Continuing my critique of your publication, in the section pasted below, please specify where the 9.3ohm value comes from. I see 11.3ohm for the load in the oscillograms.
We can calculate the current on the bulb using Ohm's law I = U/R = 3.4V / 9.3Ohm = 0.36A.
If we apply the general formula for determining the current in a closed circuit, we will not get equality with the measurement of the idle EMF.
I = (E-U)/(R+r) = (9,4V-3,4V)/ (9,3Ω+12,1Ω) = 0,129A
Because the current in the anchor, when the load is closed, contributes to the amplification of the magnetic flux in the circuit (which is visible on the second screen of the FEMM program). Self-excitation occurs, due to which the resulting EMF increases.
We can only calculate this level of the resulting EMF, it will be 26.5 volts.
Let's check: I = (E-U)/(R+r) = (26,5V-3,4V)/ (9,3Ω+12,1Ω) = 0,36A
If we apply the general formula for determining the current in a closed circuit, we will not get equality with the measurement of the idle EMF.
I = (E-U)/(R+r) = (9,4V-3,4V)/ (9,3Ω+12,1Ω) = 0,129A
Because the current in the anchor, when the load is closed, contributes to the amplification of the magnetic flux in the circuit (which is visible on the second screen of the FEMM program). Self-excitation occurs, due to which the resulting EMF increases.
We can only calculate this level of the resulting EMF, it will be 26.5 volts.
Let's check: I = (E-U)/(R+r) = (26,5V-3,4V)/ (9,3Ω+12,1Ω) = 0,36A
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