EMF expression in electrical generator

In summary, the speaker is studying electrical generators and encountered different results for the expression of EMF induced in a conductor using two different methods. They are looking for help in identifying where the error could be. The first method, using the formula e=bvl with sinusoidal B, gives an RMS value of E=2K_{f}f\phi. The second method, using e=\frac{d\Lambda}{dt}, gives E=\pi K_{f}f\phi. The speaker has considered a possible correction of \frac{2}{\pi} but the flux remains the same in both expressions. Additional notation and clarifications about the calculations are provided. The speaker is seeking suggestions to solve the issue.
  • #1
Galfer
2
0
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 [tex]e=bvl[/tex] I get an RMS value [tex]E=2K_{f}f\phi[/tex], while with [tex]e=\frac{d\Lambda}{dt}[/tex] I get [tex]E=\pi K_{f}f\phi[/tex]

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

Case 1:

p = polar pairs
r = radius
ƒ = frequency
Kf = form factor

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

Case 2:

[tex]e=\frac{d\Lambda}{dt}=\frac{1}{2}\frac{d\phi}{dt}[/tex]
[tex]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[/tex]
[tex]E=\pi K_{f}f\phi[/tex]

I thought it could be a [tex]\frac{2}{\pi}[/tex] due to a mean value, but the flux is the same in both the expressions...
 
Last edited:
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  • #2
Just adding some notation and clarifications about my calculations:

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

[tex]B_{m}=\frac{\phi}{\tau l}[/tex]

And the passage from Λ to 0.5Φ:

[tex]e=\frac{d\Lambda}{dt}=\frac{1}{2}\frac{d\phi}{dt}[/tex]

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

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

What is EMF expression in electrical generator?

EMF stands for electromotive force and is the voltage generated by an electrical generator. It is the driving force behind the flow of electric current in a circuit.

How is EMF expression measured in an electrical generator?

EMF expression is measured in volts (V) using a voltmeter. This device is connected to the terminals of the generator to measure the difference in potential between the two terminals, which represents the EMF.

What factors affect the EMF expression in an electrical generator?

The EMF expression in an electrical generator is affected by various factors such as the speed of rotation, the strength of the magnetic field, the number of turns in the coil, and the resistance of the circuit. Additionally, the type of material used in the coil and the size of the generator can also impact the EMF expression.

What is the relationship between EMF expression and electrical current in a generator?

EMF expression is directly proportional to the electrical current in a generator. This means that as the EMF increases, so does the current in the circuit. However, the current can also be affected by the resistance of the circuit, which can decrease the overall flow of current.

How can the EMF expression in an electrical generator be increased?

The EMF expression in an electrical generator can be increased by increasing the speed of rotation, using stronger magnets, increasing the number of turns in the coil, and reducing the resistance in the circuit. Additionally, using a larger generator or a different type of material for the coil can also increase the EMF expression.

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