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Derive a relation between mean or averag...

Derive a relation between mean or average value of AC and its peak value.

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i. The magnitude of an alternating current in a circuit changes from one instant to other instant and its direction also reverse for every half cycle.
ii. During positive half cycle, current is taken as positive and during negative cycle it is negative. Therefore mean or average value of symmetrical alternating current over one complete cycle is zero.
iii. Therefore the average or mean value is measured over one half of a cycle. These electrical terms. average current and average voltage can be used in both AC and DC circuit analysis and calculations.
iv. The average value of alternating current is defined as the average of all values of defined as the average of all values of current over a positive half-cycle or negative half-cycle.
v. The instantaneous value of sinusoidal alternating current is given by the equation `i=I_(m)sin omegat" or "i=I_(m)sintheta("where"theta=omegat)` whose graphical representation is given in Figure.
vi. The sum of all current over a half-cycle is given by area of positive half-cycle (or negative half-cycle). Therefore,
`I_(av)=("Area of positve half-cycle"("ornegative half -cycle"))/("Base length of half-cycle")" "...(1)`

vii. Consider an elementary strip of thickness `dtheta` in the positive half-cycle of the current wave. Let i be the mid-ordinate of that strip.
Area of the elementary strip `=id""theta`
Area of positive half-cycle
`=int_(0)^(pi)id""thetaint_(0)^(pi)\I_(m)=-I_(m)[cospi-cos0]=2I_(m)`
Substituting this equation (1) , we get (The base length of half-cycle is`pi`)
Average value of AC,`I_(av)=(2I_(m))/(pi)`
`I_(av)=0.637I_(m)`
viii. Hence the average value of AC is 0.637 times the maximum value `I_(m)` of the alternating current. For negative half-cycle, `I_(av)=-0.637I_(m).`
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