# How is current dropped through a step-up transformer, when current is dependent on resistance?

Suppose you have a 1:10 transformer. You put in 12V and you get 120V out the other end. Now the current before the transformer would be $$V=IR$$ $$I = \frac{V}{R}$$ $$I = \frac{12}{1}$$ $$I=12$$ (assuming a sample resistance of 1 ohm),

while on the other side the current would be something like $$I=\frac{120}{1}$$ given that the impedance of the load is the same.

So it appears that a step-up transformer both increases the voltage and the current. Wouldn't that just be creating free energy? Any resources I've seen tell me that a step up transformer decreases the current - but how is that shown in terms of Ohm's Law, and what's the problem in my above reasoning?

• Who says the impedance of the load looks the same on the input of the transformer? Commented Jun 14, 2023 at 4:31
• @Hearth I say it is. Don't I get to decide what load I put on the transformer, and whether its the same on both sides? Commented Jun 14, 2023 at 4:40
• You don't get to determine what the load looks like across the transformer. Think of the transformer like a lens, making a load look much bigger on one side than on the other. Connecting a 1 Ω resistor through a 1:10 transformer like that looks the same to the supply as if you'd connected a 0.01 Ω (1 Ω * (1/10)²) resistor directly to the load. Commented Jun 14, 2023 at 4:42
• There is no explanation in terms of Ohm's law because Ohm's law only applies to resistors, and a transformer isn't a resistor. (Well, okay, I admit that a real-life transformer coil does have some equivalent series resistance, but that resistance is pretty small and it's irrelevant to your question.) The voltages and currents through a transformer are determined by inductance, not resistance. Commented Jun 14, 2023 at 11:08
• @Hearth Alright, that makes more sense. Thanks Commented Jun 14, 2023 at 11:25

given that the impedance of the load is the same.

A better assumption would be that the power dissipated in the load would be the same. As the voltage has increased 10x, and the current has dropped to 1/10th, the impedance of the load has to rise 100 times.

If you load a stepup transformer with the load you suggest, that's as good as applying a short circuit to it, and an ideal transformer in that situation would draw 100x as much current. A fuse will blow, or something will smoke.

To put some figures on it, your figures from the question, let's say you have 12 V AC, with a 1 Ω load. That pulls 12 A of current, and generates 144 watts, to heat your tropical fishtank perhaps.

You still want to heat the tank, but the only heater you have available is rated for 144 watts at 120 V (you have misplaced the 12 V one in your recent move). It will draw 1.2 A at 120 V, meaning that its resistance is 100 Ω. If you connect that directly to 12 V, it will draw 0.12 A, generating 1.44 watts. So you get a 1:10 step up transformer, and transform your 12 V AC supply up to 120 V to power the high voltage heater.

Now you suddenly find the 12 V heater again, and connect it to the 120 V output of your transformer.

With an ideal transformer, the heater's 1 Ω load would draw 120 A, and it would generate 14400 watts. Obviously, not for long. Which would die first? The power supply, the transformer, the heater? Hopefully, the fuse protecting the circuit would do its job and open.

With real components, your fishtank heater transformer might only be rated for a few hundred watts, with losses at that rating of a few watts so it will stay cool. Its internal I2R heating will rise as the square of the current, so would overheat real fast if you connected it to a 1 ohm load, while the output voltage sagged to near-zero due to the voltage drop in the losses of the transformer.

• So... where is this 100x load coming from? Does the transformer have 100x impedance? What would happen if I loaded the transformer with a circuit without 100x impedance? Also you can ignore the value of 1 ohm resistance, it's just an example, you can substitute it for whatever resistance make more sense practically. Commented Jun 14, 2023 at 4:39
• @s where's the load coming from? It's the one you want to power. I've added a load-centric scenario to my answer. Commented Jun 14, 2023 at 4:57
• @Neil_UK I do like these evocative scenario examples. Have an upvote! Commented Jun 14, 2023 at 4:58
• "Which would die first? The power supply, the transformer, the heater? Hopefully, the fuse protecting the circuit" -- and hopefully not the poor fish! Commented Jun 14, 2023 at 13:42
• Be careful with the assumptions. The load characteristics can be EITHER constant impedance or constant power. Raising the voltage drops the current ONLY for constant-power loads (think motors). Raising the voltage raises the power for constant-impedance loads (think incandescent lights). Of course there can be and invariably is a mixture and the results depend on the proportions of the two. The described problem seems more the constant-impedance assumption. Commented Jun 14, 2023 at 19:19

A transformer transforms voltage by the turns ratio, current by the inverse of the turns ration, and impedance by the square of the turns ration.

If you have a 1:10 turns ratio transformer and you put 12 VAC into the primary, the voltage from the secondary will be 120 VAC. Now suppose you have a 120$$\\Omega\$$ resistor connected across the secondary, the current through it would be 1 A and the power would be 120 W.

The primary side would see the 120$$\\Omega\$$ resistance as $$Z_{pri} = Z_{sec}\times \left(\frac{N_{pri}}{N_{sec}}\right)^2 = 120\Omega\times \left(\frac{1}{10}\right)^2 = 1.2\Omega$$

The primary current would be $$12 V/1.2\Omega = 10A$$

So the power in the primary would be 120 W, the same as in the secondary (ignoring efficiency losses) and Newton is happy.

The load impedance is not the same on both sides of the transformer. The transformer imposes a certain restriction on the voltage and current, and you cannot determine the load seen on both sides independently:

simulate this circuit – Schematic created using CircuitLab

As far as the voltage source is concerned, it sees a much larger (or smaller, for N < 1) resistance than what it would see if you just put the resistor across it without the transformer.

The above can be represented even more generally like so:

simulate this circuit

The equality in both of the above circuits should be taken to mean that all node voltages and branch currents other than those through the transformer and the load (i.e., those of the voltage source in the top example; and those internal to Arbitrary Circuits A and B, as well as at the Arbitrary Circuits' ports, in the second example) are exactly equal, assuming an ideal transformer.

Your thinking is wrong because you insist on using same 1 ohm load resstance before and after the transformer on different voltages.

You don't need a transformer to realize that 1 ohm load on 12V is 12A and 144W and 1 ohm load on 120V is 120A and 14400W. These don't relate to transformer in any way.

However if you have some load on a transformer, the power taken out from the transformer must equal to power going in to the transformer.

So a 14400W load takes 14400W, and if you have 12V input, that's 1200A, when output has 120V and 1 ohm 120A load.

Now, from there you see that since the transformer reduces voltage by a factor of 10, to equal the same power, current must go up by a factor of 10.

Suppose you have a 1:10 transformer. You put in 12V and you get 120V out the other end. Now the current before the transformer would be $$\I=12\$$

No, that would not be true; the current taken by the transformer primary is dependent on the secondary load and, if the load is unconnected i.e. the secondary is open-circuit, for an ideal transformer, the primary current will be zero.

However, if you connected a 120 Ω load to the secondary, 1 amp would flow in that load but, the current in the primary would be 120 amps. With a 1 Ω load on the secondary, 120$$\^2\$$ amps would flow in the primary.

So it appears that a step-up transformer both increases the voltage and the current.

No it doesn't because, as you observe, you cannot create free energy.

• Primary current should be 10 A for a 120 ohm load, 1200 A for a 1 ohm load. Commented Jun 14, 2023 at 21:20
• Oops. Going to fix. Commented Jun 14, 2023 at 23:23

From the perspective of the source, when you add a step-up transformer with a factor of 10, the otherwise constant impedance appears to the source as 1/100th of what it was without the transformer. So yes, the current on both the primary and the secondary would be 10x higher and would take 100x as much power from the source.

But it's not "free"; the source would need 10x the input it was using before in delivering the power - e.g., add steam to the turbine or deplete the battery nominally 100x faster, etc. If the source cannot increase its output, its "internal impedance" will increase and drop the voltage of its output until the power through the impedance matches the source's maximum output ... or something melts or the generator loses its synchronism and trips offline, or whatever other practicalities apply to the theoretical source.

Any resources I've seen tell me that a step up transformer decreases the current - but how is that shown in terms of Ohm's Law, and what's the problem in my above reasoning?

The "resource" likely assumed predominantly constant-power loads, as is typical for an electric utility. A typical motor is designed to deliver torque to the shaft, and will balance voltage and current to create the requisite torque. So raising the voltage to the motor decreases its current draw, while dropping the voltage increases the current draw (and can "burn out" a motor if it drops too low... NEMA standards call for a design variation of +/- 10% in the input voltage before problems arise).

Ohm's law applies in that Power = Voltage * Current, and power into the transformer must equal power out (less losses in the core and windings). So a resistive load reacts by using more power (increasing current) with increased voltage, while inductive (motor) loads react by decreasing the current draw so that power remains fixed.