% Impedance is a very important parameter of a transformer.For distribution transformers it is generally between 4% to 6% and for power transformers is in the range of 6% to 8%.

If the %impedance is more,there is more voltage drop in the transformer that means for distribution transformers from where the consumers are connected,lesser voltage drop is preferred and hence select a transformer with lesser value of % impedance.

The % impedance is useful in calculating the short circuit current on the secondary side of the transformer and is given by

Short circuit current in KVA = Transformer rating KVA/per unit impedance

For example if the rating of transformer is 500 KVA and %impedance is 5% or 0.05 p.u.

Short circuit rating in KVA = 500/0.05 =10,000 KVA or 10 MVA

Knowing the short circuit current is important to select circuit breaker breaking capacity,rating of bus bars,relay settings etc and for calculation of cable sizes.

Short answer: Each winding of a transformer has a resistance and a self-inductance. So in the phasor domain, each winding has a complex impedance. The combination of the complex impedances of the two windings, where one impedance must be referred to the other side, is known as the equivalent impedance of the transformer, and depends on the side of the transformer it is referred to. This equivalent impedance can be expressed as a per-unit of the base impedance of the transformer, and the result is also a complex number known as the per-unit equivalent impedance of the transformer, and is independent of the side of the transformer it is referred to. The magnitude of the per-unit equivalent impedance of the transformer, expressed as percentage (i.e. multiplied by 100), is known as the percentage impedance of the transformer.

Long answer: To understand the above paragraph, you need to know 1) the equivalent circuit model of a single-phase two-winding transformer, including the meaning of each element of such model; and 2) the per-unit system. Below I’ll do a review of those concepts. Then I’ll illustrate with an example.

1. The equivalent circuit of a transformer

Below I show the equivalent circuit of a practical single-phase two-winding transformer operating at low frequencies in sinusoidal steady-state. Let’s decompose the previous phrase so that we understand the limitations of the following equivalent circuit:

• The word equivalent means it is a mathematical model [with errors] of a physical system.
• The phrase single-phase means it has only one set of windings, whose voltages are either in phase or phase-shifted by 180° (depending on the reference polarity of the voltages and the relative orientation of the windings i.e. the dots).
• The phrase two-winding means it has two windings for each set of windings.
• As a side note, there are three-phase two-winding transformers, and three-phase three-winding transformers.
• The phrase at low frequencies means at 50 Hz or 60 Hz (the model of a radio-frequency transformer is different, for example it includes capacitance between both windings).
• The phrase in sinusoidal steady state means the voltages and currents waveforms are sinusoidal (the model of a transformer for harmonics analysis is different, and the model is also different during transient state i.e. during inrush current when the transformer is first connected to the power grid).

Figure 1. Equivalent circuit, complete, exact, in real values, without eliminating the ideal transformer. Source: [1].

First, let’s quickly remember each of the elements of the above model:

• The resistances [math]R_p[/math] and [math]R_s[/math] represent the AC resistance or apparent resistance of the conductors of the windings; it is higher than the DC resistance or effective resistance due to the skin effect and proximity effect.
• The inductive reactances [math]X_p[/math] and [math]X_s[/math] represent the self-inductance of each winding, which is due to the magnetic field by the current in that winding that doesn’t link to the other winding.
• The shunt branch or parallel branch has two purposes, the first being to account for the fact that in a practical transformer, one side draws current (when it is energized/connected to a voltage source) even when the other side is open-circuited, which doesn’t happen in an ideal transformer; the second purpose of the shunt branch is to account for [math]R_N[/math] and [math]X_M[/math].
• The resistance [math]R_N[/math] represents the iron core losses, that is, losses due to eddy currents and due to hysteresis in the ferromagnetic core.
• The inductive reactance [math]X_M[/math] represents the magnetizing current, which has harmonics but its fundamental components lags the applied voltage by 90°.
• Keep in mind the hysteresis of a ferromagnetic core is a non-linear phenomena; that is, the hysteresis loop relating the exciting current and the flux is non-linear. Moreover, it isn’t even a single-valued function; that is, for a given value of the exciting current, the flux can have two values, one when it is rising and another when it is decreasing. Furthermore, the shape of the hysteresis loop depends on the peak value of the exciting current and the flux, meaning we can’t determine on which hysteresis loop from a family of hysteresis loops the core is operating until we know the peak value of either the exciting current or the flux. All of these difficulties are neglected when modeling this phenomena as a constant resistance [math]R_N[/math].
• Also keep in mind the magnetizing current is non-sinusoidal even if the applied voltage is sinusoidal. So it has harmonics. But they are neglected when modeling this phenomena as a constant inductive reactance [math]X_M[/math].
• The reason why [math]R_N[/math] and [math]X_M[/math] are placed in parallel in the model is because the magnetizing current (in the non-saturated region of the core) and the core losses current are proportional to the applied voltage.

2. Simplified equivalent circuits of a transformer

Secondly, let’s remember how we manipulate the above equivalent circuit to obtain others.

We can simplify this circuit by referring one side to the other. For that, remember that if the turns ratio [math]a[/math] is defined as [math]N_p/N_s[/math], then to eliminate the transformer and refer the secondary side into the primary side, we use:

[math]Z_s’ = a^2 Z_s; {\mathbf V}_s’ = a {\mathbf V}_s; {\mathbf I}_s’ = \dfrac{{\mathbf I}_s}{a} \tag*{}[/math]

while to eliminate the transformer and refer the primary side into the secondary side, we use:

[math]Z_p’ = \dfrac{Z_p}{a^2}; {\mathbf V}_p’ = \dfrac{{\mathbf V}_p}{a}; {\mathbf I}_p’ = a {\mathbf I}_s \tag*{}[/math]

where

[math]Z_s = R_s + j X_s; Z_p = R_p + j X_p \tag*{}[/math]

With the above two rules, we can refer the impedances, voltages and currents from one side to the other in the circuit of figure 1:

Figure 2. Equivalent circuit, complete, exact, in real values, with ideal transformer eliminated and referred to the primary side. Source: [1].

Figure 3. Equivalent circuit, complete, exact, in real values, with ideal transformer eliminated and referred to the secondary side. Source: [1].

Under full-load, the primary and secondary currents of a transformer are much greater than the exciting current. So, we can move the shunt branch to the left (before the series impedance of the primary), which results in an approximated circuit as shown below, but whose results are very close to the exact model of figures 1, 2 and 3. After moving the shunt branch, we can combine the series impedances of both windings into one (or we can just combine the series resistances into one and the series reactances into one, without combining the equivalent resistance and equivalent reactance into one):

Figure 4. Equivalent circuit, complete, approximated (with excitation branch moved), in real values, with ideal transformer eliminated and referred to the primary side. Source: [1].

Figure 5. Equivalent circuit, complete, approximated (with excitation branch moved), in real values, with ideal transformer eliminated and referred to the secondary side. Source: [1].

As mentioned earlier, the exciting current is very small (less than about 5%) of the total current. In power systems analysis, the excitation branch is neglected altogether, so the complete exact circuit of figures 2 and 3 (and the complete approximated circuit of figures 4 and 5) becomes:

Figure 6. Equivalent circuit, incomplete (with excitation branch neglected) and thus approximated, in real values, with ideal transformer eliminated and referred to the primary side. Source: [1].

Figure 7. Equivalent circuit, incomplete (with excitation branch neglected) and thus approximated, in real values, with ideal transformer eliminated and referred to the secondary side. Source: [1].

In the previous figure, the equivalent series resistance can be combined with the equivalent series reactance:

[math]\begin{align} Z_{\text{eq } p} &= R_{\text{eq } p} + j X_{\text{eq } p} \\ &= (R_p + a^2 R_s) + j(X_p + a^2 X_s) \end{align} \tag*{}[/math]

[math]\begin{align} Z_{\text{eq } s} &= R_{\text{eq } s} + j X_{\text{eq } s} \\ &= \left(\dfrac{R_p}{a^2} + R_s \right) + j \left(\dfrac{X_p}{a^2} + X_s \right) \end{align} \tag*{}[/math]

In certain power systems studies, for large transformers, even the equivalent series resistance is neglected (because it is small compared to the equivalent series reactance), and the transformer is represented by a mere equivalent series reactance. But I won’t consider this to be the case, since that’s a particular case.

The takeaway of the above discussion is the equivalent circuit of figures 4 and 5, and the meaning of [math]Z_{\text{eq } p}[/math] and [math]Z_{\text{eq } s}[/math]:

• [math]Z_{\text{eq } p}[/math] is the series impedance of the primary winding combined with the series impedance of the secondary winding referred to the primary winding, expressed in real values (ohms).
• [math]Z_{\text{eq } s}[/math] is the series impedance of the primary winding referred to the secondary winding combined with the series impedance of the secondary winding, expressed in real values (ohms).

3. The per-unit measuring system for single-phase transformers

Third and lastly, let’s remember the per-unit system. The per-unit value of a quantity (current [phasor or just magnitude], voltage [phasor or just magnitude], active power, reactive power, apparent power, complex power, impedance [complex or just magnitude], admittance [complex or just magnitude]) is the real value (in amperes, volts, watts, volt-amperes, volt-amperes reactive, ohms or siemens) of the quantity expressed as a fraction of another value (with the same units). That another value is known as the base value. The result is a dimensionless number.

Notice the per-unit value corresponding to a real value depends on the value chosen as base. Any arbitrary number can be chosen as base value, but once a base value has been chosen for two electric quantities, the base value of the rest of the electric quantities is determined by the circuit equations.

In transformers, the base values for the primary and secondary voltages are chosen, and for the apparent power. These are not arbitrarily chosen as explained above, but instead are chosen to be the same as the rated primary and rated secondary voltages and the rated apparent power. The reason is that by choosing the base primary voltage and base secondary voltage such that their ratio is the same as the ratio of the rated primary voltage and rated secondary voltage, it can be mathematically shown the turns ratio in the equivalent per-unit circuit, [math]a'[/math], is equal to 1, which means the circuits of figures 4 and 5 are simplified and become the following:

Figure 8. Equivalent circuit, complete, approximated (with excitation branch moved), in per-unit values, with ideal transformer eliminated and referred to the primary side. Source: [1], edited.

Figure 9. Equivalent circuit, complete, approximated (with excitation branch moved), in per-unit values, with ideal transformer eliminated and referred to the secondary side. Source: [1], edited.

After that, the base values for the per-unit currents and per-unit impedances are calculated using the base values for the per-unit voltages and per-unit apparent power and using the circuit equations. And after that, the equivalent series impedance of the transformer is converted from a real value to a per-unit value by dividing it by the computed base impedance.

Example: Consider a single-phase two-winding power transformer whose ratings are 20 kVA, 8 000 V/240 V. After performing the short-circuit test and open-circuit test, it is found the equivalent series impedance referred to the primary side (high-voltage side) is [math]Z_{\text{eq } p} = (38.4 + j 192) \, \Omega[/math] (that is, [math]R_{\text{eq } p} = 38.4 \, \Omega[/math] and [math]X_{\text{eq } p} = 192 \, \Omega[/math]). Determine the series impedance in per-unit, and the percentage impedance and [math]X/R[/math] ratio of the transformer.

Solution: The base value for the primary and secondary voltages are chosen to be the same as the rated voltages, that is:

[math]V_\text{b,p} = 8 \, 000 \text{ V} \tag*{}[/math]

[math]V_\text{b,s} = 240 \text{ V} \tag*{}[/math]

The base value for the apparent power is chosen to be the same as the rated power, that is:

[math]S_\text{b} = 20 \, 000 \text{ VA} \tag*{}[/math]

Now we use the circuit equations to obtain the base value of the other electrical quantities. For the base value of the impedance in the primary side:

[math]\begin{align} S &= \dfrac{V^2}{Z} \\ \implies S_\text{b} &= \dfrac{V_\text{b,p}^2}{Z_\text{b,p}} \\ \implies Z_\text{b,p} &= \dfrac{V_\text{b,p}^2}{S_\text{b}} \\ &= \dfrac{(8 \, 000 \text{ V})^2}{(20 \, 000 \text{ VA})} = 3 \, 200 \, \Omega \end{align} \tag*{}[/math]

and for the base value of the impedance in the secondary side:

[math]\begin{align} Z_\text{b,s} &= \dfrac{V_\text{b,s}^2}{S_\text{b}} \\ &= \dfrac{(240 \text{ V})^2}{(20 \, 000 \text{ VA})} = 2.88 \, \Omega \end{align} \tag*{}[/math]

Now that we’ve determined the base values, we can compute the per-unit values by dividing the real values by the base values. In this example, it is stated the given equivalent series impedance is referred to the primary side, so we must use the base value of the impedance in the primary side to find the per-unit value of the equivalent series impedance:

[math]\begin{align} Z_{\text{eq } p \text{,pu}} &= \dfrac{Z_{\text{eq } p}}{Z_\text{b,p}} \\ &= \dfrac{(38.4 + j 192) \, \Omega}{3 \, 200 \, \Omega} = (0.012 + j0.06) \text{ pu} \end{align} \tag*{}[/math]

So, 0.012 + j0.06 is the per-unit equivalent series impedance of the transformer referred to the primary side.

And what about the per-unit series impedance referred to the secondary side? For that, we must first refer the given series impedance to the secondary side:

[math]\begin{align} Z_{\text{eq } s} &= \dfrac{Z_{\text{eq } p}}{a^2} \\ &= \dfrac{(38.4 + j 192) \, \Omega}{ \left( \dfrac{8 \, 000 \text{ V}}{240 \text{ V}} \right)^2} = (0.03456 + j0.1728) \, \Omega \end{align} \tag*{}[/math]

thus, the per-unit equivalent series impedance of the transformer referred to the secondary side is:

[math]\begin{align} Z_{\text{eq } s \text{,pu}} &= \dfrac{Z_{\text{eq } s}}{Z_\text{b,s}} \\ &= \dfrac{(0.03456 + j0.1728) \, \Omega}{2.88 \, \Omega} = (0.012 + j0.06) \text{ pu} \end{align} \tag*{}[/math]

which is the same as the per-unit equivalent series impedance of the transformer referred to the primary side. This shows the per-unit equivalent series impedance of a transformer is the same irrespective of the side. So, when specifying the equivalent series impedance of a transformer in per-unit, it’s not necessary to state if it is referred to the high-voltage or low-voltage side, or the primary or secondary side. So we can drop the subscript “p” and “s” in the per-unit equivalent series impedance.

For the given transformer, it is possible to find the percentage impedance and the [math]X/R[/math] ratio. For the percentage impedance, first we express the per-unit impedance ([0.012 + j 0.06] pu) in polar form:

[math]\begin{align} Z_{\text{eq} \text{,pu}} &= \sqrt{0.012^2 + 0.06^2} \, \angle \tan^{-1}{(0.06/0.012)} \text{ pu} \\ &\approx 0.0612 \angle 78.69^\circ \text{ pu} \end{align} \tag*{}[/math]

Now we multiply the magnitude of the per-unit impedance by 100 to express it as a percentage. The result will be the percentage impedance of the transformer:

[math]\%|Z_\text{eq}| = |Z_{\text{eq} \text{,pu}}| \cdot 100 = 0.0612 \text{ pu} \cdot 100 = 6.12\% \tag*{}[/math]

The [math]X/R[/math] ratio of the transformer is:

[math]\dfrac{X_{\text{eq} \text{,pu}}}{R_{\text{eq} \text{,pu}}} = \dfrac{0.06\text{ pu}}{0.012\text{ pu}} = 5 \tag*{}[/math]

4. References

[1] Chapman, Stephen. Electric Machinery Fundamentals (4th edition). Chapter 2.

[2] Massachusetts Institute of Technology. Magnetic Circuits and Transformers: a first course for power and communication engineers. Chapter VI.

[3] Alexander, Charles K.; O. Sadiku, Matthew N. Fundamentals of Electric Circuits (5th edition). Chapter 13.

[4] Kothari, D. P.; Nagrath, I. J. Electric Machines (4th edition). Chapter 3.

We have Short circuit test to calculate the impedance of the transformer.

For this, low voltage side of the transformer is short circuited, an auto transformer and voltmeter is connected to higher voltage side.

An ammeter is connected to the lower voltage side(short circuited side).

Initially the auto transformer is gives ZERO voltage.

Now voltage is increased gradually.

The ammeter in the secondary side ( Low voltage side in our case) will will sense some current. Now voltage is increased by adjusting the auto transformer till the ammeter on other side shows the normal full load current.

Now this is the point where we can calculate the impedance.

Lets have an example.-

Transformer rating 10 KVA

Single phase primary voltage- 480 V.

Secondary voltage - 240 V

Full load current - (10 kVA X 1000)/(240 V)

- 41.67 A

Now the experiment- Assume the primary voltage injected is 24 V at which the ammeter reads 41.7 A (Transformer full load current).

Impedance is ratio of 24 V to 480 V,

i.e. Impedance Z = 24/480 =0.05.

But the impedance of transformer is expressed in %, So the

% impedance is 0.05 X 100 =5%.

Hope it helps.

In general, it increases but it depends.

A transformer’s impedance as given on the nameplate is actually a measure of the magnitude of the winding resistance in series with the inductive reactance caused by the leakage flux. There are also core loss and magnetizing components but I’m skipping a discussion of those for the purpose of this answer.

An ideal transformer has perfect magnetic coupling between all windings with no leakage flux. The impedance would, therefore, only be the winding resistance.

For a non-ideal transformer, the magnetic coupling is not perfect due to leakage flux and this causes the magnitude of the impedance to be greater. How much greater depends on many factors like the type of core, how the windings are arranged, the geometry of the coils, the shape of the wire, the amount and type of insulation used, and other factors.

For large power transformers that operate at high voltages, the impedance tends to increase with the physical size of the coils. For example, small distribution transformer banks like this

will typically have a small impedance of about 2%.

For large substation transformers that operate at high voltage levels,

the impedance will be much greater, typically over 9%, and will increase as the geometry and size of the coils increase.

In the power industry, the percent-impedance as stamped on the transformer nameplate is actually known as the impedance-voltage and is defined to be the voltage necessary to cause rated current to flow in the high-voltage winding when a short-circuit is placed on the low-voltage winding.

For example, given a full-load apparent power rating of 100 MVA and 138 kV line-to-line, the full load current will be 418 A in the high-voltage winding. If, during the short-circuit test, a voltage of 13.8 kV was required to achieve 418 A, the impedance-voltage would be 10%. Given the base impedance of 190 ohms as calculated from the nameplate ratings above, the transformer impedance would be 19 ohms (i.e., 10% of the base). This impedance is a single-phase equivalent. It is equal to the magnitude of the equivalent resistance in series with the inductive reactance of the leakage flux of both the primary and secondary windings. This is typically given by an x/r ratio. For small transformers, the x/r ratio is very low like 0.9 or less. For large transformers, the x/r ratio is high like 10 or more. So, for the large substation transformers, the leakage reactance is approximately equal to the impedance magnitude.

Footnotes

Impedance of transformer, expressed as the percentage of rated impedance of load at full rated current, is called percentage impedance. This is a very useful quantity for transformer calculations.

This impedance is calculated from short circuit test data, when a small primary voltage produces full load current in secondary of transformer. This small voltage on primary side, expressed as percentage of full rated voltage , is called percentage impedance.

This is useful quantity from which short circuit KVA or MVA can be easily calculated. In case of fault, maximum MVA that can flow in the cable is transformer rating divided by short circuit impedance.

If the transformer is applied to a audio speaker, which connected in the secondary winding, the impedance seen into the transformer at the secondary winding should be similar to that of the speaker. That is impedance matching. It was very important in audio amplifying with vacuum tubes.

If [math]n1[/math] is the number of turns in the primary of the transformer, and [math]n2[/math] the number of secondary turns, then the impedance [math]Z2[/math] seen at the secondary is

[math]Z2=(n2/n1)^2Z_{out}[/math]

where [math]Z_{out}[/math] is the output impedance of the amplifier driving the primary.

Seeing the impedance [math]Z1[/math] in the primary, and if [math]Z_S[/math] is the speaker’s impedance:

[math]Z1=(n1/n2)^2Z_S[/math]

Below is an example. The primary has 180 turns. The secondary, as connected, has 60 turns applied to the speaker (there are several terminals in the auto-transformer to conform to several speaker impedances). The speaker has 1.77 Ohm; the impedance seen at the primary is

[math]Z1=(180/60)^2 * 1.77=9*1.77=15.93\approx 16[/math] Ohm

which is what we need in the primary (see the legend).

In other power transformer applications, such as in power supplies, we are interested usually in voltage levels, which are calculated with the simple ratios [math]n1/n2[/math] or [math]n2/n1[/math], depending on the transformer port we use as the primary.

In these applications, impedance is not important, because the “internal” impedance of a mains connection 220V/50Hz or 110V/60Hz is virtually zero, so the impedance seen at the secondary will be also (almost) zero (copper lines have finite resistance…) and this “zero” is independent of the ratio of the turns.

HTH

It should be percentage impedances on their respective bases should be same. The following explanation may take some time as we go from most simple situation with our imagination until the mathematical proofs for transformers through a series of analysis. And I'm sorry for those quality less photographs being uploaded.

Every practical electrical device has some resistance for dc (electrons traveling in a conducting path are always opposed by static atoms in the material)and some impedance for ac (charge movement is always associated with production of some magnetic field around and if it has time varying in nature this flux opposes its own cause resulting in inductance or reactance (impedance =resistance + j*reactance ; reactance =2πf*inductance))

This resistance or impedance is always considered in series with the source.

Consider two batteries are connected in parallel to feed a single load. In parallel operation the voltages are to be same (to avoid circulating currents).

Case 1: Both the batteries (identical)have same internal resistance then the load current would be shared equally between the two i.e., I1=I2=Il/2.

Case 2: Let R1>R2 then I1<I2 i.e., battery with higher internal resistance feeds less load current. This is a disadvantageous as one battery drains out earlier than the other.

Case 3: Suppose consider a situation where battery1 has higher ampere-hour rating than the battery2. In this condition for both the batteries to drain at the same time battery with higher rating should feed more current than the other matching their proportion to the respective power or ampere-hour ratings. This can be achieved using different internal resistances as shown below

I1/I2 = P1/P2

=> {(E-Vo)/R1}÷{(E-Vo)/R2} = P1/P2

=> R2/R1 = P1/P2

=> R1*P1 = R2*P2

Dividing both sides with voltage sqaure

=>{ R1*P1}÷E^2 = {R2*P2}÷E^2

=> R1 ÷ {E^2/P1} = R2 ÷ {E^2/P2}

=> R1 ÷ (R1 base) = R2 ÷ (R2 base).

=> R1 (pu) = R2 (pu)

Case 4: Two transformers of different rating are operated in parallel.

The above analyses can be related with some considerations.

Note: Electrical devices like transformers should not be operated beyond their rated capacities as their life reduces drastically during that operation.

Suppose transformers with different ratings and different pu impedances are operated in parallel then the transformer with lesser pu impedance may get overloaded much before the set raching their sum of rated capacities. Hence the transformer set is derated (as the combined rating is lesser than the sum of their individual ratings)

Read this case 4 for 3–4 times if you don't understand at once.