# 3 phase transformers and equivalent circuits

There is an exercise in my textbook that goes like this:

3 single phase transformers are connected in wye-delta configuration. Each transformer has the following parameters : R1, X1, Rc = $\infty$, Xm = $\infty$, R2, X2. (R1, X1, R2, X2 being the copper resistance/reactance of the primary and secondary). The primary is connected with a 3-phase line with line-line voltage Vs. Secondary is connected to a balanced load composed of 3 impedances Z=R+jX connected in delta fashion.

So we're asked to draw the equivalent single phase circuit of

Which is, according to solutions, this : So what the primary sees is this:

Now, I understand why $$a' = \frac{V_{AN1}}{V_{AN2}} = \frac{N_1}{N_2} \sqrt{3} = a \sqrt{3}$$ and why how Z is divided by 3 to convert the impedance to the wye configuration.

What confuses me is: why are $R_2$ and $X_2$ divided by 3 on the secondary side? I would think those impedances are in series with the secondary voltage, thus I doubt we could apply $Z_Y = Z_\Delta / 3$ here.

Secondly, what would happen with $R_2$ and $X_2$ if we had this circuit made of 3 single phase transformers like this when we derive the equivalent single phase circuit:

Would $R_2$ and $X_2$ be multiplied by 3 or would it be $R_1$ and $X_1$ that would be divided by 3 in the equivalent single phase circuit?

why are $$\R_2\$$ and $$\X_2\$$ divided by 3 on the secondary side? I would think those impedances are in series with the secondary voltage, thus I doubt we could apply $$\Z_Y=Z_Δ/3\$$ here.

We divide $$\R_2\$$ and $$\X_2\$$ by 3 because the secondary is in delta but in the per-phase circuit we use equivalent wye connections. It is of course assumed the series impedance of each winding of the secondary is the same (i.e. the secondary is balanced).

Would $$\R_2\$$ and $$\X_2\$$ be multiplied by 3 or would it be $$\R_1\$$ and $$\X_1\$$ that would be divided by 3 in the equivalent single phase circuit?

We would divide $$\R_1\$$ and $$\X_1\$$ by 3 for the same reason as above: because now the primary is in delta but in the single-phase circuit we use equivalent wye connections.

Before answering, I need you to know 1) how to convert a delta-connected practical generator into a wye-connected practical generator, and 2) that ideal two-winding single-phase transformers can be modeled with dependent sources.

### 1. Converting a delta-connected practical generator into a wye-connected practical generator

The word practical (i.e. non-ideal) just means that for each phase there is an ideal voltage source with a series impedance.

Consider the following network. Just for clarification on the notation, for phase A, B and C, the line phasor currents are respectively $$\\tilde{I}_\text{A}\$$, $$\\tilde{I}_\text{B}\$$, $$\\tilde{I}_\text{C}\$$. The phase phasor currents are respectively $$\\tilde{I}_\text{BA}\$$, $$\\tilde{I}_\text{CB}\$$, $$\\tilde{I}_\text{AC}\$$. The terminal line (or line-to-line) phasor voltages are respectively $$\\tilde{V}_\text{AB}\$$, $$\\tilde{V}_\text{BC}\$$, $$\\tilde{V}_\text{CA}\$$. The terminal phase phasor voltages are equal to the respective terminal line phasor voltages because the generator is in delta. The internal line (or line-to-line) phasor voltages are respectively $$\\tilde{V}_\text{A'B}=\tilde{V}_\text{s1}\$$, $$\\tilde{V}_\text{B'C}=\tilde{V}_\text{s2}\$$, $$\\tilde{V}_\text{C'A}=\tilde{V}_\text{s3}\$$. The internal phase phasor voltages are equal to the respective terminal line phasor voltages because the generator is in delta. The internal series impedances of each armature winding are respectively $$\{\hat Z}_\text{s1}\$$, $$\{\hat Z}_\text{s2}\$$, $$\{\hat Z}_\text{s3}\$$.

Figure 1. Practical generator connected in delta/triangle.

First we apply source transformation to transform the practical voltage sources into practical current sources:

$$\{\tilde I}_\text{s1} = \dfrac{{\tilde V}_\text{s1}}{{\hat Z}_\text{s1}} \tag*{}\$$

$$\{\tilde I}_\text{s2} = \dfrac{{\tilde V}_\text{s2}}{{\hat Z}_\text{s2}} \tag*{}\$$

$$\{\tilde I}_\text{s3} = \dfrac{{\tilde V}_\text{s3}}{{\hat Z}_\text{s3}} \tag*{}\$$

This results in the following circuit diagram.

Figure 2.

Now we transform the delta-connected impedances into wye-connected impedances. Taking into account the labeling of the impedances in the previous figure and the labeling of the impedances in the following figure, we have:

$$\{\hat Z}_\text{s1Y} = \dfrac{{\hat Z}_\text{s1} \, {\hat Z}_\text{s3}}{{\hat Z}_\text{s1} + {\hat Z}_\text{s2} + {\hat Z}_\text{s3}} \tag 1\$$

$$\{\hat Z}_\text{s2Y} = \dfrac{{\hat Z}_\text{s1} \, {\hat Z}_\text{s2}}{{\hat Z}_\text{s1} + {\hat Z}_\text{s2} + {\hat Z}_\text{s3}} \tag 2\$$

$$\{\hat Z}_\text{s3Y} = \dfrac{{\hat Z}_\text{s2} \, {\hat Z}_\text{s3}}{{\hat Z}_\text{s1} + {\hat Z}_\text{s2} + {\hat Z}_\text{s3}} \tag 3\$$

This results in the following circuit diagram.

Figure 3.

Now we apply a technique known as source splitting and source shifting. For example, we can shift or move the current source $$\{\tilde I}_\text{s2}\$$ that is located between nodes $$\C\$$ and $$\B\$$ into the nodes $$\C\$$ and $$\N\$$ and also into the nodes $$\N\$$ and $$\B\$$. In this way, the KCL equation at node $$\C\$$, at node $$\N\$$, and at node $$\B\$$ is not altered: at node $$\C\$$ there still leaves a current $$\{\tilde I}_\text{s2}\$$ (that now enters to node $$\N\$$), at node $$\B\$$ there still enters a current $$\{\tilde I}_\text{s2}\$$ (that now leaves from node $$\N\$$), and at node $$\N\$$ no net current $$\{\tilde I}_\text{s2}\$$ enters or leaves (because now the current $$\{\tilde I}_\text{s2}\$$ enters and leaves the node $$\N\$$, resulting in no net current). The previous procedure is also applied for the current sources $$\{\tilde I}_\text{s1}\$$ and $$\{\tilde I}_\text{s3}\$$. This results in the following circuit diagram.

Figure 4.

If you are still in doubt as to whether the circuit of figure 4 is indeed equivalent to the one of figure 3, you can apply KCL at the nodes $$\A\$$, $$\B\$$, $$\C\$$ and $$\N\$$ of both figures, and you’ll find out they are the same equations, meaning those two circuits are indeed equivalent.

Now we combine the current sources in parallel by applying KCL. Taking into account the direction of the arrow of the current sources in the previous figure and the direction of the arrow of the current source in the following figure, we have:

$$\{\tilde I}_\text{s1T} = {\tilde I}_\text{s1} - {\tilde I}_\text{s3} \tag*{}\$$

$$\{\tilde I}_\text{s2T} = {\tilde I}_\text{s2} - {\tilde I}_\text{s1} \tag*{}\$$

$$\{\tilde I}_\text{s3T} = {\tilde I}_\text{s3} - {\tilde I}_\text{s2} \tag*{}\$$

This results in the following circuit diagram.

Figure 5.

Lastly we apply source transformation to transform the practical current sources into practical voltage sources:

$$\{\tilde V}_\text{s1Y} = {\tilde I}_\text{s1T} \, {\hat Z}_\text{s1Y} \tag*{}\$$

$$\{\tilde V}_\text{s2Y} = {\tilde I}_\text{s2T} \, {\hat Z}_\text{s2Y} \tag*{}\$$

$$\{\tilde V}_\text{s3Y} = {\tilde I}_\text{s3T} \, {\hat Z}_\text{s3Y} \tag*{}\$$

Taking into account the direction of the arrow of the current sources in the previous figure, this results in the following circuit diagram.

Figure 6. Equivalent generator connected in wye/star.

Now we revert the change of variables we did throughout the derivation, going from the last equations to the first ones. After some math, we get:

$$\{\tilde V}_\text{s1Y} = \left( \dfrac{{\tilde V}_\text{s1}}{{\hat Z}_\text{s1}} - \dfrac{{\tilde V}_\text{s3}}{{\hat Z}_\text{s3}} \right) \, {\hat Z}_\text{s1Y} \tag 4\$$

$$\{\tilde V}_\text{s2Y} = \left( \dfrac{{\tilde V}_\text{s2}}{{\hat Z}_\text{s2}} - \dfrac{{\tilde V}_\text{s1}}{{\hat Z}_\text{s1}} \right) \, {\hat Z}_\text{s2Y} \tag 5\$$

$$\{\tilde V}_\text{s3Y} = \left( \dfrac{{\tilde V}_\text{s3}}{{\hat Z}_\text{s3}} - \dfrac{{\tilde V}_\text{s2}}{{\hat Z}_\text{s2}} \right) \, {\hat Z}_\text{s3Y} \tag 6\$$

In this way, we’ve shown the delta-connected generator of figure 1 is equivalent to the wye-connected generator of figure 6; the relationship between them is given by equations (1) to (6).

—It shall be noted the neutral node of the equivalent network (figure 6) is fictitious or imaginary, it doesn’t exist in real life. The terminal fictitious line-to-neutral phasor voltages are $$\{\tilde V}_\text{AN}\$$, $$\{\tilde V}_\text{BN}\$$, $$\{\tilde V}_\text{CN}\$$. The internal fictitious line-to-neutral phasor voltages are $$\{\tilde V}_\text{A''N} = {\tilde V}_\text{s1Y}\$$, $$\{\tilde V}_\text{B''N} = {\tilde V}_\text{s2Y}\$$, $$\{\tilde V}_\text{C''N} = {\tilde V}_\text{s3Y}\$$.

Notice in figure 6 the new internal nodes of the equivalent/fictitious wye generator are $$\A''\$$, $$\B''\$$ and $$\C''\$$, while in figure 1 the original internal nodes of the actual/real delta generator are $$\A'\$$, $$\B'\$$ and $$\C'\$$.

It shall also be stated those networks are equivalent even if the generator is unbalanced and/or the rest of the circuit/power system is unbalanced, since the expressions that relate these networks don’t depend on the line or phase currents, which indeed depend on whether the system is balanced or not.

Also notice the terminal line (or line-to-line) phasor voltages and the line phasor currents of both networks (figure 1 and 6) are not affected after the transformation.

—If the delta-connected generator of figure 1 is balanced (which means the magnitude of the voltage sources is the same, they’re phase-shifted by 120° between each other, and the impedances are the same), then equations (1) to (6) reduce to the following:

$$\{\hat Z}_\text{sY} = \dfrac{{\hat Z}_\text{s}}{3} \tag 7\$$

$$\{\tilde V}_\text{s1Y} = \dfrac{{\tilde V}_\text{s1}}{\sqrt{3} \, \angle \pm 30°} \tag 8\$$

$$\{\tilde V}_\text{s2Y} = \dfrac{{\tilde V}_\text{s2}}{\sqrt{3} \, \angle \pm 30°} \tag 9\$$

$$\{\tilde V}_\text{s3Y} = \dfrac{{\tilde V}_\text{s3}}{{\sqrt{3} \, \angle \pm 30°}} \tag {10}\$$

where for equations (8) to (10) you use the first sign ($$\+\$$) if the phase sequence is positive/abc or the second sign ($$\-\$$) if the phase sequence is negative/acb.

We won't use this transformation for the generator of your example, since it is already in wye. Instead, we'll use it for the delta winding of the three-phase transformer. That is valid because, as I'll show in the following section, we can model the ideal transformer with dependent sources.

### 2. Modeling an ideal two-winding single-phase transformer as dependent sources

As you know, a practical (i.e. non-ideal) transformer can be modeled as an ideal transformer with series impedance (representing the resistance of the wires and the leakage self-inductance) and the exciting branch (which is neglected in your problem).

As you also know, a three-phase transformer can be modeled as three single-phase transformers. I think this is valid only in balanced conditions, although I'm not completely sure. However our case is a balanced system and so we need not worry.

You should also know that an ideal two-winding single-phase transformer can be modeled as a current-dependent current source in one side and a voltage-dependent voltage source in the other side. This is shown in Figure 11.10-2 of the textbook Introduction to Electric Circuits (9th edition) by Richard Dorf and James Svoboda. Below I show a screenshot.

Figure 7. a) Ideal two-winding single-phase transformer with additive spatial polarity. b) An equivalent circuit using dependent sources. Taken from Dorf & Svoboda's book.

### 3. Deriving the equations that represent the actual three-phase circuit

The procedure to derive an equivalent single-phase circuit will be as follows. First, we obtain equations in the actual three-phase circuit that describe it. Then, we propose a circuit and obtain the equations (by applying KVL, KCL, Ohm's law, etc.) that describe it. Our objective will be find a circuit such that their equations are the same as those from the three-phase circuit, which would mean the former is the equivalent single-phase circuit of the latter.

Let's consider your first example. It is a power system consisting of an ideal wye generator, a practical wye (primary)-delta (secondary) transformer, and a delta load. All are balanced. The three-phase circuit is shown below.

Figure 8. Three-phase circuit of the power system of the first example.

We can convert the delta load into an equivalent wye load by applying a delta-wye transformation. Since the load is balanced, we simply divide the given impedance by 3:

$$\{\hat Z}_\text{LY} = \dfrac{{\hat Z}_\text{L}}{3} \tag {11}\$$

Also, we substitute the ideal transformer of the practical transformer with dependent sources, as it was shown in figure 7. Take into account the reference polarity of the voltages and the reference direction of the currents in that figure: the primary and secondary currents enter the dotted terminals, and the primary and secondary voltages are positive at the dotted terminals. In figure 8 I chose the same polarities and directions for the voltages and currents of the individual transformer of phase A (the primary current is $$\{\tilde I}_\text{A}\$$, the primary voltage is $$\{\tilde V}_\text{A'N'}\$$, the secondary current is $$\{\tilde I}_\text{aa'}\$$, and the secondary voltage $$\{\tilde V}_\text{aa'}\$$).

Notice in figure 7b) the authors used the turns of the individual windings. In three-phase transformers the manufacturer usually doesn't indicate the turns, but the rated line-to-line voltages (in RMS value) at no load. With the rated line-to-line voltages, we can compute the respective rated phase voltages (not to be confused with the line-to-neutral voltages). And with the phase voltages we can compute the turns ratio $$\a\$$, which is equal to the ratio of the phase voltages:

$$\a = \dfrac{N_\text{p}}{N_\text{s}} = \dfrac{V_\text{A'N'}}{V_\text{aa'}} = \dfrac{V_\text{LL,p,rated}}{\sqrt{3} V_\text{LL,s,rated}} \tag{12}\$$

This results in the following circuit diagram.

Figure 9.

Notice the delta winding consists of (dependent) voltage sources with serie impedance, connected in delta. Therefore, we can transform it into a wye equivalent, as we saw in section 2. If you wonder, yes, source transformation is valid for independent and dependent sources. We use equations (7) to (10). However since we're interested only on phase A, we don't have to use equations (9) and (10). So, we have:

$$\{\hat Z}_\text{2Y} = \dfrac{{\hat Z}_\text{2}}{3} \tag {13}\$$

$$\{\tilde V}_\text{aa'} = \dfrac{\left(\dfrac{{\tilde V}_\text{A'N'}}{a}\right)}{\sqrt{3} \, \angle \pm 30°} \implies {\tilde V}_\text{aa'} = \dfrac{{\tilde V}_\text{A'N'}}{a \sqrt{3} \, \angle \pm 30°} \tag {14}\$$

This results in the following circuit diagram.

Figure 10.

Notice the dependent current source of phase A depends on the phasor current $$\{\tilde I}_\text{aa'}\$$. However, we lost that current after transforming the delta winding to a wye winding. Thus we must express $$\{\tilde I}_\text{aa'}\$$ in terms of the currents in the previous figure. To do that, notice $$\{\tilde I}_\text{aa'}\$$ is a phase phasor current of a delta connection, so we can transform it to the respective line phasor current, which is $$\-{\tilde I}_\text{a}\$$. Remember that the magnitude of the line phasor current is $$\sqrt{3}\$$ times the magnitude of the respective phase phasor current, and the former lags the latter by 30° for positive phase sequence or leads it by 30° for negative phase sequence. So we have:

\ \begin{align} -{\tilde I}_\text{a} &= {\tilde I}_\text{aa'} \cdot \sqrt{3} \, \angle \mp 30° \\ \implies {\tilde I}_\text{a} &= {\tilde I}_\text{aa'} \cdot \sqrt{3} \, \angle (\mp 30° + 180°) \\ &= {\tilde I}_\text{aa'} \cdot \sqrt{3} \, \angle \pm 150° \\ \implies {\tilde I}_\text{aa'} &= \dfrac{{\tilde I}_\text{a}}{\sqrt{3} \, \angle \pm 150°} \tag {15} \end{align}\

Notice everything is now in wye, so we can finally start analyzing the three-phase circuit. I hope you know that in a balanced system, the voltage between the neutral points is zero, which means in the previous figure that $$\{\tilde V}_\text{NN'} = {\tilde V}_\text{nn'} = 0\$$.

Applying KVL around the loop $$\NAA'N'N\$$:

$$\ - {\tilde V}_\text{s1} + {\hat Z}_1 \, {\tilde I}_\text{A} + {\tilde V}_\text{A'N'} - {\tilde V}_\text{NN'} = 0 \tag*{} \$$

$$\ \implies - {\tilde V}_\text{s1} + {\hat Z}_1 \, {\tilde I}_\text{A} + {\tilde V}_\text{A'N'} = 0 \tag {16}\$$

Applying KVL around the loop $$\na''n'n\$$:

$$\ - \dfrac{{\tilde V}_\text{A'N'}}{a \sqrt{3} \, \angle \pm 30°} + ({\hat Z}_2 /3) \, {\tilde I}_\text{a} + ({\hat Z}_\text{L} /3) \, {\tilde I}_\text{a} - {\tilde V}_\text{nn'} = 0 \tag*{} \$$

$$\ \implies - \dfrac{{\tilde V}_\text{A'N'}}{a \sqrt{3} \, \angle \pm 30°} + ({\hat Z}_2 /3 + {\hat Z}_\text{L} /3) \, {\tilde I}_\text{a} = 0 \tag {17}\$$

By KCL, the equation relating the primary and secondary currents is:

\ \begin{align} {\tilde I}_\text{A} &= -\dfrac{{\tilde I}_\text{aa'}}{a} \\ &= -\dfrac{\left(\dfrac{{\tilde I}_\text{a}}{\sqrt{3} \, \angle \pm 150°}\right)}{a} \\ \implies {\tilde I}_\text{A} &= \dfrac{{\tilde I}_\text{a}}{a \sqrt{3} \, \angle \mp 30°} \tag {18} \end{align} \

In this way, equations (16), (17) and (18) describe the three-phase equivalent circuit of figure 10. If the phase shift of the transformer is neglected, then equations (17) and (18) become:

$$\ - \dfrac{{\tilde V}_\text{A'N'}}{a \sqrt{3}} + ({\hat Z}_2 /3 + {\hat Z}_\text{L} /3) \, {\tilde I}_\text{a} \approx 0 \tag {19}\$$

$$\ {\tilde I}_\text{A} \approx \dfrac{{\tilde I}_\text{a}}{a \sqrt{3}} \tag {20}\$$

### 4. Deriving the single-phase equivalent circuit

Now, let's consider the following circuit.

Figure 11.

Applying KVL around the left loop yields:

$$\ - {\tilde V} + {\hat Z}_x \, {\tilde I}_x + {\tilde V}_x = 0 \tag {21}\$$

Applying KVL around the right loop yields:

$$\ - {\tilde V}_y + ({\hat Z}_y + {\hat Z}_u) \, {\tilde I}_y = 0 \tag {22}\$$

The equations that relate the transformer voltages and currents, taking into account the reference polarity and reference direction, are:

$$\a' = + \dfrac{{\tilde V}_x}{{\tilde V}_y} \tag {23}\$$

$$\a' = + \dfrac{{\tilde I}_y}{{\tilde I}_x} \tag {24}\$$

Solving for $$\{\tilde V}_y\$$ in eq. (23) and substituing it into eq. (22) yields:

$$\ - \dfrac{{\tilde V}_x}{a'} + ({\hat Z}_y + {\hat Z}_u) \, {\tilde I}_y = 0 \tag {25}\$$

In this way, equations (21), (24) and (25) describe the circuit of figure 11.

If we let $$\ {\tilde V} = {\tilde V}_\text{s1}\$$, $$\{\hat Z}_x = {\hat Z}_1\$$, $$\{\hat Z}_y = {\hat Z}_2 /3\$$, $$\{\hat Z}_u = {\hat Z}_\text{L} /3\$$, and $$\a' = \sqrt{3} a\$$, then equations (21), (24) and (25) become:

$$\ - {\tilde V}_\text{s1} + {\hat Z}_1 \, {\tilde I}_x + {\tilde V}_x = 0 \tag {26}\$$

$$\\sqrt{3} a = \dfrac{{\tilde I}_y}{{\tilde I}_x} \implies {\tilde I}_x = \dfrac{{\tilde I}_y}{\sqrt{3} a} \tag {27}\$$

$$\ - \dfrac{{\tilde V}_x}{\sqrt{3} a} + ({\hat Z}_2 /3 + {\hat Z}_\text{L} /3) \, {\tilde I}_y = 0 \tag {28}\$$

But comparing the three previous equations ((26), (27), (28)) with equations (16), (20), (19) respectively we see that $$\{\tilde I}_x = {\tilde I}_\text{A}\$$, $$\{\tilde V}_x = {\tilde V}_\text{A'N'}\$$, and $$\{\tilde I}_y = {\tilde I}_\text{a}\$$.

In other words, if in the circuit of figure 11 we choose $$\ {\tilde V} = {\tilde V}_\text{s1}\$$, $$\{\hat Z}_x = {\hat Z}_1\$$, $$\{\hat Z}_y = {\hat Z}_2 /3\$$, $$\{\hat Z}_u = {\hat Z}_\text{L} /3\$$, and $$\a' = \sqrt{3} a\$$, then $$\{\tilde I}_x = {\tilde I}_\text{A}\$$, $$\{\tilde V}_x = {\tilde V}_\text{A'N'}\$$, and $$\{\tilde I}_y = {\tilde I}_\text{a}\$$. Therefore, we can re-label the parameters, variables and nodes of the circuit of figure 11 as shown below.

Figure 12. Equivalent single-phase circuit of the power system of figure 8 (your first example), neglecting the phase shift due to the transformer.

In summary, the circuit of figure 12 is the equivalent single-phase circuit of the power system of figure 8 (which is your first example), if it is acceptable to neglect the phase shift due to the transformer. And such circuit is the proposed by your textbook (second image you posted), which means their solution is correct.

While I just proved the equivalent single-phase circuit of your first example, the same logic applies to your second example.

• "[...] if it is acceptable to neglect the phase shift due to the transformer" // Actually, I think per-phase equivalent circuits never consider the phase shift due to transformers. I wonder if this only affects the angle of the phasors (which isn't a big deal), or if it affects the magnitude of of the phasors (which would be a big deal). I'd appreciate if someone could confirm this. Jan 14, 2021 at 16:42