# How do transformers comply with Ohm's law and KVL?

I understand how to use a transformer in practice, and I know that transformers obey the law of conservation of energy, but I'm confused as to how step-up and step-down transformers obey Ohm's law and KVL. This is an example of why I get confused given my current understanding of electricity:

Say there's an ideal step-down transformer that converts 10V at 1A to 1V at 10A, and the secondary of the transformer has a 0.1Ω load across it. According to KVL, there should be 1V dropped across the load because the secondary produces 1V, and because the current is the same everywhere in a series circuit, there should be 10A through the load. Ohm's law holds because the 1V dropped across the load is equal to the 10A through the load times the 0.1Ω resistance of the load. But if you replace the load with a different resistance one, how does Ohm's law hold? Based on my current knowledge, there would still be 1V dropped across the resistor and 10A through it, but V ≠ IR.

I know that something has to be wrong with my current understanding, but I'm not sure what. Does the current through the secondary change when you change the load? Does the voltage across it change? Is the current through the secondary not equal to the current through the series load? Or does it cause a change in the current/voltage through/across the primary? I've been trying to think of an answer to this question myself and I've gone back and re-read a lot of material on inductors and transformers but I still can't figure it out, so if anyone thinks they have a good explanation, I would really appreciate it. Thank you!

• Why do you think the primary side current would still be one amp? Sep 6, 2022 at 4:30
• If you have 1 ohm load, and drive 1V into it, it consumes 1A, which is 1 watts. So there can't be 10A in the secondary, as it's 1A. Why there would be 10A if load takes only 1A? Sep 6, 2022 at 4:42
• @Hearth That was part of my question. I'm still trying to learn how transformers work and that was one of the possibilities I was wondering about. Sep 6, 2022 at 5:05
• @Justme Sorry, that was a typo for the resistance value, I meant 0.1Ω. Sep 6, 2022 at 5:06
• Transform the impedance by the square of the turns ratio and you should find Ohm's Law etc work perfectly.
– user16324
Sep 6, 2022 at 11:56

You're assuming that the currents are determined by the transformer, but they're not, the voltage ratio is determined by the transformer, and the currents are then determined by the load impedance.

So, if you have a transformer with a 10:1 turns ratio, and you put 10 V into the primary (you don't specify the current here) you will get 1 V at the secondary. The 1 V will give you 10 A through the 0.1$$\\Omega\$$ load, and because of the turns ratio the primary current will be 1/10th of that, or 1 A.

When you replace the load with a different impedance, Ohm's Law holds because the currents will change, for example if you made it a 2 $$\\Omega\$$ load the secondary current would be $$\frac{1V}{2\Omega} = 0.5A$$ and the primary current would be 1/10th of that, 50 mA.

• Thanks for the clarification! Does the loaded secondary change the impedance of the primary through back EMF or some other magnetic field? The concept makes sense now but I guess it's still a difficult subject for me to visualize. Also, sorry about the math mistake, I changed the numbers I was using and forgot to change the resistance. Sep 6, 2022 at 5:04
• Voltage and current are transformed as the turns ratio, impedance is transformed as the square of the turns ratio. So a 10:1 transformer would make a 1 ohm load look like 100 ohms at the primary. 10V/100 ohms = 100mA, so it works out. Sep 6, 2022 at 5:09
• Thank you so much, that's much clearer! Sep 6, 2022 at 5:11

Transformers driven by an ideal AC voltage source just convert the voltage as per the turns ratio, and the currents will change to match the circuit.

First you fix a scenario where secondary output is defined to be 1V and 10A. This will mean the load must be a 0.1 ohm resistor.

Then you define another scenario where the secondary now has a 1 ohm load. So the currents and voltages of first scenario do not apply any more, as 1 ohm load with 1V takes only 1A.

So from voltage, resistance and current, you can define only two out of the three, because the third parameter is defined by two of the parameters.