# Transformer - Flux density

Discussion in 'Physics' started by Skeebopstop, Mar 19, 2009.

1. ### Skeebopstop Thread Starter Active Member

Jan 9, 2009
358
9
Hi all,

I am working through the physics of transformer design, and on the attached slide, hit an inquiry of where the divide by 2 factor came from?

Any insight? I understand the rest.

Regards,

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2. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
726
Most calculations involve the angular velocity of the AC waveform - ω - radians/second.

ω=2 π FHz

An easier to understand explanation is below, from This site, towards bottom of page

3. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
522
The direct answer to your question is that your extract specifically states "the positive peak". You need to double this to get the negative peak as well.

You should observe that the frequency of the power waveform is twice that of the voltage, current or flux waveform.

I will use sinusoids.

If $\Phi$ is the maximum value of the flux and f the frequency then the flux changes from +$\Phi$ to -$\Phi$ in half a cycle i.e. 1/2f seconds.

So the average rate of change of flux is 2$\Phi$ /0.5f = 4f$\Phi$

Note the 2$\Phi$ is the distance between positive and negative peaks.

But the average rate of change of flux = average emf induced.

Thus induced emf = 4f$\Phi$ volts per turn

Multiply this by the form factor (1.11 for sinusoids) to get the RMS value and we have the well known equation

EMF induced per turn = 4.44Nf$\Phi$ where N is number of turns.

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A calculus method is as follows; If $\phi$ is the instantaneous flux then

$\phi$ = $\Phi$sin2π ft

Instantaneous induced voltage per turn is

-d$\phi$/dt volts

= -$\Phi$ x cos2πft volts (note the minus sign)

= 2πf$\Phi$ x sin(2πft -π/2)

Thus the max value occurs when the sin term is 1 and equals 2πf$\Phi$ volts per turn.

Thus the RMS value = √2 x 2πf$\Phi$ volts per turn = 4.44f$\Phi$ as before.

This is a shortened version of thatoneguys's link.

Jan 9, 2009
358
9
Thanks guys