In an inductor, a current waveform 90 degrees out of phase with the voltage waveform creates a condition where power is alternately absorbed and returned to the circuit by the inductor. If the inductor is perfect (no wire resistance, no magnetic core losses, etc.), it will dissipate zero power.

I wonder how this will change in the case of transformer. Suppose the primary and secondary coils are of zero resistance and the core is perfectly ferromagnetic, what should be the phase different between the voltage and current in the primary coil? Power is delivered to the secondary circuit so it should not be absorbed by the primary voltage source as in the case of pure inductor.


Short answer: it depends on the impedance of the load attached to the secondary coil.

A perfect transformer can be modeled as a pair of inductors with mutual inductance $M$ and self-inductances $L_1$ and $L_2$, with $M^2 = L_1 L_2$. Denote the voltage across coil 1 and the current through it as $V_1(t) = \tilde{V}_1 e^{i \omega t}$ and $I_1(t) = \tilde{I}_1 e^{i \omega t}$, and similarly for coil 2. Assume that we hook up a load of impedance $Z$ across coil 2, so that $\tilde{V}_2 = Z \tilde{I}_2$. Then from Kirchoff's laws we end up with the equations \begin{align*} -i \omega L_1 \tilde{I}_1 - i \omega M \tilde{I}_2 &= \tilde{V}_1 && \text{(coil 1)} \\ - i \omega M \tilde{I}_1 - i \omega L_2 \tilde{I}_2 &= \tilde{V}_2 = Z \tilde{I}_2 && \text{(coil 2)} \end{align*} This can be solved for $\tilde{I}_1$ and $\tilde{I}_2$, to yield \begin{align*} \tilde{I}_1 &= -\frac{i Z - L_2 \omega}{L_1 Z \omega} \tilde{V}_1 & \tilde{I}_2 &= - \frac{M}{L_1 Z} \tilde{V}_1 \end{align*}

In the limit that the load impedance is large compared to the inductance of the secondary coil, we recover the result that $\tilde{I}_1$ is 90° out of phase with $\tilde{V}_2$, and so very little power is dissipated in this case. (It's not hard to see that the power dissipated in coil 2, which is something like $\frac{1}{2} \Re ( Z \tilde{I}_2^* \tilde{I}_2)$, is also very small in this limit, so that's a good sanity check.) However, if the load impedance is comparable to the impedance of the secondary coil, then $\tilde{I}_1$ and $\tilde{V}_1$ will have a more complicated phase relationship.

  • $\begingroup$ I think current and voltage in coil 1 generally should have phase difference less than 90 degrees otherwise how the circuit 1 "knows" it is losing energy to circuit 2? $\endgroup$ – Kelvin S May 13 '15 at 16:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.