It is true that an ideal transformer can not exist (but we can approach it as closely as want by improving the engineering behind it), but I think they may be mathematically inconsistent. To illustrate, consider these two circuits:

enter image description here

In case-1, I have a \$V\$ volt battery attached to a resistor, then the power drawn is given as \$P=VI\$, this is pretty standard and one of the first things any one who has done introductory circuit theory will learn.

Now, in case-2, we have a transformer consisting of two inductors and metal rod interior (the thing you use to keep their fluxs linked), so the coil in second circuit acts as a battery for the resistor R. Now, for an ideal transformer, the power in the input circuit is the same power in the output circuit. Hence, the voltage and current are predetermined by the power conservation equation.

If we say \$I_1\$ is the current in circuit on left in case-2, and \$I_2\$ as current with \$V_2\$ as voltage for circuit on right, then we have the equation:

$$ I_1 V_1 = V_2 I_2$$

For the resistor in the second circuit, it must be that the current is determined by voltage through it (ohms law ) but we already have a predetermined current by transformer power conservation. Now, this leads to two possible current values! Which is right?[ we can isolate equation for current by relating ratio of voltage with number of turns of coil]

And, what was the mistake in my logic which lead to this contradiction? It seems like even if we account for power losses of the transformer, that is introduce a multiplicative factor \$\kappa \$ in to the power conservation equation as so:

$$I_1 V_1 = \kappa I_2 V_2$$

We still have the same contradiction.

Related question

An explicit calculation:

Suppose that we have by turn ratio:

$$ \frac{V_1}{V_2} = \nu$$

Then by the power loss accounted transformer equation: $$ I_1 \frac{\nu}{\kappa}=I_2$$

Now, for he is the 'strange' part, by Ohm's law , for the resistor: $$ V_2 = I_2 R$$

Hence, we get:

$$ \frac{V_1}{\nu} = I_1 \frac{\nu}{\kappa} R$$


$$ R = \frac{\kappa }{\nu^2 I_1} V_1$$ This equation we can isolate for \$R\$, now as V and I changes it seems that R must take multiple value but it is that R is only one single value!

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    \$\begingroup\$ now as V and I changes - you seem to think that you can set both V and I independently as you please. You cannot, You set V, and I is then determined by the R of the circuit you're applying V to. Similarly, if you set I, then V is determined by the R of the circuit you're applying V to. You can't have it both ways. \$\endgroup\$ – brhans Mar 7 at 1:58
  • \$\begingroup\$ I'm not an expert, but I think you'd benefit from examining how the output current of the transformer actually affects the input current. In a no load operating state, inductance is so high that current flow on the primary is negligible. Current flow on the secondary is negligible because of open circuit. When a load is connected, the current flowing on the secondary cancels out some of the inductance affecting the primary, allowing more current to flow on the primary. This is how the primary "sees" a different "effective resistance" than that you see on the secondary. \$\endgroup\$ – K H Mar 7 at 3:26
  • \$\begingroup\$ Ah yes, I understand now. One determines the other @brhans \$\endgroup\$ – Buraian Mar 7 at 6:44

For any given AC input voltage, the current an ideal transformer draws from the power supply is related to the load on the secondary. It's zero for no load.

If the primary voltage is Vin, then the secondary voltage is Vin*(n2/n1) ...where n2 is the secondary turns and n1 is the primary turns.

So the power at the secondary is Vin*(n2/n1)^2/RLoad

And the power at the primary is the same.

We say that the load resistance (or impedance) is reflected through the transformer by the square of the turns ratio \$(n_2/n_1)^2\$.

So a load of 100 ohms on the secondary of a 1:2 ideal transformer acts the same as 25 ohms directly on the primary voltage. Apply 10V and the power is 100/25 = 4W in either case. There will be 20V on the secondary of the transformer since it's stepping up the voltage 2:1.


simulate this circuit – Schematic created using CircuitLab

There is no contradiction. Once you fix the voltage the circuit determines the current drawn.

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    \$\begingroup\$ This is the best answer, the last statement is the crux of what I didn't understand \$\endgroup\$ – Buraian Mar 7 at 6:46

Disregarding that transformers don't work with DC voltages but AC sine waves, the problem is that you assume fixed current on the primary.

For an ideal transformer it converts voltage higher or lower based on the turns ratio. Power taken from AC source is the same as power provided to load, but because voltage is converted, so is current.

So if the voltages are different because of turns ratio not being 1:1, the currents will be different too.

  • \$\begingroup\$ Ok fixed the issue about the AC vs DC thing but this answer doesn't explain to me how there is two current values found \$\endgroup\$ – Buraian Mar 6 at 22:43
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    \$\begingroup\$ Which part of my answer that explains it you did not understand? If transformer changes voltage, it must change current too, to keep power same. \$\endgroup\$ – Justme Mar 6 at 22:55
  • \$\begingroup\$ So, we have the equation $I_1 V_1 = I_2 V_2$, we can sub in ratio of voltage as some constant, then we have I_2. According to KVL the voltage of inductor coil= voltage of resistor, hence $V_2/I_2 = R$, now this is inconsistent as R as a fixed number while V_2 and I_2 can vary freely \$\endgroup\$ – Buraian Mar 6 at 22:57
  • \$\begingroup\$ Power being equal is 'fine' but it is about how we can get two different resistance depending on how you calculate it \$\endgroup\$ – Buraian Mar 6 at 22:57
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    \$\begingroup\$ No they can't vary freely as V1:V2 is mandated by turns ratio N1:N2 which will define also I1 and I2. Nothing happens to the resistance. The resiatance just looks different when it is looked through the transformer, as transformer converts the impedance because it converts the voltage and current. \$\endgroup\$ – Justme Mar 6 at 22:59

The ideal scenario can be understood once we understand the practical case. This can be understood by using the equivalent circuit model of the transformer. When the transformer is on no load, there is a small current flowing through the primary windings. This current has two components, a magnetising current and core-loss current. The magnetising component is the one that sets up the flux in the primary. This flux gets linked with the secondary winding and induces a voltage depending on the turns ratio.

Now, if you connect a load on the secondary, circuit will be complete and hence current starts to flow. Now according to Lenz's law, this current will set up a flux opposing the flux that was set up by the magnetising current. Thus the original flux reduces. So to compensate for this reduction, the primary side draws more current from the supply such that it restores the flux to its original value. Thus, the primary current varies with the secondary current. In short, Ohm's law holds. So if we neglect the core loss component of the current, we have the ideal scenario and it works basically the same way.


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