# Homework Help: EMF expression in electrical generator

1. Aug 25, 2012

### Galfer

Hi all!

I'm studying electrical generators, and while I was trying to come up to the expression of EMF induced in a conductor I got two different results following two different ways, and I can't find the error.

With sinusoidal B, with $$e=bvl$$ I get an RMS value $$E=2K_{f}f\phi$$, while with $$e=\frac{d\Lambda}{dt}$$ I get $$E=\pi K_{f}f\phi$$

Following the calculations; can you please tell me where the error is?

Case 1:

p = polar pairs
ƒ = frequency
Kf = form factor

$$\tau=\frac{\pi r}{p}$$
$$v=2f\tau$$
$$E=K_{f}B_{m}lv=K_{f}\frac{\phi}{\tau l}l2f \tau=2K_{f}f\phi$$

Case 2:

$$e=\frac{d\Lambda}{dt}=\frac{1}{2}\frac{d\phi}{dt}$$
$$E_{M}=\sqrt{2}E=\frac{1}{2}\phi_{M}\omega=\frac{1}{2}\sqrt{2}K_{f}\phi2\pi f=\pi f\sqrt{2}K_{f}\phi$$
$$E=\pi K_{f}f\phi$$

I thought it could be a $$\frac{2}{\pi}$$ due to a mean value, but the flux is the same in both the expressions...

Last edited: Aug 25, 2012
2. Aug 28, 2012

### Galfer

$$v = conductor\; velocity$$
$$\tau = polar\; pitch$$
$$l=rotor\; and\; conductor\; length$$
$$\Lambda=magnetic\; flux\; of\; B\; linked\; with\; the\; coils$$
$$\phi = polar\; magnetic\; flux$$
$$B_{m} = mean\; value\; of\; B$$

$$B_{m}=\frac{\phi}{\tau l}$$

And the passage from Λ to 0.5Φ:

$$e=\frac{d\Lambda}{dt}=\frac{1}{2}\frac{d\phi}{dt}$$

is because with one conductor I've half a coil.

I can't solve it, if you've any ideas it'd be great.