# How can I calculate the inductance of the primary of a transformer given a specific load on the secondary?

Inductance in the primary of a transformer decreases as the load on the secondary increases. How can I calculate the inductance of the primary coil? What if it was step up or one to one ratio?

Inductance in the primary of a transformer decreases as the load on the secondary increases.

No it doesn't. It may seem like it does (because when loaded your transformer takes more current into the primary) but just imagine that the load you put on the secondary (say 1:1) ratio were applied to the primary - the current in the load would be the same (1:1 ratio) and the small current that goes into the transformer primary (when off load) will still be going into the primary (this small current is the magnetizing inductance and remains intact with varying load conditions).

If your ratio was (say) 10:1, and you connected a 10ohm resistor on the secondary, this is equivalent to connecting a 10ohm x $$\{(\frac{N_P}{N_S})}^2\$$ resistor on the primary i.e. 1000 ohms. Np and Ns are primary and secondary turns and your equivalent primary load is the turns ratio squared.

With the "equivalent" load connected on the primary, the transformer inductance may "appear" to have changed but it's still there and in parallel with the "equivalent" load.

EDIT I'm adding a picture below showing a 1:1 transformer with windings that are 10mH each. On the secondary there is a 2.53uF capacitor and the primary is excited with $$\10V_{RMS}\$$ at 1kHz: -

Conclusion, adding a load of any description does not affect the primary magnetizing inductance of a transformer. If you put an inductor on the secondary you might think the primary inductance has reduced but in fact the added secondary inductor becomes in parallel with the never-changing primary magnetizing inductance.

• If the primary takes more current, then the only variable that decreases is impedance which is dependent on reactance (resistance of coil is constant) which is dependent on frequency and inductance (2PIfl). Since frequency is constant, the only thing left is inductance. Also, when I use my LC meter to measure inductance of my transformers when unloaded, it shows the inductance which is normal. But then when I short circuit the secondary wires, the primary reads an inductance the goes towards zero (i.e from 1H to 0.002H). I'm think this formula has something to do with it: L=N(dPHI/di)
– dude
Aug 22, 2013 at 15:23
• @dude. I hear what you say but the primary magnetizing inductance doesn't change - the load you put on the secondary takes current and this current (via the turns ratio modifier) is taken from the input wires to the transformer primary - in effect you are putting a load in parallel with the primary coil. If you had a loudspeaker and cemented the cone, you would expect to see much more current drawn when you power it with a sinewave - does the inductance change? If you stall a motor, it takes more current because the "load" is transferred, via magnetic coupling to the power source. Aug 22, 2013 at 17:18
• @dude - I've added a drawing that might help you understand Aug 22, 2013 at 19:33

I disagree with Andy aka’s interpretation. The math is correct, but the interpretation of that math is one that is non-physical and clearly cannot be true.

This should be self-evident: if you connect something across the secondary winding, it does not magically appear in two places at once so as to also be across the primary. The simple, visibly obvious reality is that whatever is connected across the secondary is, in fact, only connected across the secondary. Saying it is in parallel with the primary is useful as a theoretical equivalence, but not as a true physical description of what is actually occurring. Yes, the numbers work out the same as if the load was in parallel with the primary, but that doesn’t mean it actually is.

Clearly the load isn’t in parallel with the primary because it isn’t connected in parallel with the primary. It’s that simple. One can pretend it is, the math works out to allow this, but that is just a theoretical contrivance and has no bearing on what is actually occurring physically.

Here is what actually occurs physically:

First, we need to consider why inductors have reactance in the first place. Their reactance is ultimately a result of self-induction. Changing current causes a changing magnetic field. What do changing magnetic fields do? They induce a voltage, as per Faraday’s law of induction.

Just because an inductor is the source of this changing magnetic field doesn’t make this changing field stop behaving like a changing magnetic field. Any changing magnetic flux through a given inductor will induce a voltage across it - even if the inductor itself is the source of that changing flux.

What this means is that inductor’s reactance is due to their stored energy in their magnetic field. If the current through an inductor rises, this makes the flux increase which induces a voltage through the inductor, one that opposes sign of the change in voltage. So if voltage is increasing, this opposing voltage will be the opposite polarity, resulting in a voltage drop across the inductor.

All voltage drops are due to impedance. Impedance represents something that absorbs energy - it can be dissipative (leaves the circuit permanently), which we call resistance, or it can be due to energy storage instead, which we call reactance.

In this case, the voltage drop supplies the energy to increase the magnetic field. Likewise, when the current falls, energy stored in that field is released back into the inductor as the flux decreases, inducing a voltage that again opposes the change in voltage - so now the inductor acts as a voltage source as the flux collapses.

The important thing to understand here is that it is this voltage, or EMF, which occurs due to an inductor’s own self-induction, that is ultimately the cause of an inductor’s reactance. The faster you try to change the magnetic flux, the larger this back EMF due to self-induction will be, limiting the rate at which current can rise and fall and thus reducing the peak current more and more for AC as the frequency increases. This is true for any AC component and occurs even if DC biased (so-called ripple current).

So the impedance of an inductor isn’t strictly a result of its inductance, but rather the back emf that results from that inductance.

In the specific case of a single inductor in isolation, this is dependent on the inductance alone.

In the case of a transformer (or any arrangement of coupled inductors), this is no longer the case. This is because the back emf through a given winding is no longer dependent just on that winding’s inductance. The current through other windings results in flux as well. This is reflected back into the primary winding as voltage/emf, but this emf opposes the primary’s own back emf from its self-inductance. Put another way, the flux from the current through the secondary acts to oppose the flux from the current through the primary.

This is known as mutual inductance. A better way to visualize it is energy that is transferred via changing magnetic flux, rather than stored.

The relationship is fairly straightforward.

The back emf that results on the primary winding is equal to:

$$\ V_{p} = L_{p}*\frac{di_{p}}{dt} - M*\frac{di_{s}}{dt} \$$

Where Vp is the emf across the primary, ip is primary current, is is the secondary current, and M is the mutual inductance.

When the secondary is open circuit, the mutual inductance has no effect and all of the expected back emf appears on the primary to oppose the change in current. When the secondary is loaded, however, it causes a reduction in the back emf in the primary. It is actively counteracting some of the self-induction of the primary, and as we understand from earlier, this is fundamentally the source of inductive reactance.

In summary, the inductance of either winding never changes, but the reactance of coupled inductors depends on the current through the other, not just the current through itself. So the reactance of the primary falls as the secondary is increasingly loaded because the secondary winding current opposes some of the flux of the primary, limiting the back emf induced and thus the fundamental source of the primary’s reactance.

When you measure the primary inductance, most meters are actually measuring the reactance by measuring current at a known frequency, or measuring the resonant frequency formed by the LC tank of a known capacitance and the inductor under test. The resonant frequency is ultimately a result of the frequency where impedances are the lowest. In either case, you’re really measuring impedance and the meter is calculating the inductance based on that. This only works if there are no coupled inductors or if there are, they are all open circuit. If they aren’t, these methods of measuring inductance are no longer valid. You would need to measure the mutual inductance and other winding currents at the same time to correctly determine any particular winding’s true inductance.

Is that transformer the ideal transformer? Then the answer is simple. The capacitance of the secondary side is impedance converted and observed from the primary side. Then how about with an actual transformer? I did the experiment. The photos are as follows. First, it is inductive at low frequencie, change to capacitive at higher frequencies from antiresonant frequency, and change to inductive again at series resonance frequency. The anti-resonance frequency is the resonance frequency of the self-inductance on the secondary side and the capacitance. The series resonance frequency is the resonance frequency of the short-circuit inductance on the secondary side and the capacitance.

As the load on the secondary increases, i.e. the resistance or impedance, whichever you prefer, goes down, then the reflected impedance (and inductance) of the primary goes down as well all the way to shorting the secondary is where we measure the leakage inductance. In a perfect transformer, a short on the secondary is reflected as a short on the primary. If you want to directly measure the inductance of the primary you do so with the secondary open and vise versa. From your original question, there are some methods of direct measurement of iron core transformers with inductance as high as 50H or more that work quite well but you do have to take into account the vectors and not just scalar values if you want a reasonably accurate answer. Here is a link to a nice article that does work. You have to use peak voltages and currents and your angles must be in radians. https://meettechniek.info/passive/inductance.html

• Do you know why "a short on the secondary is reflected as a short on the primary" I'm trying to understand the reason behind it
– VMMF
Apr 5, 2019 at 18:11

This is fairly easy: just Thevenin transform it. Take your transformer+load, transform your nonideal transformer into a series and parallel impedance + ideal transformer, and then transform that again into a thevenin equivalent without the transformer entirely.

http://en.wikipedia.org/wiki/Equivalent_impedance_transforms

Inductance in the primary of a transformer decreases as the load on the secondary increases.

No it doesn't.

Actually, it does with non-air-core transformers. The permeability of the iron changes with flux density, which means the inductance changes when the amount of current through the transformer changes. Loading down the secondary increases the current through the primary / the flux density in the iron core, and the inductance of both the primary and secondary changes.

This is easily demonstrated by driving any old iron-core transformer / inductor with a variac and watching the current and voltage through the coil as you slowly turn up the variac. Use ohm's law to give the inductive reactance and you will see it fall by a factor of 2-4 as the current through the coil increases - with the "other" side of the transformer open.

Example - primary coil of a microwave oven transformer, magnetic shunts and secondary coil removed. Here is the inductive reactance of that coil at various currents...

460 ma ---> 109 ohms. 1.3 A ---> 70 ohms. 2.2 A ---> 45 ohms. 5.0 A ---> 22 ohms.

• Kevin
• This is wrong - the secondary ampere turns due to the load and those ampere turns on the primary also due to the load are exactly equal and opposite. This means the flux density in the core doesn't change. Your second paragraph is correct but this is all about an open circuit secondary and contradicts the question which is to do with secondary loading effects. Jun 30, 2016 at 11:35

Ohm resistors (resistance) in the secondary create active currents both in the primary and in the secondary. Coils in the other hand create imaginary current (current is late by 90 degree from voltage) so there is no way that adding load in the secondary is effecting the inductance of the primary, only the Ohmic resistance.

Power is wasted on resistance devices with (real resistance) in the formula of I^2*R

• You are confused here, Daniel. Coils to not "create imaginary current". Coils cause current to flow out of phase but these currents are not imaginary, they are real, measurable and cause heating in wires, etc. You might be confused because we use complex numbers in their calculation but we use j rather than i, probably to attempt to avoid that confusion. Aug 24, 2018 at 17:23