# Practical limits of Ohm's law

1. Feb 24, 2017

### bbrianc

Simple premise: take ohms law with a purely resistive ac circuit. 1 volt through in series with a 2 ohm resistor, I have 0.5 amps of current. I want to maintain constant power. In practice, "how low can I go" with the voltage to increase the current? There's probably many variables I'm missing, so please feel free to mention them. If it's more straight forward to use rms for voltage that's fine. Thank you for your responses!

Last edited: Feb 24, 2017
2. Feb 24, 2017

### magoo

You should be saying how high can you go with voltage to increase the current!

Ohms Law E = I R or I = E / R

3. Feb 24, 2017

### bbrianc

Thanks for the correction! I forgot to add the goal of keeping constant power my mistake. Original question edited. Thank you

4. Feb 24, 2017

### cnh1995

Well, you can't reduce the applied voltage and have a higher current through a fixed resistance as it contradicts Ohm's law.
For changing voltage and current levels while maintaining constant power, a transformer is used.

5. Feb 24, 2017

### phinds

Forgetting your mistake about lower voltage meaning higher current, since I understand what you mean, you could, in theory, probably go a lot lower than it would be possible to measure, given the sensitivity and precision of actual voltage and current meters, so it's interesting only in a theoretical way. In practical terms you would also run into problems trying to create a voltage source that would produce extremely low voltages and extremely high currents. This too, would be a limiting factor in practical terms.

6. Feb 24, 2017

### Staff: Mentor

you're right, many factors. High on the list is the circuit components melting.

But there are some good physics lessons in the question. Some laws, like Newton's Laws are derived from theory. Their range of applicability is much broader than an empirical law like Ohms Law.

It is valuable to keep in mind that both theoretical and empirical approaches to physics contribute useful knowledge.

7. Feb 24, 2017

### magoo

I = E / R

multiply numerator and denominator by I and get
I = (E I) / (R I)

since E I is a constant and R I equals E, you get

I = constant / E which applies for constant power loads

This explains why increasing the voltage lowers the current.

A lot of work on energy conservation in the U.S. can be found under the heading "conservation voltage reduction". Some electric utilities have even lowered their average service voltage to reduce energy usage on circuit. Their data also reveals that circuits with a lot of motor load don't behave as well as those with more resistive loads, which ties in with this discussion.

8. Feb 24, 2017

### NTL2009

I think the OP is asking something like - does the formula fall apart near the electron-volt level? Or, can we even have a current if the source were below the electron-volt level?

9. Feb 24, 2017

### magoo

If that's the case, then you can ignore my post. Sorry if I misunderstood.

10. Feb 24, 2017

### NTL2009

Well I'm just guessing as well. Hopefully the OP will get back and clarify the direction, it was pretty open ended I think.

11. Feb 24, 2017

### bbrianc

Yes you have the right idea. Maybe not that low, but more like in terms of components and ampacity of whatever conductor is being used... coupled with a voltage level too low for a real source to generate.

Great discussion so far thank you all for contributing.

12. Feb 24, 2017

### NTL2009

Well, using slightly easier numbers as an example, say 1V, 1 A and 1 Ohm for 1 Watt of power, and if you want to maintain that power with a 1mV source, you need 1000 Amps, and therefore 1 milli-Ohm R. Let's also assume some real-world physical set up, say on a lab bench, with something like 6" conductors between your source and load. And also arbitrarily say you want your conductors to provide no more than 1% error, then your conductors must have less than 0.01 milli-OHM per foot. I'll leave it to you to look up a conductor size that would reach that level.

And you could take it another factor of 1000, to 1 uV, 1 Mega-Amp and 1 u-OHM, and conductors of 0.01 u-OHM per foot. I'm guessing it won't fit on a bench at that point.

13. Feb 24, 2017

### Staff: Mentor

That's how I interpreted it. R=V/I expresses a linear relationship between V and I. It is empirical, and obviously valid only with certain ranges of V and I. In this case the nonlinearities (such as flashover, or melting) occur at high values of V and I, more easily than small ones. But even at the small end we have QED (quantum electrodynamics) instead of Ohm's Law.

14. Feb 24, 2017

### Baluncore

Constant power makes the product of load V and I constant. Resistance is the ratio of V / I.

Can the resistance of a resistive load always be further reduced? Yes, but it is not practical. How will you pay for the conductor material and the impedance transformation required to drive such a low resistance load?

High voltage transmission lines use low currents to reduce the I2R losses, that increases transmission efficiency. Why are you trying to make your resistive load really difficult to drive?

15. Feb 24, 2017

### bbrianc

Thank you all these are excellent answers pretty much exactly what I was looking for. My basic thought experiment is related to magnetism actually. (More current gives you a stronger B field)

16. Feb 24, 2017

### Baluncore

Winding n turns on a coil gives n times the magnetic field of one turn. The wire is n times longer and the wire cross-section is proportional to 1/n, so the wire weighs the same. If the voltage available is known, then the current will be limited by resistance. You choose the wire size, and the number of turns, so as to match the coil resistance to the power supply voltage.

Last edited: Feb 25, 2017