Usually, it is said that step-up transformers are used to transmit power at high voltage and low current, so that $I^{2}R$ losses are minimum. But what does this "low" current actually mean? Yes, it is low compared to what is in the primary circuit, but the primary current is in fact dependent on the number of turns and load in the secondary. So is it correct to say step-up transformers reduce current?

To make things clear, suppose the primary voltage is 10V and the secondary voltage is 100V. If there is a 10 ohm resistor in the secondary, then the secondary current is 10A. Since the secondary current is 10A, the primary current should be 100A. (Right?). If the resistance of the wire (in secondary) is 'R', then the power loss in the secondary circuit is $10^{2}R = 100R.$ Now, if you just remove the transformer and directly connect the power source to the resistor, then you have a 10V source and 10 ohm resistor, which means the current in the circuit is just 1A. That would imply that the power loss is just $1^{2}R = R$, 100 times less than what was when the transformer was used.

What is it that I'm missing here?

  • 1
    $\begingroup$ Step up is used to get it ready for transmission. It is stepped down before you finally use it. Your example it not perfect because the load is changing. When 10 ohm is connected to 100 V output, the load is 1000 W. When 10 ohm is connected to 10 V, the load is just 10 W. We need to keep the load same for the comparison to be fair. :) $\endgroup$ Commented Jan 3, 2022 at 7:47
  • $\begingroup$ @Whiskeyjack Can you elaborate and write an answer? I think this is the point I'm having trouble with. What should I do to make the comparison fair in this specific example? $\endgroup$
    – Sasikuttan
    Commented Jan 3, 2022 at 8:00
  • $\begingroup$ And what is the “power source “?In the secondary circuit your R is the winding resistance. What is the R in your second example? $\endgroup$
    – Bob D
    Commented Jan 3, 2022 at 9:32

4 Answers 4


The step-up transformer boosts voltage on the secondary. Since the power must be preserved on primary and secondary, the current is reduced by the same factor:

$$V_P \cdot I_P = V_S \cdot I_S$$

The loads are rarely characterized as resistances. In reality they have some complex impedance. For example, motors are modelled as inductances, air conditioning devices even have negative impedance etc.

For transmission you should think of it like this: there is some power $P$ required by loads; if the voltage on the transmission line is $V$ then the current will be $I = \frac{P}{V}$. The total power output equals loads plus ohmic losses in the transmission line.

If the load is purely resistive as in your example, then power increases with voltage since $P_R = \frac{U^2}{R}$.

See related discussion about power transmission: https://physics.stackexchange.com/a/682024/149541


Assume that we have a generator or a power plant that generates electrical energy to power up an appliance located far away. I'll make these assumptions:

The working voltage required by the appliance: 100 V

Distance between the appliance and the power plant: 1 km

Transmission wire resistance: 1 ohm

Appliance load: 10,000 watts (if the load is resistive, it means the load resistance is 1 ohm for the rated voltage of 100 V)

I am not fixing the voltage that the power plant can generate. Deliberately keeping it flexible to simplify things.

Case 1:

We don't use a step-up transformer. The load will require 10000/100 amps = 100 amps of current. Voltage drop in transmission line = 100 amps x 1 ohm = 100 V.

Thus power plant needs to generate 200 V and 100 amps to deliver full power to the load.

Wasted power in transmission line = 100 A x 100 A x 1 ohm = 10,000 watts. Thus almost half the power gets wasted in the transmission line itself. I deliver 10,000 watts to the load and waste 10,000 watts in the transmission wires.

If we assume that the power plant can only generate 100 V. Then we can't even supply full working voltage to the load because of transmission voltage drop.

Assuming 100 V as power plant voltage and the load being a resistive load (Rload = 1 ohm), the situation becomes somewhat like this:

Total circuit resistance (load + transmission lines) = 2 ohms

Current = 100V/2ohms = 50 amps

Power wasted in transmission lines = 5,000 watts.

Power delivered to the load = 5,000 watts.

Thus we can't deliver the full power to the load. Also, my load will get only 50 V instead of 100 V. In practice, many loads are non-resistive and the way household appliances are made, it might not even turn ON at half the working voltage.

Case 2:

Let's have a step-up transformer before transmission lines and a step-down after the transmission line. Let's generate 100 V from the power generator and step up this voltage 100 times.

On the input side near the power plant, we feed 100 V and get 10000 V out. Current at this voltage = 10000 / 10000 A = 1 A.

I just need to send 1 A through my 1 km long transmission cable. Voltage drop = 1 x 1 = 1 V which is negligible.

Power loss in the transmission line = 1 x 1 x 1 = 1 W.

I will be able to step this voltage down near the load using a step-down transformer and feed to the load. My load gets full power (10,000 W) while I lose only 1 W in the transmission lines.


You are missing the point.

The step-up is between the originally produced electricity in the primary, and the voltage in the secondary which is transmitted in the transmission lines.

When you finally use it, just before the electricity is delivered to your home, it is stepped down by a very large factor. The "arrival primary" has the same voltage as the "departure secondary", very high. Well, not exactly the same because of ohmic losses in the transmission line, but the whole point is to keep those as small as possible. The "arrival secondary" has much lower voltage. Thus the current in the transmission line is much less than the current used at the endpoint.

I don't know the exact voltage when the electricity is produced, and thus how the current that is produced is compared to what is actually used at the other end. There will be ohmic losses in the production, because the current is higher than in the transmission lines, but much fewer losses than what would occur in the transmission lines if the voltage had not been stepped up (and thus the current stepped down) for the transport.

  • $\begingroup$ I know all that you stated, but I'm not able to connect it with my specific example. @Whiskeyjack has made a good point on 'load', and I think that is where I'm having trouble. Would you mind elaborating on that? $\endgroup$
    – Sasikuttan
    Commented Jan 3, 2022 at 8:31
  • $\begingroup$ @Curiouserandcuriouser OK, so suppose the voltage in your home is 100V and you use a bulb of 100 ohm, so that the current is 1 A and your bulb uses 100 W. The load is 100 W. But the lines that transport the electricity from the power station may be at 100.000 V. The contribution of your bulb to the consumption in your city will be just 1/1000 of 1 A. That would not contribute such an enormous loss on the lines. Now in the power station itself, it is possible that the initial voltage is, say 1000 V (I have no idea of what it really is). There, your bulb adds 1/10 of 1 A, more than on the lines. $\endgroup$
    – Alfred
    Commented Jan 3, 2022 at 8:41
  • $\begingroup$ @Curiouserandcuriouser But 1/10 of 1 A is acceptable in the power station, which is "small". But it has to be transported on very long distances on the transmission lines. Such a current, multiplied by the number of bulbs in your home and the number of houses in your city, would cause an enormous loss on long-distance transport. Hence the need to reduce to by a first voltage step-up, before the final voltage step-down. $\endgroup$
    – Alfred
    Commented Jan 3, 2022 at 8:48

If you connect a one ohm resistor to a one volt AC power supply, you get one amp (rms) of current flow (and dissipate one watt of power). If you insert (between the power and the resistor) an ideal transformer which steps the voltage up to ten volts, then the resistor will draw ten amps of current and dissipate 100 watts of power. This power must be delivered to the primary of the transformer as 100 amps from the power supply. So the current coming out of a step-up transformer is less than what goes in, but is more than what would floe without the transformer.


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