Knee Point Voltage from a Test Report

The knee point in a current transformer (CT) is the specific voltage level on its excitation curve where the iron core begins to saturate, meaning a small 10% increase in secondary voltage requires a 50% increase in magnetizing current.

From the data, it appears that at 0.05A (and above) the core is already saturated.

An increase of 100% in current (0.05 -> 0.1 A) results in only a few percent in voltage.

Just a thought... these numbers appear to be made up --- is this a homework or test question?

This is from actual "test report" from the vendor; got it from the manufacturer after so many requests.

The knee point in a current transformer (CT) is the specific voltage level on its excitation curve where the iron core begins to saturate, meaning a small 10% increase in secondary voltage requires a 50% increase in magnetizing current.

We have 5 data points from the data provided. The current I should increase 5 times the increase in voltage V.

I take the 5 data points provided, combine into 4 segments, calculate change in voltage and current, normalize to value of voltage and current, and then compare these normalized values.

Five samples

I = 0.050000 0.100000 0.200000 0.500000 1.000000 (Amps)

V = 780 800 820 840 860 (Volts)

Difference Four segments between 5 samples

dI = 0.050000 0.100000 0.300000 0.500000 (change in Amps)

dV = 20 20 20 20 (change in Volts)

Resample I and V to 4 samples to line up with dI and dV

I2 = 0.075, 0.15, 0.35, 0.75

V2 = 790, 810, 830, 850

Divide changes (dI and dV) into current and voltage (I2 and V2)

dI/I2 = 0.6667 0.6667 0.8571 0.6667

dV/V2 = 0.025316 0.024691 0.024096 0.023529

Find ratios of (change in current) / (change in voltage)

R = (dI/I2) / (dV/V2) = 26.333 27.000 35.571 28.333

R should be about 5

From the data supplied, it appears that the core is saturated

https://www.linkedin.com/pulse/knee-point-voltagevk-ps-class-current-transformer-sudhir-tiwari/