Physics Q37 – JEE-MAIN 2025

Home Physics Alternating CurrentQuestion #14

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 :

Clue 1 of 3: Understand Bulb Ratings

The power and voltage rating of a bulb are typically RMS values and can be used to determine its resistance.

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Ninja Strategy•RMS vs. Peak Current Relationship

First, calculate the RMS current, then multiply by $\sqrt{2}$ to find the peak current. Eliminate options that are clearly RMS current or other derived values.

🧠 Full Solution Path

Step 1: Calculate the resistance of the bulb✦ Active

The resistance of the bulb can be determined from its rated power and voltage.

R = \frac{V_{r a t e d}^{2}}{P_{r a t e d}} = \frac{(220 V)^{2}}{100 W} = \frac{48400}{100} = 484 Ω

Step 2: Calculate the RMS current○ Expand

Since the bulb is connected to an AC source of RMS voltage $220 V$ , the RMS current flowing through it is:

I_{r m s} = \frac{V_{r m s}}{R} = \frac{220 V}{484 Ω} = \frac{1}{2.2} A = \frac{5}{11} A \approx 0.4545 A

Step 3: Calculate the peak current○ Expand

The peak value of current ( $I_{p e a k}$ ) is related to the RMS value ( $I_{r m s}$ ) by the formula:

I_{p e a k} = I_{r m s} \times \sqrt{2} = \frac{5}{11} \times \sqrt{2} A \approx 0.4545 \times 1.414 A \approx 0.6427 A

Rounding to two decimal places, the peak current is approximately $0.64 A$ .

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Physics Question 37 – JEE-MAIN 2025

Clue 1 of 3: Understand Bulb Ratings

🧠 Full Solution Path