# Confusion - ideal transformer on load

When there is no load connected to the secondary, E is equal to U, and no current flows in an ideal transformer. Now if we connect a load on the secondary of a transformer, the primary will draw a current because E<U (temporarily), why when E gets equal to U the transformer will still be drawing a current "I1" from the mains, doesn't that violate the reason current started flowing in the first place?

I know that the secondary field must be counterbalanced at all times by the primary field, but I have a confusion about the fact that even though there is no potential difference because E=U there is still a current flowing.

Can anyone explain that? And please correct me if I am making any thinking errors.

Image from: DC Machines and Transformers by K Murugesh Kumar

I have a confusion about the fact that even though there is no potential difference because E=U there is still a current flowing.

The induced voltage is in series with the inductance of the winding and, given that the induced voltage is exactly the same as the applied voltage, there is, in effect, zero volts across the primary inductance. Given that an inductor's inductive reactance is zero when 0 volts is applied to it, the current that flows is this: -

$$\color{red}{\boxed{\dfrac{0 \text{ volts}}{0 \text{ ohms}}}}$$

This is an indeterminate quantity and rightly so.

In other words, for an inductor, you cannot say that zero amps flows just because there is zero volts across it. Another example is the current in an inductor when the applied voltage is a sinewave: -

I've drawn purple circles on the above graph (taken from here). Those purple circles are placed to coincide with the applied voltage across the inductor equalling zero - what do you notice about the current? It's not zero of course.

Another example; take an inductor and applied (say) 1 volt to it then remove it from the supply and instantaneously short it out - does the current fall to zero or does it continue to flow. It's the latter of course and this serves to remind us that the current flowing in an inductor has little to do with the applied voltage currently present.

This however, is always true: -

$$\boxed{V = L\cdot\dfrac{di}{dt}}$$

• Your answer was very helpful indeed. But if we follow that principle there arises a conflict because when there is no load connected to the secondary of the transformer, there will be no current flowing in the primary E=U, but if we connect the load then even though E=U there will be a current flowing. So we have two cases where E=U but in the first case current is not flowing but in the second case current flows. Keep in mind that the reason current starts to flow is that E gets smaller than the applied voltage U temporarily.
– ell
Commented Mar 15, 2021 at 8:15
• @ell there is no conflict at all. Current starts to flow because a load is connected to the secondary and that causes secondary ampere turns (MMF) which might either increase or decrease the magnetization flux in the core; this resolves itself with an equal-in-magnitude primary MMF that opposes (Lenz) the secondary MMF in order to maintain Faraday's law of induction. Commented Mar 15, 2021 at 9:32
• Its all clear now.
– ell
Commented Mar 15, 2021 at 11:12
• @ell to understand why people give free help please take 2 minutes to read this. Commented Mar 15, 2021 at 11:14
• I did read it but I couldn't find any information regarding your point. Please be more clear.
– ell
Commented Mar 15, 2021 at 13:40

TL;DR

There is always a primary current with no load due the fact that the mutual coupling requires primary inductance current of about 10% of rated load. Then a unity turns ratio transformer can match voltages. But at full load the core and winding losses will result in a 10% secondary loss from no load with a resistive load.

I think of it as the Mutual coupling is dependent on the existence of primary flux to couple, C to the secondary to overcome hysteresis on magnetic AC fluxes.

• That doesn't explain much.
– ell
Commented Mar 15, 2021 at 8:20