Question

A light bulb rated 100 W is connected to an AC source of 220 V, 50 Hz. The rms current through the bulb is

• A. 0.454 A
• B. 0.545 A
• C. 2.20 A
• D. 0.22 A

Hint:

For AC circuits containing only resistance, the RMS values of current and voltage behave exactly like DC values. You can use Ohm's law (\( V_{\text{rms}} = I_{\text{rms}}R \)) and standard power formulas (\( P = V_{\text{rms}}I_{\text{rms}} = I_{\text{rms}}^2 R \)) directly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the root mean square (rms) current flowing through a light bulb, given its power rating and the rms voltage of the AC supply.
Step 2: Key Formula or Approach:
For a purely resistive load like an incandescent light bulb, the average power \( P \) in an AC circuit is given by the product of the rms voltage (\( V_{\text{rms}} \)) and the rms current (\( I_{\text{rms}} \)):
[ P = V_{\text{rms}} \times I_{\text{rms}} ]
Rearranging to find the rms current:
[ I_{\text{rms}} = \frac{P}{V_{\text{rms}}} ]
Step 3: Detailed Explanation:
Given parameters:
Power rating, \( P = 100 \text{ W} \)
RMS Voltage, \( V_{\text{rms}} = 220 \text{ V} \)
Frequency, \( f = 50 \text{ Hz} \) (Note: frequency does not affect current in a pure resistor).
Calculating the current:
[ I_{\text{rms}} = \frac{100}{220} \text{ A} ]
[ I_{\text{rms}} = \frac{10}{22} \text{ A} = \frac{5}{11} \text{ A} ]
[ I_{\text{rms}} \approx 0.4545... \text{ A} ]
Rounding to three decimal places, we get approximately 0.454 A.
Step 4: Final Answer:
The rms current through the bulb is 0.454 A, which corresponds to option (1).