I have to correct you on your misconception here. A ring core does not suffer from the geometric demagnetization factor so the full effect of the permeability comes into place. With 200 turns at 1 amp wound onto a core having mu = 1000 the internal field is equivalent to having that 200 turns wound on air and carrying 200 kiloamps.
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I agree, thank you, the permeability counts. So in this case as already said "a strong variable flux also modifies the permeability", and so the toroidal inductance depends on time.
If this is really the case and we inject a sine signal in the toroidal coil then the current should not be sinusoidal anymore, and we should see new frequency components appear. It is to be checked.
The toroidal core sinks partially into the cylindrical coil so the inductance of this one will also depend on time. We will assume that this variation remains negligible, otherwise the calculation will be complicated.
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Anyway, i am lousy with math, but when you say "adding 0.5*L*I(t)² in the sum" (magnetization), i guess you mean with "L" the inductance of the air coil (18mH), with "I" the current at magnetization (261.3uA) and with "(t)" the magnetization time (148.4us).
Putting that into your formula, i get: (0.5 x 0.018 x 0.0002613 x 0.0001484)² which equals to 1.2e-19 which is very small.
Itsu
Itsu
I don't think so. L is the inductance of the air coil, but U(t) and I(t) are the instantaneous values of each point measured by the scope. To approximate an integral, we make a sum of a calculus with discrete values taken at constant interval (which corresponds to "dt"). So we have to take the values U(t), I(t) over the time period P that we are interested in. If you can not directly provide a calculation formula to the scope, I suppose that if it is connected to a PC in ethernet or USB, the scope can provide the measurements, each at a time interval T corresponding probably to its sampling frequency (that in MS/s) or to a chosen value (I have never tried).
We then transfer them to Excel, with U in column A, and I in column B, we just have to add in column C the formula =A1*B1+0.5*0.636*B1^2 (0.636 is the inductance value L). This gives us the instantaneous power U*I+0.5*L*I² for the first measurement.
Then we copy the formula of column C on all the lines, each line corresponding to values measured by the scope, so we have in column C all the instantaneous powers measured.
And at the bottom of column C, if we have for example 400 measurements over the period P, we add the formula =SUM(C1:C400)*T where T is the time interval between each measurement, which gives the energy over the period P of 400 measurements (If the scope provides the measurements at a sampling rate of 5 MS/s, we will have T=1/5000000).
Without guarantee, I may have forgotten something (I count on Smudge for the corrections
). I think I'm getting tired of math... I've lost the habit.