All Electric Insulations are dielectric in Nature and have a Capacitive Property.

During application of Voltage across the Electrical Insulation, initially there will be a charging current After few instants, the Insulation becomes totally charged and the Capacitive Chrging Current becomes zero.

Hence, it is recommended tp measure insulation resistance after 1 Minute from instant of application of Voltage.

When we apply a Voltage across an Insulation, there will be a corresponding current through it. Although this current is very small (in milliampres).

It has mainly 4 components:

1) Ic, Capacitive Current
2) IR, Conductive Component (Resistive/Leakage Current)
3) IS, Surface Leakage Current (Due to deposition of the dust on the Insulator)
4) IP, Polarisation component (Polarisation Current)

Significance of PI Test:

Let I be the Total Initial Current the PI Test.

Value of the IR just after 1 minute of the IR Test is, R1= V
IR + IS + IP

Megger Value after the 10 minutes Test is: R10= V

PI = V / (IR + IS) = 1 + IP
V / (IR + IS + IP) IR + IS

If (IR + IS) >> IP, the PI of the Insulator approches to 1 and large IR or IS or both ndicate unhealthiness f the insulation.

The Value of the PI becomes high, if (IR + IS) is very small compared to IP. Thus high PI implies that the given insulator is Healthy.

For good insulator, Resistive Leakage current, IR is very small.

PI should be more than 2 and it is very hazardous if it becomes less than 1.5.

Similar Posts

Leave a Reply