I'm reading an old military electronics manual and there is a chapter on transformers which has some confusing information.

There is a schematic of a step-up transformer similar to this:


simulate this circuit – Schematic created using CircuitLab

So it's basically just a transformer with an AC voltage on the primary and a load resistor on the secondary.

The text associated with this picture is lengthy but at the end there this paragraph:

(The text is not in English and the translation is mine. Some liberty is taken in translation but it is perfectly accurate)

"As the secondary side of the transformer (one with the load resistor) has a higher number of turns around the core, this is classified as a step-up transfomer. This means that if we divide the number of turns on the secondary by the number of turn on the primary, we get the ratio of voltages across the windings. Also, to conserve energy, the current in the secondary is stepped down with respect to the current in the primary by the ratio of turns to make the power (product of current and voltage) equal in both sides of the transformer. Therefore, the current on the secondary is easy to calculate as the ratio of the secondary current to the primary current is always the same ratio as the ratio of the number of turns."

This last line in bold is what noticed. Shouldn't it be "the maximum current that the secondary side can output is equal to the ratio of the turns". Surely the real current on the secondary side is dependent on the resistor?

A little thought experiment: I have 5 volts and 1 amp on the primary. This means that the power is 5 watts and so the power on the secondary has 5 watts as maximum value. If the secondary has 10 times more turns, the voltage across the secondary coil (and therefore across the load resistor) is now 50 volts. Now I apply Ohm's law: Divide the voltage (50 volts) by the resistance (1000 ohms) to get 50 / 1000 = 0.05 or 50 milliamps of current flowing across the load. The maximum possible current across the load would be 100 milliamps (the voltage is multiplied by 10, so maximum current is the current on the primary divided by 10 to keep the product of voltage and current the same as it is on the primary side). The calculated 50 milliamps is less than the maximum 100 milliamps to everything is fine.

But the text seems to somehow imply that the transformer forces the maximum amount of current across the load always, regardless of the resistance of the load! Clearly this makes no sense, as then we would have 50 volts and 100 milliamps across the resistor, and this would contradict Ohm's law (dividing 50 volts by 100 milliamps to get the "resistance" gives 500 ohms, which is not the value of the resistor).

The text is associated with the picture that has a resistor on the secondary. But surely the text is written with a transformer in mind that has perhaps a short circuit on the secondary. Is this a mistake by the author or is there something deep that I have misunderstood about transformers?

  • 1
    \$\begingroup\$ This goes both ways. The only way you'd have 1 amp on the primary is if you had 100ma on the load. The load on the secondary affects the current drawn on the primary. If you calculate the power in and power out, you will see this happening. In a lossless transformer, the primary:secondary current ratio is always the same as the turns ratio \$\endgroup\$
    – BeB00
    Apr 25, 2017 at 19:49
  • \$\begingroup\$ We seem to be having a lot of similar questions at the moment, here's another \$\endgroup\$
    – Finbarr
    Apr 25, 2017 at 19:53
  • 1
    \$\begingroup\$ Just as " the real current on the secondary side is dependent on the resistor", the real current in the primary is dependent on the current in the secondary. The ratio holds - you just need to apply cause & effect correctly. \$\endgroup\$
    – brhans
    Apr 25, 2017 at 20:08
  • \$\begingroup\$ the text is accurate; your "max" interpretation is the practical effect. outside of talking about measurements and loads, amperage is usually talked about in terms of potential/max; a 20A circuit breaker doesn't always burn 20A, a 24V/3A PSU doesn't always force 3A, etc... \$\endgroup\$
    – dandavis
    Apr 25, 2017 at 20:57

2 Answers 2


No, the text is exactly right.

The product of voltage and current on both sides - power - is the same. So, knowing one current, you know the other (thanks to this relationship).

So ... do you know the primary current?

No, but you know the secondary current and it IS determined by the resistor - exactly as you say.

Now, THAT determines the primary current - it's the turns ratio * the secondary current.

Now put these two relationships together and see that the 1 kilohm resistor has been transformed down to a 1 kilohm/N^2 load on the primary. N from the step-up in the voltage from primary to secondary, and N more from the step-up in current back to the primary.

(In reality there's also a small primary current, always there regardless to the load, and in addition to it - called the magnetizing current - powering the magnetic fields in the transformer)

EDIT for clarification :

1) yes, you can make an equivalent circuit with a 1/N^2 load across the primary instead. i.e. 10 ohms in parallel with the primary, if it was a 10:1 step-up transformer, instead of a 1K load on the secondary.

2) Yes, with no load, no current flows in the primary ... except for the magnetizing current mentioned above. Remember you're supplying it with AC : the primary is a very large inductance, which is a very high impedance.

This means you can use transformers for "impedance matching" in audio and radio circuits, as well as AC power conversion.

  • \$\begingroup\$ Interesting, could you please elaborate more on ".. 1 kilohm resistor has been transformed down to a 1 kilohm/N^2 load on the primary." ? I don't understand what do you mean, do you mean that we can somehow create an equivalent circuit with a load on the primary? \$\endgroup\$
    – S. Rotos
    Apr 25, 2017 at 20:33
  • \$\begingroup\$ And if the ratio holds, wouldn't this mean that if the secondary is left open (to current flows), no current flows in the primary either; but I am still aplying a voltage to the primary so how come no current flows through the winding? \$\endgroup\$
    – S. Rotos
    Apr 25, 2017 at 20:40
  • \$\begingroup\$ Current in the primary is dependent on current in the secondary, which depends on the voltage in the primary and the load in the secondary. So the primary "sees" the load in the secondary, since that load induces current in the primary. The load that it sees is proportional to the number of turns. \$\endgroup\$
    – Chris M.
    Apr 25, 2017 at 20:44
  • \$\begingroup\$ As to the open secondary, you'll have a voltage in both windings. Without load, there can be no current. V = IR rearranges to I = V/R. With an open circuit, R is given to be infinite, so I becomes 0. \$\endgroup\$
    – Chris M.
    Apr 25, 2017 at 20:45
  • \$\begingroup\$ @ChrisM. Thank you, you have been very helpful. One more thing: Let's say I have a soldering iron which needs 30W. We'll imagine this iron is a resistor with a value say, 10000 ohms (completely arbitrary but it doesn't matter) connected across the secondary. I have a 12 VDC battery. I cannot just connect the battery across the load because the power would be (voltage ^ 2) / resistance = (12 ^ 2) / 10000 = 0.0144W which clearly is nowhere near enough. But let's say I have an inverter-like device that turns the 12 VDC into 12 VAC and steps the voltage up using a transformer. (continues) \$\endgroup\$
    – S. Rotos
    Apr 25, 2017 at 22:07

Here is how I view Transformer action, as of today


simulate this circuit – Schematic created using CircuitLab


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