As we know alternating current changes direction and we can represent this graphically by a pure sine wave as shown in figure-1 below

**Fig-1: Waveform of Alternation Current (AC)**

In the figure you can observe the wave is changing its amplitude from 0 to a positive maximum point and then again comes back to 0. But instead of going back to positive maximum point again, it changes its direction with same amplitude to the opposite to reach maximum negative point. After reaching the maximum point in opposite direction it again comes back to zero. And it keeps on changing in the same manner. Here positive and negative points only show the direction of the wave. It doesn’t mean that maximum positive point has highest amplitude and maximum negative point has lowest amplitude. But they only show the direction of the flow of current.

If you start adding the instantaneous amplitude values of both cycles (i.e +ve and -ve), the end result will be zero. Hence the average would be zero. **See Fig-2**

**Fig-2: Amplitude points taken at various instants**

In the above figure we take three instantaneous amplitude values I_{1}, I_{2} and I_{3 }from the positive cycle. Whereas –I_{1}, -I_{2} and -I_{3 }are taken from negative cycle. In this case average value is:

In reality we use many machines which run on AC. If we apply above logic to calculate the power its mean no machine should work. But this is not the case. Its mean machines are getting energy from AC. At some instants they are getting zero energy and at some instants they are getting maximum (peak) energy. But at other instants they are getting some energy values between the peak and minimum values. In short the amplitude of AC is continuously changing but delivering energy too. So in order to explain how much energy these machines are getting to run we have to devise some other method to calculate the average value of AC.

So instead of taking instantaneous values we use a very small time interval to note the amplitude values. **See fig-3**

**Fig-3: Small time intervals at various locations of one AC cycle**

Now the amplitudes I_{1}, I_{2} and I_{3} are the mid points of the yellow strips. These strips are so small that values of I_{1}, I_{2} and I_{3 }are almost constant (**See Fig-4)**. The AC is completing a one full cycle in time interval of 2.

**Fig-4: Almost constant amplitude between two close points A & B**

So area under a small rectangular yellow strip is **“***I x dt***”** . If** I_{p}** is the peak amplitude of AC, then value of AC at any instant is given by

*I = I _{p}*

*Sint*

Now we can calculate the average value by finding the area under any one half cycle divided by time. We take positive half cycle. So

Putting the value if ´** I**´ in above

Note: The average value is 63.7% of the peak value

Tayyab Raza Cricketer

Sir please share some basic info regular useful articles about electrical like that what is voltage watts ohms current etc.

B’coz we can easily stand ur method. Plzzzz

Asif Haroon

ok i will upload soon