Transformer equivalent circuit

A real transformer is two windings coupled by a shared iron core — resistance, leakage flux, and core loss all tangled together with a magnetic linkage that ordinary circuit theory does not know how to handle. The equivalent circuit cuts through this: it replaces the magnetic coupling with an ideal turns-ratio transformer plus a handful of ordinary resistors and reactances, then eliminates even the ideal transformer by referring every secondary quantity to the primary side (or vice versa). What is left is a plain phasor circuit — a series impedance and a shunt branch — that you can analyse with the same Kirchhoff's-law tools used for any R-L-C network. This is not a side topic: every later calculation in the unit, from voltage regulation to interpreting the open-circuit and short-circuit tests, is built directly on the referred impedances $R_{01}, X_{01}, Z_{01}$ derived here.

The core insight

Referring the secondary's resistance and leakage reactance to the primary by $a^2$ (turns ratio $a = N_1/N_2$ ) collapses the magnetically coupled transformer into a single-sided series impedance $Z_{01}=R_{01}+jX_{01}$ plus a shunt magnetising branch, turning it into an ordinary AC circuit.

The core idea

What each element models. Every winding has ohmic resistance ( $R_1$ primary, $R_2$ secondary) that dissipates $I^2R$ heat, and leakage reactance ( $X_1$ , $X_2$ ) that represents the small fraction of each winding's flux that fails to link the other winding — it produces a voltage drop proportional to current but contributes nothing to the mutual EMF. The core itself is represented by a shunt branch across the primary EMF point: a resistance $R_0$ carrying the core-loss (working) current $I_w$ , in parallel with a reactance $X_0$ carrying the magnetising current $I_\mu$ . Between the two windings' leakage impedances sits an ideal transformer of turns ratio $N_1:N_2$ — pure ratio, no losses, no leakage.

Eliminating the ideal transformer. The ideal-transformer block can be removed entirely by referring all secondary quantities to the primary side, using $a = N_1/N_2$ :

R_2' = a^2R_2,\qquad X_2' = a^2X_2,\qquad Z_2' = a^2Z_2

where: impedances (resistance and reactance) referred from secondary to primary scale by $a^2$ , voltages scale by $a$ , and currents scale by $1/a$ — the inverse of the impedance scaling, since impedance is voltage over current. Referring the other way (primary to secondary) simply divides by $a^2$ instead. This single rule is what lets a magnetically coupled device be redrawn as one ordinary loop.

Combined referred impedance. For regulation, efficiency, and short-circuit-test calculations it is standard to lump the primary leakage impedance and the referred secondary leakage impedance into one series impedance, referred to the primary:

R_{01} = R_1 + a^2R_2, \qquad X_{01} = X_1 + a^2X_2, \qquad Z_{01} = R_{01}+jX_{01},\ \ |Z_{01}|=\sqrt{R_{01}^2+X_{01}^2}

where: $R_{01}, X_{01}$ are the total resistance and leakage reactance of both windings, expressed entirely in primary-side ohms, and $Z_{01}$ is the phasor sum — resistance and reactance are in quadrature, so they combine by Pythagoras, not by plain addition. The secondary-referred set is obtained by dividing through by $a^2$ : $R_{02}=R_{01}/a^2$ , $X_{02}=X_{01}/a^2$ , $Z_{02}=Z_{01}/a^2$ .

Figure

Transformer equivalent circuit showing primary resistance and reactance in series then a shunt magnetising branch then the referred secondary impedance leading to the load

Derivation of referred and combined impedances

Derivation AReferring impedance across the ideal transformer to build R01, X01, Z01

The goal is to show, starting from the ideal-transformer relations between primary and secondary voltage and current, exactly why an impedance sitting on the secondary side must be multiplied by $a^2$ when it is redrawn on the primary side, and then to combine the two windings' impedances into the single referred quantity $R_{01}, X_{01}, Z_{01}$ that every later regulation and OC/SC-test calculation depends on.

Step 1. Start from the ideal-transformer relations. For an ideal transformer of turns ratio $a = N_1/N_2$ , primary and secondary EMFs are related by $E_1 = aE_2$ , and primary and secondary currents (ideal, lossless) by $I_1 = I_2/a$ — voltage steps by $a$ , current steps by $1/a$ , so that power $E_1I_1 = E_2I_2$ is conserved across the ideal block.

Step 2. Write the secondary leakage impedance as seen from the secondary terminals. The secondary winding's own impedance is $Z_2 = R_2 + jX_2$ , and the voltage it drops when carrying $I_2$ is $I_2Z_2$ .

Step 3. Reflect that drop to the primary side. Because the ideal transformer scales voltage by $a$ and current by $1/a$ , an impedance $Z_2$ on the secondary appears, from the primary's viewpoint, as $Z_2' = (aE_2\text{-side voltage change})/(I_2/a\text{-side current})$ , which works out to $Z_2' = a^2Z_2$ — the same relation used for $R_2'=a^2R_2$ and $X_2'=a^2X_2$ above. The $a^2$ arises because voltage and current both scale, once in the numerator and once (inverted) in the denominator of $Z=V/I$ .

Step 4. Add the referred secondary impedance to the primary's own impedance, since after referring, both sit on the same side and in series along the same current path: $R_{01} = R_1 + a^2R_2$ and $X_{01} = X_1 + a^2X_2$ .

Step 5. Combine resistance and reactance as a phasor, not arithmetically, because they act at 90 degrees to each other on the current: $Z_{01} = \sqrt{R_{01}^2+X_{01}^2}$ . Check: dimensionally $a^2$ is a pure number (ratio of turns squared), so $R_{01}, X_{01}$ carry ohms just like $R_1, X_1$ — consistent with the worked example below, where $R_{01}=0.38\ \Omega$ sits between $R_1=0.2\ \Omega$ and the referred $R_2'=0.18\ \Omega$ that were added to make it.

Worked example

A transformer has primary resistance $R_1 = 0.2\,\Omega$ , primary leakage reactance $X_1 = 0.5\,\Omega$ , secondary resistance $R_2 = 0.005\,\Omega$ , secondary leakage reactance $X_2 = 0.013\,\Omega$ , and turns ratio $a = N_1/N_2 = 6$ . Find $R_{01}, X_{01}, Z_{01}$ (referred to primary) and $R_{02}, X_{02}$ (referred to secondary).

Step 1. Refer the secondary resistance and reactance to the primary by multiplying by $a^2 = 36$ :

R_2' = 36\times0.005 = 0.18\ \Omega, \qquad X_2' = 36\times0.013 = 0.468\ \Omega

Step 2. Add to the primary's own winding values to get the total primary-referred series impedance:

R_{01} = 0.2+0.18 = 0.38\ \Omega, \qquad X_{01} = 0.5+0.468 = 0.968\ \Omega

Z_{01} = \sqrt{0.38^2+0.968^2} = \sqrt{0.1444+0.9370} = \sqrt{1.0814} \approx 1.040\ \Omega

Step 3. Convert to secondary-referred values by dividing by $a^2=36$ :

R_{02} = \frac{0.38}{36} \approx 0.01056\ \Omega, \qquad X_{02} = \frac{0.968}{36} \approx 0.02689\ \Omega

The referred series impedance is $R_{01}=0.38\ \Omega$ , $X_{01}=0.968\ \Omega$ , $Z_{01}\approx 1.040\ \Omega$ (equivalently $R_{02}\approx0.01056\ \Omega$ , $X_{02}\approx0.02689\ \Omega$ referred to secondary). The small secondary-referred values are typical of a step-down transformer, whose secondary has fewer turns and hence smaller ohmic impedance in absolute terms.

Competitive stretch

A 20 kVA, 2000 V/200 V single-phase transformer has $R_{01} = 2.0\,\Omega$ , $X_{01} = 3.0\,\Omega$ . At rated full-load current and unity power factor, find the primary-side voltage drop across $Z_{01}$ , in volts and as a percentage of rated primary voltage. Rated primary current is $I_{1,rated} = S/V_1 = 20{,}000/2000 = 10\ \text{A}$ . At unity pf ( $\cos\theta=1,\sin\theta=0$ ) the approximate drop $I(R_{01}\cos\theta+X_{01}\sin\theta)$ reduces to $IR_{01} = 10\times2.0 = 20\ \text{V}$ , which is $20/2000\times100 = 1\%$ of rated primary voltage — a small drop that feeds directly into the voltage-regulation calculation covered next.

Common traps

Forgetting to square the turns ratio for resistance/reactance ( $\times a^2$ ) while voltage refers by $\times a$ and current by $\div a$ — mixing these three rules up is the single most common error.
Referring in the wrong direction — sanity-check: referring secondary quantities to the primary of a step-down transformer ( $a>1$ ) should make them larger, not smaller.
Adding $R_{01}$ and $X_{01}$ arithmetically instead of combining them as $\sqrt{R_{01}^2+X_{01}^2}$ — they are in quadrature, not in series algebraically.
Assuming the magnetising branch can always be dropped — it is a fair approximation for full-load series-drop calculations, but it is exactly what the open-circuit test measures, and it cannot be ignored for core loss or no-load current.

Ask the tutor: If the $a$ versus $a^2$ referral rule or the step from $R_{01}, X_{01}$ to the voltage-regulation formula is still unclear, ask the AI tutor to walk through a few quick referral drills or connect $Z_{01}$ to how the short-circuit test measures it directly.

Why does referring an impedance use a squared, but referring voltage and current use a to the first power?

Impedance is voltage divided by current, $Z=V/I$ . Voltage referred from secondary to primary scales by $a$ (multiply), while current scales by $1/a$ (divide). So an impedance built from those two scales as $a/(1/a) = a^2$ . It is not an extra rule — it falls straight out of Ohm's law applied to quantities that already scale by $a$ and $1/a$ .

Can I just ignore the magnetising branch R0, X0 and only use R01, X01 for every calculation?

Only for full-load series-drop type calculations such as regulation, where the no-load magnetising current is small compared to full-load current, so dropping the shunt branch barely changes the answer. But the magnetising branch is exactly what the open-circuit test measures (core loss and no-load current), so for OC-test interpretation, no-load performance, or core-loss questions you must keep R0 and X0 in the picture.

Does it matter whether I refer everything to the primary or to the secondary?

No — both give the same physical circuit, just with the numbers scaled by $a^2$ (impedances), $a$ (voltages) or $1/a$ (currents). Referring to primary is the more common convention in this course because supply-side quantities (like rated primary voltage) are usually the given data, but the secondary-referred set $R_{02}, X_{02}, Z_{02}$ is equally valid and is simply $R_{01}, X_{01}, Z_{01}$ divided by $a^2$ .