JEE MainPhysicsAlternating Current
An alternating voltage source starting from zero reaches its maximum amplitude in 5 ms . The frequency of the AC source is:
Options
- A100 Hz
- B50 Hz
- C200 Hz
- D25 Hz
Correct answer
B. 50 Hz
Step-by-step solution
The time taken by an alternating quantity to reach its peak value from zero is exactly one quarter of its time period T . Given that the time to reach the peak value is t = 5 ms . Therefore, T 4 = 5 ms T = 20 ms = 20 10⁻³ s The frequency f of the AC source is given by: f = 1 T = 1 20 10⁻³ = 1000 20 = 50 Hz Answer: 50 Hz