## What is the Percentage Impedance?

The percentage impedance of a transformer is marked on most nameplates – but what is it and what does the Z% figure mean?

The impedance of a transformer is the total opposition offered to alternating current. This may be calculated for each winding.

However, a rather simple test provides a practical method of measuring the equivalent impedance of a transformer without separating the impedance of the windings.

When referring to the impedance of a transformer, it is the equivalent impedance that is meant.

### Definition

The** percentage impedance** **of a transformer** is the volt drop on full load due to the winding resistance and leakage reactance expressed as a percentage of the rated voltage.

It is also the percentage of the normal terminal voltage required to circulate full-load current under short circuit conditions.

In other words, **percentage impedance of a transformer** is the percentage of the rated voltage applied at one side (primary winding) to circulate rated current on transformer keeping its other side (secondary winding) under short circuit conditions.

It is marked on transformer nameplate in every electrical substation.

## Explanation of Percentage Impedance

If we apply rated voltage at primary winding of a transformer keeping its secondary winding short-circuited, then amount of current at both windings will be extremely high as compared to the rated current.

This current is called short circuit current and its magnitude is very high due to zero impedance offered by the load (secondary winding is short-circuited).

Now, if we reduce the applied voltage on the transformer primary i.e. we apply a percentage of rated voltage in transformer primary, current on both windings will also reduce.

At a particular percentage of rated voltage, the rated current will flow on transformer windings. This percentage of rated voltage at one side of the transformer which circulates rated current on transformer windings keeping its other side winding short-circuited is called **percentage impedance of the transformer**.

## Calculation of Percentage Impedance

In order to determine equivalent impedance, one winding of the transformer is short-circuited, and just enough voltage is applied to the other winding to create full load current to flow in the short-circuited winding.

This voltage is known as the **impedance voltage**.

Either winding may be short-circuited for this test, but it is usually more convenient to short circuit the low-voltage winding.

*The transformer impedance value is given on the nameplate in percent. This means that the voltage drop due to the impedance is expressed as a percent of rated voltage. *

### Example Calculation

For example, if a 2,400/240-volt transformer has a measured impedance voltage of 72 volts on the high voltage windings, its impedance (Z), expressed as a percent, is:

Z% = (Impedance Voltage / Rated Voltage) x 100

percent Z = (72/2400)*100 = 3 percent

*This means there would be a 72-volt drop in the high-voltage winding at full load due to losses in the windings and core. Only 1 or 2% of the losses are due to the core; about 98% is due to the winding impedance. *

Z = V/I

If the full load current is 10 amps:

Z = 72V/10A = 7.2 Ohms

### Changing the Percentage Impedance Value

The leakage flux is a function of winding ampere-turns and the area and length of the leakage flux path.

These can be varied at the design stage by changing the volts per turn and the geometric relationship of the windings.

### The Effect of Higher and Lower Percentage Impedances

It is easy to calculate the maximum current that a transformer can deliver under symmetrical fault conditions.

By way of example, consider a 2 MVA transformer with an impedance of 5%. The maximum fault level available on the secondary side is:

2 MVA x 100/5 = 40 MVA

and from this figure, the equivalent primary and secondary fault currents can be calculated.

Dylan says

May 18, 2019 at 6:57 amPlease rephrase the beginning sentence “the total opposition offered an alternating current. ” I believe it does not make grammatical sense.

Using above recommendation to short circuit the secondary side, the percentage impedance is calculated to be 3%. Will this number also applies to the primary side?

When carrying out this test, will we expect also full load current on the short circuited secondary side? How long should this test last in order to measure the full load current on the primary side?

Thank you.