Concepts Of Physics MCQ Edition [Volume 2]PhysicsAlternating Current
A bulb with a rating of 60 W at 220 V is connected to a household alternating voltage supply of 220 V . Determine the maximum instantaneous current passing through the filament.
Options
- A0.39 A
- B0.27 A
- C0.55 A
- D0.19 A
Correct answer
A. 0.39 A
Step-by-step solution
The power rating of the bulb is P = 60 W and the rms voltage is V_ rms = 220 V . The rms current flowing through the bulb is given by: I_ rms = P V_ rms = 60 220 = 3 11 A The maximum instantaneous current (peak current) for an alternating current is: I₀ = 2 I_ rms I₀ = 2 3 11 1.414 0.2727 0.385 A Rounding off to two decimal places, we get 0.39 A .