IEEE 43-20139 min read

Worked Example: Insulation Resistance and Polarization Index Acceptance Test on a 6.6 kV Motor

Two related IEEE 43 condition-assessment checks on the same winding — a minimum-IR pass/fail and a polarization index that clears its minimum by a much narrower margin.

Scenario

Motor rated voltage6.6 kV
Winding typeForm-wound stator, built after 1970
1-minute IR reading150 MΩ
10-minute IR reading330 MΩ
Test temperature40°C (reference temperature — no correction needed)
Insulation classB/F/H

Step-by-step calculation

Step 1: Look up the minimum insulation resistance for this winding type (IEEE 43 Table 3)

IEEE 43-2013 sets the minimum 1-minute IR at 40°C by winding type: 100 MΩ for most ac windings built after about 1970 (form-wound coils), 5 MΩ for random-wound and sub-1 kV form-wound coils, and kV + 1 MΩ only for windings made before about 1970 and field windings.

minRequired = 100 MΩ (form-wound, post-1970)
the older kV + 1 rule would give only 6.6 + 1 = 7.6 MΩ, which is far too lenient for this motor
minRequired = 100 MΩ

Step 2: Correct the measured IR to the 40°C reference (not needed here)

The widely used rule of thumb is that IR roughly halves for every 10°C above 40°C, or doubles for every 10°C below — since this test was already run at exactly 40°C, no correction is applied.

Step 3: Check the corrected IR against the minimum

correctedIR ≥ minRequired?
150 MΩ ≥ 100 MΩ
Passes, with a 1.5x margin

Step 4: Compute the polarization index from the two timed readings

PI = IR(10 min) / IR(1 min)
330 / 150
PI = 2.20

Step 5: Compare PI against the Class B/F/H minimum

PI ≥ 2.0 (Class B/F/H minimum)?
2.20 ≥ 2.0
Passes, but with a much narrower margin (10%) than the IR check itself

Result summary

CheckRequirementActualStatus
1-minute IR (corrected to 40°C) vs. IEEE 43 Table 3≥ 100 MΩ150 MΩ✓ PASS
Polarization Index vs. Class B/F/H minimum≥ 2.02.20✓ PASS
This winding passes both checks — a 1.5x margin on the basic IR test, but only a 10% margin on polarization index, which the IEEE 43 'Good' band (2.0-4.0) confirms is an acceptable but not exceptional result. The PI check is the tighter of the two constraints here.

Key insight: IR and PI test two different things and can disagree — a winding can pass a healthy absolute IR reading while its PI trend (how much IR rises over the 10-minute test as the insulation polarizes) is comparatively weak, or vice versa. PI specifically screens for moisture or contamination, which affects how the reading develops over time rather than its absolute level at any one instant, which is why both checks are run together rather than relying on either alone.

Try it with your own numbers

Every input in this example is editable in the live calculator — free, no signup.

Open Insulation Resistance Checker calculator →

Frequently asked questions

What does a low PI (close to 1.0) actually indicate if the absolute IR reading is still high?

A PI close to 1.0 despite an adequate absolute IR value can indicate surface contamination, moisture absorption, or leakage current paths that don't fully 'polarize' the way clean, dry insulation does — the IR reading barely climbs from 1 minute to 10 minutes because the current flowing is dominated by a steady leakage/surface-conduction component rather than the capacitive charging and dielectric absorption current that should decay over the test.

Why does insulation class change the minimum PI requirement?

Class A insulation (the oldest, least thermally robust classification) has a lower IEEE 43 minimum PI of 1.5, while the more common modern Class B/F/H insulation systems are held to a stricter 2.0 minimum — reflecting that better-quality modern insulation systems are expected to show a stronger polarization trend when healthy, so a weaker trend is a more meaningful red flag for them.

More in Transformers, CT/VT & Condition Testing

← Back to all worked examples