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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?

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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 magentizing 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: -

enter image description here

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.

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  • \$\begingroup\$ 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) \$\endgroup\$ – dude Aug 22 '13 at 15:23
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    \$\begingroup\$ @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. \$\endgroup\$ – Andy aka Aug 22 '13 at 17:18
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    \$\begingroup\$ @dude - I've added a drawing that might help you understand \$\endgroup\$ – Andy aka Aug 22 '13 at 19:33
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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.

enter image description here

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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
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    \$\begingroup\$ 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. \$\endgroup\$ – Andy aka Jun 30 '16 at 11:35
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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

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  • \$\begingroup\$ 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 \$\endgroup\$ – VMMF Apr 5 at 18:11
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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

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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

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    \$\begingroup\$ 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. \$\endgroup\$ – Transistor Aug 24 '18 at 17:23

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