In India we use electric supply at 230 volts / 50 Hz whereas in US it is 110 V / 60 Hz. Since power loss in an equipment (say resistive heating element) is directly proportional to the voltage, therefore use of 230 V should be less energy sufficient. Why isn't so?

  • 2
    \$\begingroup\$ Uh, wouldn't you want to use the power in a resistive heating element? \$\endgroup\$ Aug 21, 2014 at 3:21
  • 1
    \$\begingroup\$ Do you mean "less energy efficient"? And, could you use a different example? Because with resistive heating element, it's confusing what is the power loss. \$\endgroup\$ Aug 21, 2014 at 7:18
  • \$\begingroup\$ Yes it is true for DC, but look at your data: 50 Hz, so you have an AC voltage. Alternate current avoid the problem of power loosing. That is why AC is used instead of DC in eletric supply. \$\endgroup\$
    – albanx
    Aug 21, 2014 at 10:59
  • 1
    \$\begingroup\$ @albanx: P=IV and V=IR apply to both DC and AC. There's nothing about AC that makes it ignore resistive power losses. In fact, electric stoves, toasters, and incandescent light bulbs all work on this principle, turning AC voltage dropped across a resistor into heat and light. You may have been told that AC voltage conversion in a transformer is "lossless," and if you squint, that is true, though there are IR and other losses here, too. \$\endgroup\$ Aug 21, 2014 at 11:47
  • \$\begingroup\$ 230V? Uhhh Here at Nagpur I've been testing the reading on the supply line with my DMM at random over the last year. Typical reading is 249V ... in the evenings it drops to around 246V \$\endgroup\$
    – Everyone
    Oct 10, 2014 at 6:52

3 Answers 3


Power is the product of voltage and current. (P=IV) If you double the voltage, you halve the amount of current required to deliver the same amount of power.

Voltage drop in a conductor is the product of the current running through it and the conductor's resistance. (V=IR)

That's the simple answer to your question: lower current through a fixed resistance equals lower voltage drop, so a higher proportion of the voltage gets through to the point of load.

Say we want to deliver 1 W to a device. We could do it any number of ways, such as 1 V @ 1 A or 1000 V @ 1 mA. Let us say there is 1 mΩ of resistance between that power source and the point of load. What are the consequences of each choice?

  1. 1 V @ 1 A: The voltage drop from 1 ampere through 1 milliohm is 1 millivolt. The power lost is thus 1 mV × 1 A = 1 mW.

  2. 1000 V @ 1 mA: The voltage drop from 1 milliamp through 1 milliohm is 1 microvolt. The power lost is thus 1 μV × 1 mA = 1 nW.

The first scenario literally wastes a million times more power.

There's no free lunch, though. Higher voltage requires better insulation, greater spacing between conductors, or both.1 It's also harder to make high-voltage semiconductors. This is why the supercapacitor energy density problem hasn't been solved by "just" charging them to a megavolt.2

You may then ask, if the costs from insulation and such go up as a function of the voltage, doesn't that eat up the benefits? The simple answer is "no," which is obvious from the fact that power generation companies do everything they reasonably can to keep their costs down. If they're going to higher and higher voltages, they must have a good reason. We can infer that the cost of insulation as a function of voltage goes up slower than the cost of lost power as the inverse function of voltage.

The 115/230 volt thing is small potatoes. Except in very large buildings, that difference only shows up within shouting distance of the outlet that you're measuring it on.

From the power pole transformer back to the power generation station, international power systems are more similar than different. The power generation station typically boosts the generated power up to tens or hundreds of thousands of volts to take advantage of the lower power loss. They primarily use free air as their insulator, which is of course cheap, as long as you can afford the space it takes, which you can at the top of an electrical distribution tower.

The world's power distribution system does of course use things other than free air as an insulator. Glass and ceramic insulators have been developed to the point of art.

A vast amount of knowledge and expertise goes into dealing with the insulation problem, every day. For example, when a big coil in a motor-generator set arcs over, copper vaporizes and windings can go flying. There are companies that do nothing other than cope with such incidents, skillfully re-winding the coils to bring the M-G set back into operation.


  1. Increasing the space between two conductors increases the amount of insulation but does not increase the quality of that insulation, which is why I talk about these two aspects separately.

  2. In the energy pseudoscience world, you'll find people pointing out that the energy density in a capacitor is a function of the square of the voltage. That is, charging a capacitor to twice the voltage requires storing four times the energy. So, the logic goes, the energy storage problem is easy, innit? Just charge a capacitor to a billion volts, and you'll have all the power on tap you could want.3

    That's one problem solved; let's move on to world peace.

    Too bad it's only easy when you ignore the costs and difficulties involved in creating and using such capacitors.

  3. 1 farad at 1×109 volts is approximately sufficient to power Norway for a year. Just scale it up a bit and you can vaporize the moon, no problem.

  • \$\begingroup\$ Maybe you can help me understand then: why do countries use this voltage as the "standard" as opposed to 110V, if higher voltages require more hardy (and therefore likely more expensive) infrastructure? \$\endgroup\$
    – Asa Graf
    Aug 21, 2014 at 3:45
  • \$\begingroup\$ @AsaGraf: I've answered that now in the second half of the answer. \$\endgroup\$ Aug 21, 2014 at 4:12
  • \$\begingroup\$ @Asa - insulation is far less expensive than copper (or other metal), especially when you add up tens of thousands of miles of it. Higher voltage means they can use much smaller wire to carry it. A wire carrying 100 volts at X current (=100X watts) would also be able to carry X current at 100,000 volts (=100,000X watts). \$\endgroup\$ Aug 21, 2014 at 4:30
  • \$\begingroup\$ @Warren: Thanks for great additional information ! \$\endgroup\$
    – bhavesh
    Aug 21, 2014 at 5:38
  • 2
    \$\begingroup\$ @albanx: What AC consideration do you refer to? If you are simply making the same comment as above, then there is nothing to consider. P=IV and V=IR apply to AC just as well as to DC. \$\endgroup\$ Aug 21, 2014 at 12:15

Higher voltages make power transmission easier!

Ideally, the wires we use to transmit electricity would have no resistance. Unfortunately, they do, and the resistance is proportional to the length of the wire and inversely proportional to the cross sectional area of the wire.

According to Ohm's law, the voltage drop across a resistive load is V = IR, and the power delivered to such a resistor is P=VI, or if we substitute IR for V as in the previous equation, we get P=RI^2.

In power transmission, power delivered to a wire (which is essentially a resistive load) is bad. That power does nothing but generate heat in the wire. So it becomes necessary to try to minimize the losses in the wire.

From the above equations, we can see that the power lost to a wire is proportional to voltage, but proportional to the SQUARE of current. So if we want to get the same power at the end of the wire, we can send a low voltage and a high current, or a high voltage and a low current.

Minimizing current and maximizing voltage leads to the least power lost in power lines!

Additionally, we could make wires fatter to reduce their resistance, but that comes at a very high cost. Doubling ALL the wires in a power grid would likely cost far more than doubling the voltage, and both would reduce power losses by a factor of 1/2.


hertz also plays a role 50 vs 60 hertz...acft generators run at 400 hertz and is much more efficient than 50 or 60 hertz. hertz = cycles per second. the slower the hertz the slower the generator has to turn. cheaper generators. that's why we started at 60 hertz here and 50 hertz overseas. most company's would like to go to 400 hertz but the cost of doing so now would be too big the consumer would need to change almost if not all electrical devices in the home. ac is easy to transmit down power lines but dc is more efficient at running household devices. ac could be converted to dc but would cost power company's to install and use converters. again your household devices would need changed to dc devices. there are always trade offs when using electricity and we are stuck with what we have. be glad we don't need to do without running a fridge by hand would not be fun.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.