I see every day high voltage, high current; high voltage, low current. I rarely see low voltage, high current; why?

I know that I can take a high voltage, high current signal, send it through a step-down transformer and theoretically get a low voltage high current (with the same power output, of course), but I never see this done.

Can someone give me a reason as to why I wouldn't want a low-voltage high-current scenario?


There's a perfect example of a low voltage, high current device: a welder. Voltage as low as a few tens of volts, currents over 100A. That's kiloWatts.
The reason you don't see the combination low voltage/high current that often is that it's a rather inefficient way to get the power required: you need thick cables to carry the high current. It's much more practical to run the 1.5kW water kettle on 230V AC, so that it only needs 6.5A, than to run it on 20V, where you would need a thick cable to carry the 75A. Remember that we get the high voltage of 230V/115V from the grid already.

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    \$\begingroup\$ Another example for high current at low voltage are desktop CPUs, at say 1V and 100A. Though this is converted down from higher voltages by converters very close to the CPU. \$\endgroup\$ – starblue Oct 19 '11 at 14:12
  • \$\begingroup\$ Thanks for the replies. Can you followup with this question? : Why do I need to convert my 230V or 120V to a lower voltage, if the device is just going to pull the current required to reach its desired Power? i.e. I can't use an 18v DC power supply to power a 5V DC device \$\endgroup\$ – IDLacrosseplayer Oct 19 '11 at 14:36
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    \$\begingroup\$ @IDLacrosseplayer - Power consumption. If the load is resistive, say 1k\$\Omega\$, at 5V it will consume 25mW, at 18V this would be 324mW, or 13 times as much. Even with a fixed current load the 5V device will consume 72% less than the 18V device. In the kettle case you want high power consumption, the same kettle at 5V will never bring the water to cooking temperatures. \$\endgroup\$ – stevenvh Oct 19 '11 at 15:14

First, it may help to think of voltage as pressure and current as flow. Various combinations of high/low voltage/current are used. For example, low voltage and high current is used to run the starter motor in your car. It's only 12V, but can exceed 100A. However note the drawback, which is thick cables going to the starter.

From a pure power standpoint any product of voltage and current that results in the same value is equivalent. However, losses in real systems can be proportional to current, especially when the electric power has to be moved some distance.

Let's say you had to deliver 1kW 100m away. In theory, 10A at 100V and 100A at 10V are equivalent. However, back in reality we're stuck with real cable, not theoretical ones. The voltage a cable has to be able to handle effects how it is insulated. The current a cable has to be able to handle effects how thick the conductor is. Insulating to 100V is trivial. Just about any layer of insulation you put on the cable will do 100V. You wouldn't save any money trying to make a cable only good for withstanding 10V. Just to overcome external mechanical stresses and normal abrasion, the insulation will be thick enough to withstand 100V without much additional cost. However, the thicker conductor costs real money. Copper or whatever you use as the conductor will have a significant cost to carry 100A for 100m. Now consider that the power lost in the cable (a resistor for this purpose) is proportional to the square of the current thru it. To get the same total power loss in the cable at 100A as you did at 10A, you actually need 100 times as much conductor material.

Now think about transferring 100s of MW between power generating stations and cities. The economics favor making the voltage as high as possible and thereby decreasing the current for the same level of power. Such transmission lines usually work in the 500kV to 1MV range. There are costs at really high voltages too. Power companies carefully study the tradeoffs in deciding what voltage to run a transmission line at.

  • \$\begingroup\$ Whoever downvoted this, what exactly do you think is wrong? \$\endgroup\$ – Olin Lathrop Apr 27 '18 at 10:55

I always like to give the water and plumbing analogy here.

If you think of the voltage as the diameter of a pipe, and the current as force of the pump moving the water. The water coming out of the end of the pipe, say in gallons per minute, is the power.

Now, if you increase the diameter of the pipe (the voltage), you can get more water through at once - thus increasing the "power".

Also, if you increase the force of the pump (the current), you force the water through faster. This however increases the pressure inside the pipe, so you need to use a pipe with a thicker wall. This is equivalent to the size of the conductor. Bigger conductors (or thicker walled pipes) can hold more current without "bursting" (melting).

Of course, making the pipe too big (too high a voltage) and the eco-protestors all turn up and start complaining about the damage laying the pipe does to the countryside, and people start to get hurt (electrocution), so you need to build bigger and better barriers to keep people away (insulation).

So, you can see that there is always a balancing act between the thickness of the pipe wall and the diameter of the pipe in order to get the amount of water through it that you need.

Yes, I know the analogy isn't quite 100% right, but it's good for getting an understanding of the relationships between them.

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    \$\begingroup\$ Your plumbing analogy in the second paragraph is completely wrong. The diameter of the pipe is more analagous to the inverse of resistance, voltage is the force moving the water, and gallons/minute is analagous to current not power. The remaining errors in this post seem to follow from your original basic misconceptions. \$\endgroup\$ – Olin Lathrop Oct 19 '11 at 16:18
  • \$\begingroup\$ Like I said, it's not 100% right, and you may equate them to something different to me. This is how I equate them. Regardless of what you equate them to, this helps me to teach other people successfully about how the different values interact. \$\endgroup\$ – Majenko Oct 19 '11 at 16:22
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    \$\begingroup\$ The water analogy can be useful, but I'm sorry, you just have it wrong. Voltage is pressure, current is flow, power is pressure times flow, just like it's voltage times current in electronics. That's how the physics works. You can't just make up what you want and still stick to the physics. \$\endgroup\$ – Olin Lathrop Oct 19 '11 at 16:37
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    \$\begingroup\$ Power is pressure times flow. In the metric system, pressure is in Pa (N/m^2, kg/ms^2), flow is in m^3/s, power is in J (Nm/s, kgm/s^2). "How can power be pressure times flow when the pressure sets the flow?" First, you're ignoring the resistance [friction, in the fluid model], Second, it'd be more like saying A = BB, and power is indeed proportional to voltage squared given a constant resistance. \$\endgroup\$ – Random832 Oct 19 '11 at 18:17
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    \$\begingroup\$ I would happily sort these disagreements out in chat, but do you ever see Olin in chat? No. Anyway, I agree that yes, there may be "better" analogies than this one, but just because it isn't the same analogy as Olin's doesn't make it "wrong". Of course, putting my variables in Olin's analogy won't work. Neither will putting Olin's variables into my analogy. The two analogies are different. There is no such thing as a "perfect" analogy. yes, some analogies may correlate more variables than another, but that doesn't immediately make all other analogies wrong. \$\endgroup\$ – Majenko Oct 20 '11 at 8:52

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