questions about ideal transformer

You rather have the polarity backwards. You could say that application of a secondary current load causes an incremental reduction in the core magnetic flux. This causes an incremental reduction in the primary back EMF, causing the primary current to increase until the flux level is back up to the required value to give a back EMF equal to the primary voltage.

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Sorry, I should have made the current reference directions clear at the first place.

I have found another problem, my English ablibity. I'm not sure if I understand your description. I'm not sure if I have used the terms mutual-induced, self-induced EMFs, or back EMF correctly as you do. So please allow me to start over again, and please check if I get the correct concept.

For the circuit attached, lst's focus on the time period where the voltage source v1 is increasing from zero.
One thing is for sure, that is, i1 is increasing, i2 is also increasing but in the opposite direction to the secondary current reference direction, correct?

Based on
v1(t) = L1*(di1/dt) + M*(di2/dt)

Assuming L1 and M are positive, since (di2/dt) is now negative, v1(t) is fixed at the moment, the current i1 must be increading more rapidly to make the whole term [ L1*(di1/dt) + M*(di2/dt) ] be equal v1(t).
More rapid increase in i1 means v1(t) has to increase in order to overcome the 'Back EMF' due to the increasing i1.
Do I understand you correctly?

Thank you!

MrAl said:

Hello again,


Well thanks for doing the core drawing over. That's much better.


Here's an expression for the output voltage (Ro is the load resistance):
Vout=(Ro*M)/sqrt((w*M^2-w*L1*L2)^2+Ro^2*L1^2)

and the phase shift is:
Phi=-atan2((Ro*M*(w*L1*L2-w*M^2))/((w*M^2-w*L1*L2)^2+Ro^2*L1^2),(Ro^2*L1*M)/((w*M^2-w*L1*L2)^2+Ro^2*L1^2))


For an ideal transformer, this simplifies to:
Vout=sqrt(L2/L1)

and

Phi=0

and of course these are both independent of the load Ro.


The initial voltage at the primary causes an increase in current which causes the flux to increase. The increasing flux generates a back emf in both windings of course.

Click to expand...

Thank you very much, MrAl.

The output voltage and the phase shift formulas are useful, but I haven't reached that yet.

I need to 'digest' some materials first. Thanks!

Hi again Heidi,

Ok sure. My main point was to show that when the transformer is ideal the coupling factor is 1 and that means a special relationship between the mutual inductance and the two other inductors must exist. This condition causes a simple form for the output voltage, and also more toward what you were looking for the phase shift is then zero.