Physics · Alternating Current · NEET
AC follows I = I0 sin(wt). In the first half-cycle (0 to T/2), sin is positive, so current is positive. In the second half-cycle (T/2 to T), sin is negative, so current is negative with the same shape. When you add up (integrate) all the values over one full cycle, the positive part and the negative part are equal in size but opposite in sign, so they cancel. Mathematically, the average of sin(wt) over one full period is zero, so I_avg = 0.
No. Current is definitely flowing every instant — it is just changing direction. Zero average only means the charge pushed forward in the first half is exactly pulled back in the second half, so the NET charge transported over a full cycle is zero. A bulb connected to AC still glows because heating depends on I squared (always positive), not on the plain average.
RMS first squares the current. Squaring turns every negative value into a positive one (because a negative number squared is positive). So there is nothing left to cancel — the squared graph is always above zero. That is why RMS (root mean square) is non-zero and is the value we actually use, while the plain average is zero.
Because over a full cycle the answer is trivially zero and useless. Over just the positive half-cycle there is no cancellation, so the average is meaningful: I_avg(half cycle) = 2*I0 / pi = 0.637 I0. NEET often tests this half-cycle value, so do not confuse it with the full-cycle value (which is 0).
Yes. AC voltage V = V0 sin(wt) is also symmetric — equal positive and negative halves. So the average value of AC voltage over one full cycle is also zero, for exactly the same cancellation reason.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
It is exactly zero, because the positive half-cycle and negative half-cycle are equal and opposite, so they cancel.
I_avg = (1/T) times the integral of I0 sin(wt) over 0 to T, which equals 0 because the integral of sin over one full period is zero.
Yes. V = V0 sin(wt) has equal positive and negative halves, so its full-cycle average is also zero.
Because the average over a full cycle is zero and gives no useful information. RMS squares the current first (removing cancellation), so it gives a meaningful, non-zero value used to calculate power and heating.
I_avg(half) = 2*I0/pi = 0.637 I0. Over a half cycle there is no cancellation, so the average is non-zero.
