JEE Main 2021 · 18 March · morning · Single correct · +4 / −1
An AC source rated 220 V, 50 Hz is connected to a resistor. The time taken by the current to change from its maximum to the rms value is :
- (a) 2.5 ms
- (b) 25 ms
- (c) 2.5 s
- (d) 0.25 ms
Answer and solution
Answer (a) 2.5 ms
Solution
Physics sir · 3 steps
Start the clock at the peak: the cosine has to drop to 1/√2, and that takes one-eighth of a period.
- Start at the peak. Measure time from the instant the current is maximum, so i = i₀ cos ω t. The rms value is i₀/√2.
cos ω t = 1/√2 ⇒ ω t = π/4 ⇒ t = π/4 × T/2π = T/8
- Find the period. At 50 Hz, T = 1/50 s = 20 ms. The 220 V rating does not matter here, because only the timing of the wave is asked.
- So. t = (20 ms)/8 = 2.5 ms, which is option (a).
The idea to keep. Peak to rms is a phase change of π/4, which is always T/8 whatever the voltage. Find T from the frequency and divide by 8.
Answer from the NTA final answer key. Solution by PrepWiser.