Physics · Alternating Current · NEET
One full cycle (period T) is 360 degrees. Starting from zero, the current climbs to its peak at 90 degrees, returns to zero at 180 degrees, reaches the negative peak at 270 degrees, and comes back to zero at 360 degrees. So the very first peak is reached at 90 degrees, which is only a quarter of the cycle. Quarter of T is T/4, not T/2. T/2 (half a cycle) is when the current has already gone to peak and come back down to zero.
Time period and frequency are linked by T = 1/f. So time to peak = T/4 = 1/(4f). Example: if f = 50 Hz, then t = 1/(4 x 50) = 1/200 = 0.005 s = 5 ms. Always convert f to T first, then take a quarter of it. Do not put f directly where T should go.
Set the current equal to its peak: i0 = i0 sin(wt), so sin(wt) = 1. The smallest angle where sin = 1 is wt = pi/2 (90 degrees). Now w = 2 pi f = 2 pi / T. Substitute: (2 pi / T) t = pi/2, which gives t = T/4. So the maths and the graph agree: first peak at T/4.
Yes. Each of the four stages of a cycle takes the same time, T/4, because the sine wave is symmetric. Zero to peak = T/4, peak to zero = T/4, zero to negative peak = T/4, negative peak to zero = T/4. Adding all four gives T, the full period.
No. The time to reach the peak depends only on the frequency (through T = 1/f). Whether the peak current is 2 A or 200 A, and whatever the circuit elements are, the quarter-cycle timing t = T/4 = 1/(4f) is the same. Amplitude changes the height of the wave, not its timing.
The peak value of an alternating current is 5 A and frequency is 60 Hz. How long will the current, starting from zero, take to reach the peak value?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
t = T/4 = 1/(4f), where T is the time period and f is the frequency of the AC.
t = 1/(4 x 50) = 1/200 s = 0.005 s = 5 milliseconds.
No. It depends only on frequency. A higher or lower peak just makes the wave taller or shorter; the quarter-cycle timing stays t = 1/(4f).
Twice: once at the positive peak (at T/4) and once at the negative peak (at 3T/4). Between them the current passes through zero.
It is a quick single-step numerical that mixes T = 1/f with the quarter-cycle idea. NEET sets 1/(2f) as a tempting wrong option, so knowing peak = quarter cycle saves marks.
