step 3: Verify the Calculations
To verify the resistance, compute the power using the calculated resistance: \[ P = \frac{(220)^2}{968} = \frac{48400}{968} = 50 \, \text{W} \] This matches the given power rating, confirming the resistance is correct.
The peak voltage formula is standard for AC circuits, and $220\sqrt{2}$ V is the expected form for a 220 V RMS supply. Step 4: Analyze Options
Option (1): 968 $\Omega$, 220$\sqrt{2$ V.} Correct, as it matches our calculated resistance and peak voltage.
Option (2): 484 $\Omega$, 220 V. Incorrect, as the resistance is 968 $\Omega$, and 220 V is the RMS voltage, not the peak.
Option (3): 968 $\Omega$, 220 V. Incorrect, as the peak voltage should be $220\sqrt{2}$ V, not 220 V.
Option (4): 484 $\Omega$, 220$\sqrt{2$ V.} Incorrect, as the resistance is 968 $\Omega$, though the peak voltage is correct.