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As mentioned earlier, the core losses include Foucault and hysteresis and its model is as you can see here:
By using this relation,
we can obtain the Foucault and hysteresis resistance models which Foucault resistance model is linear and independent from the voltage while hysteresis resistance model is non-linear and dependent on the voltage.
If the CT Primary is open and voltage is fed from the secondary side,
since Rct and the current drawn from the source is clear, the voltage of the parallel branch is calculated using:
E= Vterminal –RCT * I
After calculating E, the core casualties are calculated using:
Pc= E*I *Cos (<phi)
By having the core losses value and its extended for formula
It is possible to calculate Ke, Kh and X by solving the three equations in the three variables system.
To solve these equations, it is possible to apply three different voltages.
But if the frequency is the same, its determinant turns zero and no answer can be obtained for the variables.
Therefore, it is better to apply one of the voltages with a different frequency.
In the first step of solving these equations with V1 and f1 as voltage and frequency,
Ke and (Kh*(f1-x) parameters can be obtained using a two equations in the two variables system.
Now, if one of the voltages is applied along with f2 as the frequency, “x” is obtained and by substituting it in Kh*(f1-x), Kh obtained achieved as well.
Due to voltage application limit for CTs with a very high saturation point,
it is possible to find the saturation point in the nominal frequency by decreasing the voltage and frequency with the same proportion.
Because the relation of the according flux is V/f.
But does not changing the frequency change the CT system? To explain this subject,
we need to examine those components of the CT that are affected if the frequency and voltage are changed.
According to previous analyses, Rct is dependent on the material of the wire and Reddy to the material of the coil.
In Xm branch, due to the proportional change of V and f and the stability of the flux, this branch remains unchanged as well.
Therefore, the only parameter affected by the voltage and frequency is hysteresis resistance which was obtained
Now suppose that to obtain the saturation point, the fed voltage and frequency are split in two.
Is it possible to obtain the saturation point in the nominal frequency by multiplying the voltage and current by two?
This question will be answered in the following.
Suppose that the are CT1.5  Kh=10 -3 x=1.8Ke=10 -4 knee point is obtained by decreasing the frequency to 25 Hz with 500 volts of voltage and 15 milliamps .
with a -45 the grid
EExc = Vterminal - Rct IExc
Since now we have Ke، Kh and x, by dividing EExc by Reddy and Rh, la and lh currents are calculated.
By subtracting these two from lExc, the self-inductor current is obtained. According to this relation,
Phi=L* I
when the flux is fixed, the self-inductor current is fixed as well.
Therefore, by doubling the voltage of the parallel branch, this current remains fixed.
After doubling the voltage and frequency, as it is shoawn in this formula
the Foucault and hysteresis currents are calculated and added by self-inductor current to achieve lExc using this relation.
Then, the voltage of the parallel branch which is 25Exc is added to Rct* lExc and the terminal voltage is obtained.
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As you know, the maximum output voltage of a device in one-phase mode is “450” volts.
Usually, to avoid damaging the measuring equipment, “Measuring CTs” go to saturation in low voltages.
But the saturation voltage of some “Protection CTs” such as “TPX”, “TPY” and “TPZ” is too high and is beyond the ability of the device.
To test these “CTs”, it is possible to increase the voltage until the “CT” goes to saturation but
some scientific references suggest that high voltages must not be fed into “CTs” because high voltages can damage the insulation of the equipment.
Another method for excitation test in “CTs” with too high saturation voltage is to decrease the frequency.
This method is also mentioned in the standard CT” excitation test which can be performed with an “AMT105” device.
As you saw, the knee point of a 5P10 protection CT in 50 Hz frequency was measured to be about 159 volts.
If you wish to specify the value of the final voltage and injected frequency manually, you need to uncheck “Calc.
From TestObject” option and fill “Frequency” and “V end” fields. For example, “25 Hz” and “186 volts” are entered as frequency and final voltage respectively which means “372 volts” in “50 Hz” frequency.
Note that by changing the voltage, it is so much probable that the wirings change as well.
In this method, “Demagnetize” option is checked by default so that the magnetic residue is removed and the aforementioned calculations are done.
After adjusting this settings and specifying the test time, click on “Init Test” to initiate the test.
As you can see in the “Signal View”, by decreasing the frequency and equating, the knee point is almost equal with what was measured in 50 Hz frequency.
Therefore, in cases where the CT saturation point voltage is beyond the limit of the device, by decreasing the frequency, you can measure the knee point and its saturation.
