Question

Derive that how we found the average value of ac for different values . please if u Know full derivation give the answer

Answer 1

here is your answer.....

Answer 2

I know the basics of how to find these values from a graph, but not sure how to prove that average voltage of half-wave rectifier is peak/pi and rms is peak/2. Same for full-wave rectifier where average is 2peak/pi. I understand that the full-wave rectifier equation for rms, peak/root 2, is a standard equation, would you derive the other equations from this one?
If you accept that the RMS value of full wave rectified sine is
Vpk
2
√
Vpk2 (exactly the same as the unadulterated AC signal) then you should see that the heating effect (power) for any given resistor as a load will be halved on half wave rectified voltage.
So, let's assume a 1 ohm load resistor (for mathematical convenience) and square the full wave RMS voltage to get power: -
Power =
V
2
2
V22
and half this power (half wave rectifier) is
V
2
4
V24
.
So convert back to RMS voltage by taking the square root and you get the RMS value for half wave rectified voltage is
V
2
V2
.
For average value of a FWR sine wave with peak
V
P
VP
: -
V
AVE
=
1
π
∫
π
0
V
P
sinθ dθ
VAVE=1π∫0πVP sinθ dθ
V
AVE
=
V
P
π
(−cosθ
)π0
VAVE=VPπ(−cosθ)0π
V
AVE
=
2
V
P
π
=
2
π
V
P
=0.637
V
P
VAVE=2VPπ=2πVP=0.637VP
It's all about finding the area under one half cycle so, the integral finds the total area under the sine wave and the division by
π
π
divides area by length to get average height. Pi is used because it's convenient to work in radians for
θ
θ
.
Because we are talking averages, for a HWR sine wave, it's half the average value of the FWR sine wave