enter image description here

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 L1l and L2l 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 Gc and Bm and that only the no-load current goes through Gc and Bm.

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 Bm? 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)

  • 1
    \$\begingroup\$ It's just a daft way of showing current direction. Mathematically it resolves but it paints a silly picture initially. It all began when he showed secondary current in a daft direction. Some people have strange ways of showing things. I suggest you (legitimately) redraw secondary current in the other direction and see how easy it is to analyse. \$\endgroup\$
    – Andy aka
    Commented Feb 22, 2023 at 22:21
  • \$\begingroup\$ The load part cancels out. \$\endgroup\$ Commented Feb 23, 2023 at 0:01
  • \$\begingroup\$ Not sure if I1 and I2/a cancelling out would help explain where the second term of Equation 2.22 comes from... that second term looks like it needs I2/a to pass through the parallel path, no? \$\endgroup\$ Commented Feb 23, 2023 at 0:06

2 Answers 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.

  • \$\begingroup\$ You seem to be in direct disagreement with folks who say that the combination of I1 + I2/a does not go through Gc and Bm, and that only the no-load current passes through Gc and Bm. \$\endgroup\$ Commented Feb 22, 2023 at 23:46
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    \$\begingroup\$ @artist_and_not_EE_by_training is the combination of I1+I2/a not the no-load current? \$\endgroup\$ Commented Feb 23, 2023 at 0:01
  • \$\begingroup\$ No I1 is the primary current and I2/a is the secondary current referred to the primary circuit \$\endgroup\$ Commented Feb 23, 2023 at 0:03
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    \$\begingroup\$ The combination of \$I_1 + I_2 / a \$ must go through \$G_c\$ and \$B_m\$ -- that's Kirkoff's current law, and if it weren't true then there would be a whole lot of electronic devices that would work differently than they actually do. But that does not contradict the notion that that particular sum of currents is, in fact, the no-load current. \$\endgroup\$
    – TimWescott
    Commented Feb 23, 2023 at 3:44
  • \$\begingroup\$ @TimWescott I'm going to run with your comment for now as it's the only internally consistent explanation I've seen so far. What you said accommodates for the idea that only the no-load current goes through the parallel path, and still allows for I2/a to "touch" Bm, allowing the second term of Eq 2.22 to even exist. \$\endgroup\$ Commented Feb 23, 2023 at 15:37

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

enter image description here

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.

enter image description here

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.

  • \$\begingroup\$ Your drawing definitely makes much more sense, but doesn't the voltage drop described by the second term of Eq 2.22 require I2/a to "pick up" the mutual inductance Bm by crossing the parallel path? Could you confirm that I2/a in your drawing doesn't actually need to ever "touch" Bm for the second term of Eq 2.22 to exist? If confirmed, that would blow my mind a little bit... so strange \$\endgroup\$ Commented Feb 23, 2023 at 15:27
  • \$\begingroup\$ Bm is not a mutual inductance, it's magnetising inductance. Or at least that's how it's normally drawn, though G/S seem to dimension it an admittance. 2.22 and 2.23 cannot be the whole story as they don't include a Gc term. I think you are right that Grainger/Stephenson is BS. I suspect that others, and I, were so thrown by the obviously wrong assertion that load current flows through the magnetising inductance, that we didn't look more closely. Sorry. Load current does flow through the mutual inductance, and Bm aint it. \$\endgroup\$
    – Neil_UK
    Commented Feb 23, 2023 at 15:48

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