Question 133: AC fundamentals

Question

An electric bulb rated as $100 \mathrm{~W}-220 \mathrm{~V}$ is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is :

Accepted Answer

To find the peak current through the bulb, follow these steps:
Step 1: Calculate the bulb's rms (root mean square) current:
The formula to find the rms current is:
$$I_{	ext{rms}}=\frac{P}{V}$$
Here, $P = 100 	ext{ W}$ and $V = 220 	ext{ V}$. Substitute these values into the formula:
$$I_{	ext{rms}} = \frac{100	ext{ W}}{220	ext{ V}} \approx 0.455	ext{ A}$$
Step 2: Convert the rms current to peak current:
Use the following formula to find the peak current:
$$I_{	ext{peak}} = I_{	ext{rms}} 	imes \sqrt{2}$$
Substitute the value of $I_{	ext{rms}}$ into the formula:
$$I_{	ext{peak}} = 0.455 	imes 1.414 \approx 0.64	ext{ A}$$
The correct choice is
0.64 A
(Option B).

Answer

0.32 A

Answer

0.64 A

Answer

0.45 A

Answer

2.2 A

AC fundamentals

An electric bulb rated as 100 \mathrm{~W}-220 \mathrm{~V} is connected to an ac source of rms voltage 220 V. The peak value of current through the bulb is :

Choose the correct answer: