Electronic – Current and voltage in a transformer

transformer

I would like to find currents and voltages at R1 and R2 resistors. I put sinus voltage at left , 12V for example with frequency at 50Hz. Each coil has the same number of turns N. Irons circuit have the same reluctance. I consider this exercice like a theoretical problem, no losses from copper, no losses from iron, etc.

I attempt to find all fluxes, currents and voltages. In the drawing, red arrow is the sum of flux in each iron. Magenta arrow is a voltage. Black arrow is a flux created by coil.

enter image description here

I used LTSpice and it's OK if I let for K3=K1²=K2² but current in primary is a sinusoidale shape WITH add a DC shape, so the primary consume energy and the secondary don't use it.

vV1 0 3 dc 0 ac 1 0
+ sin(0 {1200*1.414213562} 6000 0 0 0)
rR4 1 0 1e-012
rR3 5 0 1e-012
rR5 8 0 1e-012
K1 LL1 LL2 0.9
K2 LL2 LL3 0.9
K3 LL1 LL3 0.81
rR2 0 10 0.006
rR1 0 9 0.009
lL3 7 8 1
lL2 2 5 1
lL1 3 1 1
.ends

Look at current DC:

enter image description here

enter image description here

Best Answer

You should consider separating the two cores into two transformers like this: -

schematic

simulate this circuit – Schematic created using CircuitLab

R1 and the two cores looks complicated the way you have it drawn but the only thing linking the two cores is the current flowing through R1 - that's how I read the diagram and that's what I've redrawn.

Because these are 1:1 perfect transformers with no leakage, you can then move R2 from the right of XFMR2 to its primary and discard XFMR2. Now you have a 9 ohm + 6 ohm load (15 ohm) on the secondary of XFMR1 and, by the same technique this load can be assumed to be directly connected to the input supply voltage of 12V.

This gives a power of: -

P = \$\dfrac{12^2}{15}\$ = 9.6 watts

EDIT to include rationalized picture and calculation of magnetization current: -

enter image description here

Although not shown, the 6 ohm load resistor is connected to the final secondary winding and the total referred load is therefore 15 ohms causing a current of 0.8A. This is load current. Primary magnetization current is 0.6A because \$\sqrt{0.8^2 + 0.6^2}\$ = 1A.

A primary mag current of 0.6A implies the primary mag inductive reactance is \$\frac{12}{0.6}=20\Omega\$. Therefore inductance is \$\frac{20}{2\pi\cdot 50} = 0.0637 H\$. From this you can calculate flux.