# Transformer and Thevenin Theorem

Consider an ideal transformer connected to a real AC voltage source (with output resistance Rg).

Now consider to apply the Thevenin theorem between the output terminals of the secondary circuit. I was told that, when we evaluate the Thevenin voltage, it is equal to Vg because:

The current in the secondary is 0 -> the current in the primary is 0 because of the transformer equation (Is = Ns/Np Ip), in which we put Ip = 0. So we get Vp = Vg and so Vthevenin= Vs = Vg.

I did not understand the physical meaning of that thing: if I have 0 current at the secondary, why should I have 0 current at the primary, which is a closed circuit? It is simply a normal circuit with an inductor.

Moreover, if it is true that Ip = 0, the transformer is off since there is not the magnetic coupling between the inductors: so how can we say that Vs = Vg?

That's the difference between an ideal transformer and a real transformer. The transformer equation is a very crude model of a real transformer.

Whenever you combine an ideal model (transformer) with a "real" model (the voltage source) you may get results that don't make sense, particularly when you start asking about the "physical meaning".

If not current flows through the primary then no current flows through the resistor, so the voltage drop across the resistor is zero, so the voltage across the primary is equal to Vg, so the voltage across the secondary is Vg. Vg = Vs.

Plenty of errors here. Maybe somebody has told to you bullshit or you have misunderstood something.

Ideal transformer is assumed to cause no voltage drop. That doesn't mean that the secondary output voltage is =the voltage between primary terminals! The winding ratio still works. It works ideally, ie the voltage at the output is (Ns/Np) * the voltage between primary terminals. Ideality means this is true with all loads.

Ns=number of turns in secondary

Np=number of turns in primary

Because ideal transformer doesn't waste energy, the primary current is (Ns/Np)*output current. You had replaced this with something else.

Output current = zero means that primary current is zero. Rg doesn't drop voltage, so the primary voltage is =Vg. The secondary voltage isn't=Vg, you must multiply Vg by the winding ratio.

Your drawing has a hitch which suggest that the transformer isn't fully ideal. There seems to be inductances for primary and secondary (=L1 and L2). In fully ideal transformer those inductances are considered to be infinite or at least so large that the idle primary current can be considered to be zero. If inductances are given, but the transformer is said to be otherwise ideal, then the case is complex and it must be calculated with complex phasors.

I skip it.

• Thank you for the answer, but I cannot understand this sentence: "Output current = zero means that primary current is zero". Mathematically it is ok, but the primary circuit is a normal closed circuit with a voltage source and an inductor: why 0 current? – Kinka-Byo Apr 17 at 21:16
• Ip=(Ns/Np) * Is = (Ns/Np) * 0 = 0, Ideal transformer doesn't need idle current. – user287001 Apr 17 at 21:19
• From a math point of view it is ok, but I do not understand the physical reason... – Kinka-Byo Apr 17 at 21:20
• This is not real transformer, it's the limit case of an infinite sequence of transformers where the inductances grow with no limit, but all losses stay zero and the winding ratio stays constant. The behavioural equations of the ideal transformer are the limit of the behavioural equation sets in the sequence. Every transformer in the sequence has idle primary current but the limit value of the idle current is zero. Hopefully you know sequences and limits in math. – user287001 Apr 17 at 21:25
• Ok,clear! So Ip = 0 because when we use the equation Ip = (Ns/Np)Is we are considering that those components I have indicated as inductors are simply components magnetically coupled and not also inductances (only jwM and not jwM+jwL). Thank you, clear. – Kinka-Byo Apr 18 at 1:09