Physics HL IB 2008: Transformer currents question

Question

A question on the IB HL Physics paper of a couple of years back looks like this:

In the second image, taken from this video solution, you see that the given answer should be D.

Now, I have an issue with this and would like to see where I'm going wrong, in case.

First, there's always the ambiguity (typical of these tests) about how is the current in the secondary defined, but I think the standard way is opposite to what's shown in this picture:

I'm taking this pic because it neatly summarises the constitutive relationships of an ideal tranasformer, which is the object of the question.

In the question, $n = \dfrac{N_2}{N_1} = 200/100 = 2$ so that while voltage is increasing in going from primary to secondary, current is halved.

Also, depending on your convention, the ratio between primary and secondary current could entail a 180 degrees phase rotation, or a sign inversion.

However, the answer considered correct is showing some sort of 90 degrees phase shift. The video reported above talks about the output current being proportional to the induced emf and therefore to the variation of primary current, however I find this explanation sketchy and more than anything else, the relationship between currents in an ideal transformer is:

$$i_2 = - \dfrac{1}{n} i_1$$

Nothing else. So there can be a sign inversion, as said, depending on the convention, but no extra phase shift.

Are the constitutive relationships I'm using somehow incorrect for the case at hand? If so, why?

Answer 1

This just seems like a mistake, your understanding of the ideal transformer is correct. There should be no phase shift between the currents (apart from a $180^\circ$ difference depending on the sign convention). This requires that the coil inductances be very large, which is assumed in the ideal transformer. When this is the case, to keep the total flux from blowing up (due to the large inductances), the fluxes due to the primary and secondary currents (almost) cancel, which requires that the currents to be proportional.

Here are the mathematical details. Consider two coupled inductors, and suppose they are perfectly coupled for simplicity. The flux through the primary is $$\Phi_1 = L_1 i_1 + M i_2=L_1 i_1 + \sqrt{L_1 L_2}i_2= \sqrt{L_1L_2}(\sqrt{L_1/L_2}i_1+i_2)$$ where $M$ is the mutual inductance and $L_1$, $L_2$ are the self-inductances of the individual coils. Now suppose $L_1$ and $L_2$ are extremely large, approaching infinity. $\Phi_1$ is typically governed by a voltage applied to the primary (it is the time integral of the primary voltage): it isn't very large like the inductances. The only way the flux can remain small while $L_1$ and $L_2$ approach infinity is $$\sqrt{L_1/L_2}i_1+i_2\approx0$$ $$i_2 \approx -\sqrt{\frac{L_1}{L_2}}i_1=-\frac 1 n i_1.$$

The following equivalent circuit of a pair of coupled inductors is also useful to keep in mind:

$$ L_{l1} = \left(1 - k \right)L_1 $$ $$ L_{l2} = \left(1 - k \right)L_2 $$ $$ L_{Mag} = kL_1 $$ $$ N_1 : N_2 = L_1^2:L_2^2 $$ where $k=M/\sqrt{L_1L_2}$ is the coupling coefficient. You can see that if not for the magnetization inductance $L_{Mag}$, the primary and secondary currents would still be related by the ideal transformer relation. As long as $L_{Mag}$ is large, that is its impedance is large enough that its current is a small fraction of the total primary current, $i_2 = -\frac{1}{n}i_1$ is valid.