I'm trying to understand transformers and am reading this site: Mutual inductance and basic operation
I am a little confused about how to reason about the voltages involved in a transformer.
Assume that the number of windings in both coils of a transformer are the same.
Assume at time 0, a voltage of V1 starts increasing from 0 to some positive value.
This change in current creates a back emf, which as I understand studying so many years ago, is a voltage that opposes the current that created it (Lenz's law).
If this back emf is being created by the change in flux of the magnetic field, then if the second loop of the transformer is wound in the same direction as the first loop( because this determines the direction of the magnetic field) I would expect that the back emf voltage from the applied voltage in the first loop should also be induced in the 2nd loop of the transformer.
according to the linked website, the voltage induced in the second loop should be of the same magnitude as the applied voltage.
But I remember being taught in school that when current is run through a way, the induced emf from self-inductance is a voltage in the opposite direction with a smaller magnitude.
Is this incorrect? perhaps my memory is wrong?
According to the website, with the same the # of windings on both sides of a transformer the voltage through the 2nd side of the transformer is the same as the applied voltage.
But if the back emf from self-induction is a smaller voltage then I would expect the induced voltage in the 2nd loop to be a smaller voltage.
Is my memory of the magnitude of the back emf wrong?
Best Answer
Using one of Maxwell's equations, namely $$U = \oint \vec E \cdot d\vec s = - \frac{d\phi}{dt} = -N \frac{d\phi}{dt}$$ we are able to find the voltage at both ends of a transformer. All we need to know is the number of windings on both sides (in your case $N_1 = N_2 = N$) and one of the voltages. Since we assume the iron core has no stray fields, the magnetic flux from the first coil is exactly what passes through the second coil: $\phi_1 = \phi_2 = \phi$. Now, we set both equations equal: $$\frac{U_1}{N_1} = \frac{d\phi}{dt} = \frac{U_2}{N_2}$$ Since $N_1 = N_2 = N$, all we have possibly left is $U_1 = U_2 = U$. The voltage in both coils is the same if they have the same number of windings.
We know that the self-inductance of a coil is given as $$L = \frac{\phi I}{N}$$ Since both coils are the same (same geometry, same number of windings, ...), their inductances must be equal, too: $L_1 = L_2 = L$. We write: $$\frac{I_1 L}{N} = \frac{I_2 L}{N}$$ and cancel to get $I_1 = I_2 = I$, we also have the same current flowing in both coils.
Your primary coil produces a magnetic field $B = \frac{\mu_0 NI}{l}$ an therefore a magnetic flux of $\phi = B A = \frac{\mu_0 NI}{l}A$ with $A$ being the cross section of the primary coil. Using $L = \mu_0 \frac{N^2 A}{l}$ as the self-inductance for a solenoid, we find that $$B A = \frac{\mu_0 NI}{l} A = \phi = \frac{LI^*}{N} = \frac{\mu_0 \frac{N^2 A}{l}I}{N} = \frac{\mu_0 N A I^*}{l}$$ which means that $I = I^*$, where $I$ is the supplied current and $I^*$ is the induced current. Both have the same magnitude again.
So in closing, I don't think it's correct to say that you should expect a smaller voltage due to self-inductance. In fact, you can try this out yourself with a homemade transformer and test the theory. Also, an energy observation might be useful (power in = power out).