Do Grainger/Stevenson incorrectly describe an equivalent circuit for a transformer?

Question

The equivalent circuit for a transformer, as shown above, is supposed to be consistent with the following equations:

\begin{align} \small{\mathrm{Equation~2.22:}}~V_1&=(r_1+j\omega L_{11})I_{1} + (j\omega L_{12})I_{2} \end{align} \begin{align} \small{\mathrm{Equation~2.23:}}~V_2&=(j\omega L_{21})I_{1}+(r_2+j\omega L_{22})I_{2} \end{align} where \begin{align} B_m=(a\omega L_{21})^{-1} \end{align} \begin{align} \mathrm x_1=\omega L_{1l} \end{align} \begin{align} \mathrm x_2=\omega L_{2l} \end{align} with L₂₁ representing mutual inductance, and L_1l and L_2l representing leakage inductances.

"By writing Kirchoff's voltage equation around the path of each of the currents I₁ and I₂/a in Fig 2.5, the reader should find that Eqs. (2.22) and (2.23) are satisfied exactly" (Grainger & Stevenson, p50). The text in bold seems to suggest that I₁ and I₂/a take the paths that I drew in red, but authoritative voices on this site insist that load currents do not go through G_c and B_m and that only the no-load current goes through G_c and B_m.

How can the second term of Equation 2.22 \begin{align} \mathrm~(j\omega L_{12})I_{2} \end{align} exist if I₂/a doesn't pass through B_m? Is the equivalent circuit really consistent with the equations? (Please note that I am well aware of the purpose of an equivalent circuit as a mere conceptual tool)

Answer 1

When analysing a circuit, it's entirely possible to assign loop currents arbitrarily. As long as the equations are then derived from those consistently, the mathematics will work out fine.

It is mathematically correct to draw the current loops you have

One current loops through core losses and back through the power supply. Another current loops through the ideal transformer and back through the losses. This will be mathematically fine. However, it's lousy for intuition, because the currents not named well.

When we operate the transformer, we find the primary current varies with the load current. This means that we have to do mental gymnastics to see how the secondary current sums with the core losses to become primary current.

When we operate the transformer, we find the core flux, that is the volts per turn, that is the current through Bm, stays constant as the load current varies. If the load current simply flows through the core, then that's not true. So we have to do mental gymastics with the primary current differencing with the scaled load current to see why the core current stays constant with varying load.

The equations handle this just fine, it's our heads that don't.

If instead we choose different current loops, then it lines up with our intuition, and real transformer behaviour.

Now we have chosen different loops, and given them sensible names. They both loop through the power supply. All that's different is which lines have a single current flowing through them, and which have two that need to be added when we make the loop equations.

Now it's easy to see how Icore can stay constant, regarless of load current. We can see how a varying load current results directly in a varying primary current.

Answer 2

The text in bold seems to suggest that I1 and I2/a take the paths that I drew in red

Nope. The current that goes through \$G_c\$ and \$B_m\$ is -- by Kirkoff's current law -- \$I_1 + I_2 / a\$.

If I1 and I2/a do not both go through Gc and Bm, how can V1 be a function of both I1 and I2/a?

Well, first, the combination goes through \$G_c\$ and \$B_m\$. Second, the voltage at the top of \$G_c\$ and \$B_m\$ is impacted by \$I_2/a\$ going through the equivalent resistance and inductance on the right.