The total impedance is Z = jwL1 + jwL2 - 2jwM + R = R +jw(L1 + L2 - 2M)
V resistor = R(V source)/(R + jw( L1 + L2 -2M)) where w is radians per second, w= 2π(frequency in Hertz)
The mutual inductance is found from
M = k√((L1)(L2)) where k is the coupling coefficient which is 0 ≤ k ≤1
If you wind both coils on an iron core k will be close to 1, in this case L1/L2 = (N1/N2)2 where N1 is the number of turns of L1 and N2 is the number of turns of L2. Let the turns on L1 be constant
So Z = R + jwL1( 1 +( N2/N1)2 - 2(N2/N1)
Then
V resistor = V source/( R + jwL1(1 - 2(N2/N1) + (N2/N1)2) so the voltage across the resistor varies according to frequency ( w) and turns ration. I assume that you'll keep the frequency constant.
Assuming no phase shifts, then V? = Vs – Vb1 –?Vc2
But that depends on how the coils are connected, they could add instead of subtracting.
If N1 is the turns ratio for the first, and N1 is for the second, then
V? = Vs – (Vs/N1) –?(Vs/N2) = Vs (1 – (1/N1) – (1/N2))
with the signs dependent on the phase of the coils.
Post a link to the schematic.
I am afraid your question is not clear. Could you please send the diagram by link, if you can please, so that I would try to help you.Thank you.
The point is to quickly drop (or regulate) a voltage from source with the minimum of hardware. Imagine an AC generator feeding two bucking coils in series, followed by a load resistor. The bucking coils' turns ratio can be adjusted for sake of the exercise, in fact I'm trying to understand what the V will be across the load by altering the turns ratio between the two bucking coils. Also how do I determine the number of turns needed on the bucking coils to vary the voltage? (Note: I've seen tutorials dealing with auto-transformers but this is not quite the same thing, as there's no secondary.)
