# Current and voltage in a 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.

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:

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

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: -

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.

• Ok, I understood your case, but it's not the same exercice. I have problem with the central coil and the flux inside. I don't know how to calculate it. I think the input current is 1A because the flux giving is phi/2, the formula NI=RPhi must works, phi must be related with I. Apr 4 '14 at 8:44
• The central coil is in fact two series coils wound on separate iron cores - the second iron core only receives flux because of the current thru R1. The current thru R1 is also dependant on R2. I don't see how this can be incorrect from your diagram. I appreciate you want to calculate fluxes but I think you are missing the point about how the central coil works. You may also be over-complicating things by calculating magnetization flux - this doesn't contribute to current flow in a load. Apr 4 '14 at 8:54
• "The central coil is in fact two series coils wound on separate iron cores" I don't understand this point. Could you explain more please ? Why PSIM give me very strange result ? Apr 4 '14 at 9:05
• If R1 was open circuit there would be no flux in the 2nd core. OK so far? The central coil links both cores equally - ok with this? Just because each turn links both cores, it doesn't mean you can't "lump" a composite winding into two separate windings. I guess that's my simple explanation. My circuit still has the current thru R2 linking both transformers. Apr 4 '14 at 9:09
• I think I need to understand the second case first. Could you explain second case where R2 is open ? flux in central coil is phi/2 (1.5-2*0.5), so the voltage must be at 0.5V no ? Apr 4 '14 at 9:12

So assuming a lossless system the ratio of turns from the primary coil to the secondary coil will give you the "gain" or voltage amplification ratio.

So let us say you have 5 turns on the primary coil and 10 turns on the secondary coil. You then apply a voltage of 1V across the terminals of the primary coil. The potential difference across the terminals causes a current to flow through in a coiled or spiral shape around the iron core. Using the right hand rule http://en.wikipedia.org/wiki/Right-hand_rule you can see that a magnetic current (this is a very loose term I am using to make the idea simple) is induced in this iron core.

This magnetic current now flows towards the secondary coil. As this magnetic current passes through the secondary coils the "inverse" of what happened before (Electric current -> Magnetic Current) occurs. As the system is "ideal" and there are no losses the input power must equal the output power.

Hence a potential difference of 2V will exist across the secondary terminals, and depending on the impedance across the secondary terminals a current will flow. Hence I1V1=I2V2 so if we know the ratio of amplification of the voltage then we know that the current is going to be attenuated by the same factor (If we double the voltage from the primary terminals to the secondary terminals then the secondary current induced is half of the primary current which induced it).

I cannot see your picture (I think my browser is acting up) and I have talked about the simplest case possible. However if the number of turns N is the same left to right (or primary to secondary) then there is no change in the voltages and current from the primary side to the secondary side. In fact, if there are no losses, and no turns ratio, and all the impedance in the system is resistive then you can think of the whole system as a loop of wire connecting R1 and R2 which would form some sort of potential divider driven by a sinusoidal input.

I hope this helps you think.

• There are 2 irons inside central coil, so if a current pass through central coil, this create flux phi/2, like reluctance is the same for each iron, there is phi/2 in right iron too. This would say there is a voltage at right coil, so a current in R2, so a flux. This flux move inside central coil and change the voltage. Apr 4 '14 at 7:46

I understood, it's because I let K3=0 and it's not possible in physics. So K3 must be like K1*K2 not lower. LTSpice control this parameter and say if there is a problem. Multisim don't control any parameters of link selfs.