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I am currently performing the open circuit test on a transformer at my school to separate the hysteresis loss and the eddy current loss. Thus, I am changing the frequency of the AC current while keeping the flux constant by changing the applied voltage accordingly.

However, the current reading on my multimeter kept fluctuating badly. So, I wanted to know what is supposed to happen in theory and if the no load current is supposed to change when changing the frequency while keeping the flux constant.

Some more context:

The equipment my school has is not great right now. Actually its pretty terrible. The signal generator I am currently using can only give out 5V maximum, and the transformer I have is a dissectible transformer with no specifications. Due to the tiny voltage, the current reading is also pretty tiny too. Like I have to use microamps or milliamps on the multimeter to measure the current. When I said my multimeter fluctuates badly, let me just describe what it does:

Current at 35Hz: Around 3.5mA

Current at 40Hz: Around 3.2mA and it just keeps going up and down constantly

The current going down at 40Hz has screwed up my experiment as that means the power loss due to iron core losses decreased when frequency increased. However, the power loss due to iron core losses is supposed to increase with frequency when the flux is kept constant.

How do I find the no load current on a transformer?

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  • \$\begingroup\$ If it's a dissectable transformer, is it mechanically stable? Small changes in the residual airgap in the core will make large changes in the no load (magnetising) current. \$\endgroup\$
    – Neil_UK
    Commented Jun 13, 2018 at 15:33
  • \$\begingroup\$ I changed your question, revise it with a specific question, if this isn't what you desired. Please provide a schematic of your setup \$\endgroup\$
    – Voltage Spike
    Commented Jun 13, 2018 at 15:34
  • \$\begingroup\$ I expect more than 3.5mA at 5V. I expect that your signal generator is detecting a short circuit, and limiting the current. But I don't know for sure, because I don't know exactly what you have. \$\endgroup\$
    – david
    Commented Jun 14, 2018 at 5:39
  • \$\begingroup\$ Is your ammeter designed to measure 40Hz currents? In any case the current can't go up and down; it must be stable or your equipment is not working correctly. \$\endgroup\$
    – AlexVB
    Commented Apr 1, 2021 at 18:19

3 Answers 3

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DC resistance on the primary is key. A purely inductive load on an AC circuit will not consume any power (like an ideal transformer with no load on secondary). Ideal means, no DC (ohmic) resistance, only inductive reactance. Since all wires have some DC resistance, no ideal transformers exist. Then there is capacitance in the wires (and between wires) which will cause additional losses in the inductive primary.

And, although unloaded, any capacitance in the secondary needs to be considered as a minor loss.

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How do I find the no load current on a transformer?

Apply rated V , f with no load and measure current.
Primary inductance determines current vs voltage for no load current.

Do you know how to compute/measure L?
You can expect 3rd and 5th harmonics to rise above rated voltage sharply as core begins to saturate and and inductance geins to drop sharply.

Current source test

Ensure no stray voltage is induced on lines before testing. The current source is high impedance and prone to noise on this method with low currents.

Try terminating with real load Resistor current and subtract .

Before you can measure EC losses you get excitation current thus phase and amplitude must be measured.

  • For insulation failures they often sweep with expensive Omicron equip. From 0.01Hz to 1 MHz.

There are written standards for measuring all parameters if you know how to search.

You can use another better transformer to step up voltage using secondary input then measure voltage and current on a scope with a 0.1V current shunt power resistor.

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A transformer with an open secondary is basically just an inductor. If you increase the frequency, the inductive reactance will increase. For a fixed voltage, if you increase the frequency the current will decrease.

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