Let's say we have a step-up transformer which receives an input of 2 Amps at 100V. (200W) Let's assume we get a 1000V and 0.2 Amp output. (same 200W) Output amps are obviously determined by the input power and the output voltage.

Now, considering all this, can we really say that I = V / R holds under all circumstances? Because in this case, however small the resistance of the secondary circuit, amps will still be constrained by the input power of the primary circuit.

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    \$\begingroup\$ You are trying to fix (or assuming fixed) too many variables. Assuming 100% efficiency, Vin * I in = Vout * Iout. You can fix 3 of those, the fourth will follow. \$\endgroup\$ May 13, 2014 at 14:18
  • \$\begingroup\$ You are conflating several different ideas. You start by assuming that a transformer is ideal, then you start talking about Ohm's law. Ohm's Law only applies to an ideal resistor, not to any other circuit element. Now, if you want to add an ideal resistor in series with the windings of an ideal transformer, then you can ask a meaningful question. \$\endgroup\$
    – Joe Hass
    May 13, 2014 at 20:50

3 Answers 3


Because in this case, however small the resistance of the secondary circuit, amps will still be constrained by the input power of the primary circuit.

From elementary ideal transformer theory, if the secondary is loaded with a resistance \$R\$, the primary voltage and current phasors must satisfy

$$\frac{V_p}{I_p} = \frac{N^2_p}{N^2_s}R$$

So, one cannot freely specify both the primary voltage and current. If the primary is driven by a voltage source, \$V_p\$ is fixed and the primary power (and thus, secondary power) is given by

$$P_p = \frac{|V_p|^2}{\frac{N^2_p}{N^2_s}R}$$

In other words, given the primary voltage \$V_p\$ and resistance \$R\$, the power is fixed.


I = V / R always holds. With the resistance of the secondary circuit, your output voltage and current will drop accordingly. If you actually transmit all 200W over to the secondary circuit (ideal case), then the resistance in the secondary coil would act like the internal resistance in a battery or power source. The total power consumed will then be distributed between your load and your secondary coil according to standard voltage divider equations. The final result is that your output load will see less voltage and power as its resistance drops because the secondary coil resistance is consuming more and more power.

  1. Ohm’s law states that when voltage goes up, current goes up.
  2. Transformers must follow the rule Voltage1 x Current1 = Voltage2 x Current2 (Power Conservation). Thus as Voltage2 goes up (by using a step-up transformer), Current2 goes down.

One equation says that current goes up when voltage goes up, the other says that current goes down. Now, how does this make sense?

Imagine the following thought experiment: we connect a 1 volt AC generator to a 1 ohm resistor and measure the current. By Ohm’s Law, we should get 1 ampere of current. Now imagine we stuck a 1:10 transformer in the circuit, splitting our one circuit into two electrically-separate circuits. If we measured the current in the second circuit we would see that the Ohm's law still holds, we would get 10 volts(remember it's a 1:10 transformer)/1 ohm = 10 amps. Therefore, in accordance with the Ohm's law as voltage increased so did the amperage.

Now is the exciting part. In order to preserve the condition Voltage1 x Current1 = Voltage2 x Current2, the intial circuit (1 volt, 1 amp) immediately starts drawing 100 amps of current!

Therefore, you may say that the current in the second circuit is relatively lower(10 amps) than in the first one(100 amps) but higher when compared to it's previous state.(1 amp)

The initial circuit obeys Ohm’s law too: the transformer acts somewhat like a resistor, whose resistance goes down as the current demand of the right-circuit goes up.

An expanded version of this explanations is presented here

  • \$\begingroup\$ We'd rather you summarize the important points rather than just give a link. The link may go bad at any time. \$\endgroup\$
    – The Photon
    May 13, 2014 at 17:12

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