Formula for the average EMF of generator

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songoku
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Homework Statement
In my notes, the average emf of generator is given by ##\frac{4NBA}{T}##. I want to know how to derive this formula
Relevant Equations
##\varepsilon=\frac{4NBA}{T}##

##\varepsilon=NBA\omega \sin (\omega t)##
I know the formula of emf of generator is ##\varepsilon=NBA\omega \sin (\omega t)##. If I draw the graph of emf against time, the graph will be sinusoidal and if I find the average, the average will be zero.

How can the average emf of generator is ##\frac{4NBA}{T}##?

Thanks
 
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Even if the generator had a rectified output, the average fem wouldn't match your notes. Can you check them?
 
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What is (4NBA)/T suppose to mean? Peak? RMS? Half-cycle average?
 
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songoku said:
How can the average emf of generator is ##\frac{4NBA}{T}##?
The formula is (I believe) correct when finding:

a) the average emf of a simple DC generator, or

b) the average emf for the fully-rectified output of a simple AC generator.

If you know a little calculus, find the average value of ##\sin x ## over, say, the positive half of a cycle; then the rest should be easy.
 
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Gordianus said:
Even if the generator had a rectified output, the average fem wouldn't match your notes. Can you check them?
I have checked, it is what I wrote.

pbuk said:
Because we are interested in the magnitude of the EMF.
Steve4Physics said:
The formula is (I believe) correct when finding:

a) the average emf of a simple DC generator, or

b) the average emf for the fully-rectified output of a simple AC generator.

If you know a little calculus, find the average value of ##\sin x ## over, say, the positive half of a cycle; then the rest should be easy.
I understand.

Gordianus said:
What is (4NBA)/T suppose to mean? Peak? RMS? Half-cycle average?
I suppose it would be half-cycle average

Thank you very much Gordianus, pbuk, Steve4Physics
 
Gordianus said:
Even if the generator had a rectified output, the average fem wouldn't match your notes. Can you check them?
For information, the average value of ##sin(x)## over a positive half-cycle is ##\frac 2{\pi}##.

With ##\mathscr E =NBA\omega \sin (\omega t)## and ##\omega = \frac {2 \pi}T##, this gives the OP's post #1 formula for the average value of a fully rectified output.
 
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