JEE Main 2025 · 4 April · morning · Single correct · +4 / −1
An alternating current is represented by the equation, i=100 √2 sin (100 π t) ampere. The RMS value of current and the frequency of the given alternating current are
- (a) 100/√2 A, 100 Hz
- (b) 50 √2 A, 50 Hz
- (c) 100 √2 A, 100 Hz
- (d) 100 A, 50 Hz
Answer and solution
Answer (d) 100 A, 50 Hz
Solution
Physics sir · 3 steps
Read the peak off the front of the sine and ω off the inside; then divide by √2 and by 2π.
- Match the form. Compare with i = i₀ sin(ω t). The number in front is the peak current, i₀ = 100√2 A, and the number multiplying t is ω = 100π rad s⁻¹.
- RMS current. For a sinusoidal current, iᵣₘₛ = i₀/√2 = 100√2/√2 = 100 A.
f = ω/2π = 100π/2π = 50 Hz
- So. The RMS current is 100 A and the frequency is 50 Hz, which is option (d).
The idea to keep. In i = i₀ sin(ω t), the coefficient is the peak value and the number beside t is ω, not f. RMS is peak over √2; frequency is ω over 2π.
Answer from the NTA final answer key. Solution by PrepWiser.