Insulation Resistance Test and Polarization Index Test

Both Insulation Resistance Test (IR Value Test) and Polarization Index Test (PI Value Test) are conducted on HV machine to determine service condition of the insulation. In HV machines and winding are likely to be affected by moisture and contamination. IP test is conducted specially to determine the dryness and cleanliness of winding insulation.

In insulation resistance test, a high DC voltage is applied across, conductor and ground more specifically. The voltage is applied across the insulator. Due to this applied high DC voltage there will be a current through the electrical insulator. Big dividing the applied voltage by this current we get the actual resistive value of the insulator.

Say the applied high voltage is V and corresponding current through the insulator is I. Hence as per Ohm’s law the value of insulation resistance is

[math] R\;=\;\frac{V}{I}[/math] This test is generally done by means of megger.

Megger gives required direct (dc) voltage across the insulator an it also shows the resistive value of insulator directly in Mohm range.The megger are generally of 500 V, 2.5 KV an 5 KV.
500 V megger are used for insulation test upto 1.1 KV rated insulation. For high voltage transformer, other HV equipment and machines, 2.5 or 5 KV megger are used.
As all insulators are dielectric in nature they have always a capacitive property. Due to that, during application of voltage across the electrical insulator, initially there will be a charging current. But after some time when the insulator is totally charged, the capacitive changing current becomes zero and then only resistive conductive current presents in the insulator. That is why it is always recommended to do insulation resistance test at least for 1 minute as it is proved that charging current totally becomes zero after 1 minute.
Only measuring insulation resistance by megger for 1 minute does not always give reliable result. As the resistive value of an electrical insulator also varies with temperature. This difficulty is partially solved by introducing polarity index test or in short PI value test. The philosophy behind PI test is discussed below.
Actually when a voltage is applied across an insulator there will be a leakage current from line to ground. Although this leakage current is very small is in millampere or sometimes in micro ampere range, but it has mainly four components.
1) Capacitive component.
2) Resistive or conductive component.
3) Surface leakage component.
4) Polarization component.
Let us discuss one by one.

Capacitive Component

When a DC voltage is applied across in insulator, because of its dielectric nature there will be an initial high charging current through the insulator from line to ground. Although this current decays exponentially and becomes zero. Generally this current exists for initial 10 seconds of the test. But it takes nearly 60 seconds to decay totally.

Resistive or Conductive Component

This current is purely conductive in nature flows through the insulator as if the insulator is purely resistive. This is direct flow of electrons. Every insulator should have this component of electric current. Since in practice every material in this universe persists some sensitive nature.

The resistive or conductive component of insulator leakage current remains constant throughout the test.

Surface Leakage Component

Due to dust, moisture and other contaminants on the surface of the insulator, there is one small component of leakage current through the outer surface of the insulator.

Polarization Component

Every insulator is hygroscopic in nature. Some contaminant molecules and mainly moisture in insulator are very polar. When an electric field is applied across insulator the polar molecules align themselves along the direction of electric field. The energy required for this alignment of polar molecules, comes from voltage source in form of electric current. This current is called polarization current. It continues until all the polar molecules allied themselves along the direction of electric field.

It takes around 10 minutes to align the polar molecules along electric field and that is why if we take megger result for 10 minutes, there would be no effect of polarizing in megger result.
So when we take megger value of an insulator for 1 minute, the results reflects, the IR value which is free from effect of capacitive component of leakage current. Again when we take megger value of an insulator for 10 minutes, the megger result shows the 12 value, free from affects of both capacitive component and polarization component of leakage current.
Polarization index is the ratio of megger value taken for 10 minutes to the megger value taken for 1 minute.
The significance of polarization index test. Let I is the total initial current during polarization index test or PI test.
I_C is the capacitive current.
I_R is resistive or conductive current.
I_S is surface leakage current.
I_P is polarization current of the insulator.

[math] Hence,\;\;I\;=\;I_C\;+\;I_R\;+\;I_S\;+\;I_P[/math]
Value of insulation resistance test or IR value test, i.e. value megger reading just after 1 minute of the test, is-

[math] R_1\;=\;\frac{V}{IR\;+\;IS\;+IP}[/math]
Megger value of 10 minute test, is

[math] R_{10}\;=\;\frac{V}{I_R\;+I_S}[/math]
Therefore, result of polarization index test, is

[math] \frac{R_{10}}{R_1}\;=\;\frac{\frac{V}{I_R\;+I_S}}{\frac{V}{IR\;+\;IS\;+IP}}\;\;=\;\frac{I_R\;+I_S\;+I_P}{I_R\;+I_S}\;\;=\;1\;+\;\frac{I_P}{I_R\;+I_S}[/math]
From the above equation it is clear that, if the value of (I_R + I_S) >> I_P, the PI of insulator approaches to 1. And large I_R or I_S or both indicate unhealthiness of the insulation.
Th value of IP becomes high if (I_R + I_S) is very small compared to I_P. This equation indicates that high polarization index of an insulator implies healthiness of insulator. For good insulator resistive leakage current I_L is very tiny.

It is always desired to have polarization index of an electrical insulator more than 2. It is hazardous to have polarization index less than 1.5.