# How does KVL work in transformers?

I have studied ideal transformers (transfer 100% of energy) and that the power loss in the secondary coil is I^2*R..
Then I asked myself, why can't we use these equivalent formulas for power (IV and V^2/R)??
I tried to use them but they gave me different values for power loss
I searched for an answer here and I discovered that the supply voltage is not equal to the drop voltage through the resistance in the secondary coil.
So IV and V^2/R works when V is the drop voltage..

After that I had another question, How can we apply Kirchhoff's voltage law in the secondary coil?
Due to KVL the sum of voltages must equal zero.
But appearantly, that doesn't exist in the secondary coil as long as the drop voltage is not equal to the supply voltage..

Any one has an explanation?

anorlunda
Staff Emeritus
So IV and V^2/R works when V is the drop voltage..
Congratulations. You must have missed that point in a circuits lesson. But you reasoned yourself to the correct conclusion.

Due to KVL the sum of voltages must equal zero.
But appearantly, that doesn't exist in the secondary coil as long as the drop voltage is not equal to the supply voltage..
In this case, you're reasoning yourself to the wrong conclusion. KVL applies to an entire closed loop. Put a load across that secondary, then sum all the V drops around the loop and KVL works just fine.

Delta2
Homework Helper
Gold Member
I think one point you have missed for sure is that the supply voltage is applied to the primary coil therefore when we want to deal with KVL for the secondary coil the supply voltage does not appear anywhere...