I am studying electrical engineering, but I still have some doubts about Ohm's law and electrical power.

According to Ohm's law: \$I=\dfrac{V}{R}\$ So that means voltage is proportional to current.

Higher voltage = higher current.

Now I was taught that in order to transmit on power lines, the voltage needs to be transformed higher so that according to \$P = U * I\$, current decreases and there is less heat loss.

But wasn't current proportional to voltage? How can a higher voltage not induce a higher current?

For another example let's take a 100 W light bulb. I was told that if I have a 10 V Voltage, the light bulb "will drain" 10 Amps in order to function.

Is this wrong? Is 100 W the amount of maximum power the light bulb can handle or the power it needs, and how can it "take" 10 Amps if current depends on the applied voltage and bulb resistance?

I don't understand how current doesn't depend on the applied voltage according to Ohm's law.

  • \$\begingroup\$ 100W = 100V x 1A or 1V x 100A or somewhere in between. Same power, but one will have less I2R losses to transmission line. \$\endgroup\$ Apr 11, 2019 at 19:10

5 Answers 5


You are confusing some things (which i understandable).

In a simple resistive circuit, then with some voltage source V and a resistor R, the current is indeed \$I=\frac {V}{R}\$

Increasing the voltage will indeed increase the current.

If we need to do a transform to transmit power, we use a transformer:


simulate this circuit – Schematic created using CircuitLab

A transformer is passive; apart from inefficiencies, the output power and input power are the same, so if we had 1kV input voltage and a 10:1 step up transformer we would have a 10kV output. (We will get to the current shortly).

At the other end, if we had a 100:1 step down transformer, we would have 100V out. Let's say we have a 100W light bulb here, which would draw 1A.

Now work backwards to see what the input current is:

Input to the step down transformer is 100W at 10kV = 10mA and the input to the step up transformer is 100W at 1kV = 100mA.

Ohm's law still holds.

  • 1
    \$\begingroup\$ just dont put the 100W bulb rated for 100V on the output of 1kV and expect it to draw 100mA \$\endgroup\$ Apr 11, 2019 at 22:10

Because transformers aren't resistors.

V=IR only works for resistors. It is meaningless for any other sort of component, including semiconductors, transformers and the like. It's even misleading for resistors that get hot, such as tungsten filament lamps, as the resistance changes with temperature.

In the case of power lines, then the lines themselves do have a resistance. But if there is a transformer at the other end of the line, then just looking at the resistance of the line itself gives the wrong answer.

  • \$\begingroup\$ It might be useful to mention the term ohmic conductor, for which voltage and current are directly proportional. \$\endgroup\$
    – Pzy
    Apr 11, 2019 at 19:20

All of the answers above are good ones. My only contribution will be to point out that the raison d'être for a power line is to maximize power transmission and minimize losses. Ohm's law not only gives us the simple E=IR relationship, but can be extended to power. Of particular interest for us is the variant, P=I^2*R. In other words, the power lost through the transmission line is equal to the square of the current times the resistance.

In general, there's only so much that can be done to minimize the resistance of the wires. But setting that aside, the impact of reducing the current through the lines is tremendous as the power loss is reduced by the square of the current reduction.

Since P=IE, you can transmit the same power either at high current and low voltage or at low current and high voltage. So by multiplying the voltage by 100, you can reduce the current by a factor of 100, and then the lost power is reduced by a factor of 10,000 (assuming all else is kept equal. It won't be, though, because less current means you can use thinner conductors and higher voltage means you have to use thicker insulation).


If you double the transmitting voltage you only need half the current pulled through the wire in order to transmit the same power (P=VI). Half the current through the wire means the power dissipated in the wire has gone down by a factor of 4. (P=I^2.R).

Or alternatively, if the current through the wire has halved then the voltage dropped across the wire has also halved which also gives a factor of 4 reduction in power dissipation in the wire (P=V^2/R).

Your confusion is that you are thinking about the transmitting voltage as though it is all dropped across a resistor which it is not, it is just a generated voltage. And the current it supplies is that pulled by the consumer.

With regard to the light bulb,

The light bulb has a certain resistance and so at a certain supply voltage it will pull or draw the current it needs in order to satisfy Ohms Law. The power dissipated in light and heat will be V.I, I^2.R or V^2/R.


AC transformers raise the voltage as high as safely possible for the insulation of the network to withstand lightning and arc voltage. By doing this the impedance is raised by the voltage ratio N squared and the wires can handle more power transferred with more current and less loss.

Power transfer still obeys Ohm's Law ( ignoring small efficiency losses)

If a 100W load @ 100V is 1A or R=V/I= 100 Ohms , transforming this down to 1V@ 100A is 0.01 Ohms = 10 milliohms

Thus the V ratio is 100:1 and impedance Ratio is 100:0.01 = 10,000 or 100^2

Then in theory, transforming up to 10kV for 100W is what equivalent load impedance and current?

more reading material

Here are some simple design tools online at Digikey but if you understand Impedance from Ohm's Law, R=V/I or Z(f)=U(f) /I(f) .... then compare the transformed voltage and equate power to understand for example that a boost AC inverter divides the load impedance onto the drivers . Zin=Zout/N² where N=Vout/Vin , does this make sense? Any questions?

With more understanding , you can apply this to almost anything including RF from Z=R+jX using some algebra for Z=sqrt(R² + X² )

  • \$\begingroup\$ -1 voter is another example of a toxic low-contributer to this room and unable to express his comment to support his vote \$\endgroup\$ Apr 11, 2019 at 22:07
  • \$\begingroup\$ Wasn't me, but maybe it's because you threw in a load of equations without explaining any of them. \$\endgroup\$
    – Finbarr
    Apr 13, 2019 at 8:15

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