Do no-load current and load current share path in transformer?

Question

In the equivalent circuit for a transformer, does only the no-load current I₀ pass through core loss R_c and magnetizing reactance X_m? Or does load current also pass through R_c and X_m?

The figure above suggests only the no-load current passes through R_c and X_m, but my textbook shows both the no-load current and the load current I₂/a passing through the parallel path. Which depiction is correct?

Note: the textbook authors chose the direction of I₂ to yield positive coefficients in the equations \begin{align} \mathrm V_1&=(r_1+j\omega L_{11})I_{1} + (j\omega L_{12})I_{2} \end{align} \begin{align} \mathrm V_2&=(j\omega L_{21})I_{1}+(r_2+j\omega L_{22})I_{2} \end{align}

Answer 1

The first diagram shows I0 passing through Rc and Xm. The primary current Ip is the sum of I0 and the scaled output current that is not explicitly shown.

The second diagram shows the difference between I1 and I2/a passing through Gc and Bm, without calling it anything.

These two versions are equivalent, at least mathematically.

There are two specified currents at the node where Xp/X1, Rc/Gc, Xm/Bm and RsN2/x2a2 meet. The third is given by Kirchoff.

The current flowing through the Xm/Bm Rc/Gc components (let's call this the core current) is controlled essentially by the constant input voltage, as the components Rp/r1 and Xp/x1 tend to be very small. It is therefore more or less constant.

I find the first diagram easier to be intuitive about, as we can just say that the input current Ip is the sum of this core current, and the turns-scaled load current drawn by the ideal transformer.

My problem with the second one, though entirely consistent with the first, is that we have a small current drawn by the core components, which consists of the small difference between two large and variable currents I1 and I2/a. As such, I find it difficult to think other than that this core current could be very wild and variable. As it happens, I1 varies exactly as needed to be the sum of the more or less constant core current and I2/a, so thinking of the core current as a difference is unhelpful, at least to me, and it sounds like to you as well.

Answer 2

does only the no-load current I0 pass through core loss Rc and magnetizing reactance Xm?

Yes, Approximately. The coupling magnetization current, represented by Io is not very sensitive to the load current unless you consider the drop in voltage across Zm from secondary loading.

Also, the current in Rc is much lower than Xm.

As @Neil answered, the two schematics are equivalent and do not contradict each other or this answer.

Answer 3

does only the no-load current I0 pass through core loss Rc and magnetizing reactance Xm?

Correct.

does load current also pass through Rc and Xm?

No it doesn't.

my textbook shows both the no-load current and the load current I2/a passing through the parallel path

The load current does not pass through Rc/Xm aka Gc/Bm.