My question is one of physics and mathematics and sort of getting my head around something. In math, you have to convert units so that they are all the same to do any sort of calculation. For example, you cannot say:
20 meters / 5 feet = 4 meters
They’re just different units! It doesn’t work. So why can you, in physics, say:
I = V / R
How does that work? Wouldn’t the ratio change if it is, say, amperes vs. a different unit? How can formulae be non unit-specific? How does that work? And if they are, then how can you divide:
Amperes = Voltage / ohms
You’re dividing different units! How can you divide Voltage by Ohms and get Amperes? Can someone please explain? How does that work? Wouldn’t that be the equivalent of:
20 meters / 5 feet = 4 inches
All the units are different!
How does this work? This question might sound stupid, but I can’t really get my mind around it.
Thanks in advance!
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You have hit on exactly why I rebel when I can against the arcane unit definitions named after people, like volt, Newton, Hertz, Tesla, et. al. Volt is named after Volta, by the way.
Anyway, how indeed can one divide the units for V/R = I which is volts/ohms = amperes?
The answer is simple, these aren’t the real units, they are defined units based on the real ones. And the real ones can and do cancel out in division to leave you with real units for current. The arcane definitions are:
Volt = (kg.m^2)/(C.s^2) where kg is kilograms, m is meters, l is amperes, C is Coulombs, and s is seconds.
Ohm = (kg.m^2)/(C^2.s)
Ampere = C/s
So there you are in real units, I = V/R = (kg.m^2)/(C.s^2)//(kg.m^2)/(C^2.s) = C/s which is the definition of an ampere. QED.
As you can see, in real units, they do divide out: the kgs go, the m^2 go etc. That leaves just the C/s which is the definition for ampere.
Electric current is always I = Q/t.
The electric current is defined to be how many units of charge will flow per each second.
Of course, Q must be a unit of CHARGE, and t must be a unit of TIME, else the formula would make no sense at all.
For charge, we have Coulombs (C), Franklins (Fr = 0.3336 × 10^(-9) C), elementary charges (e = 1.602176487 × 10^(-19) C).
For time, we have seconds (s), minutes (min = 60 s), years (yr = 31556926 s), the list goes on.
Now, we can have currents of:
3 Fr/yr, 123 e/s, 123 e/min, 26 C/s.
Notice, thought, that since Ampere was a pretty cool guy who liked electric current, they decided to call the units for electric current, on the SI (Couloumb per second), after his name, Ampere.
So 26 C/s is really 26 A. That is just because A = C/s. Notice that THIS is unit sensitive. Well, C is an unit, and s is also a unit, so you can’t really put other units in there.
3 Fr/yr is NOT 3 A, but 3 C/s is. Both are electric currents, in anyway. It is perfectly fine to say that the flow of a current is one Franklin per each year (notice that Franklin is a unit of charge). So I = Q/t is NOT unit sensitive. You can use whatever units you want to (must be charge over time, though), but just remember that the current will only be in Amperes if the units are Coulombs and seconds.
Please edit your question if you didn’t understand, I’ll try to help!
I = V / R is unitless. It’s the same as saying
speed * time = distance.
You just have to use the correct units. If speed = miles/hour and time is hours, then the easiest thing is to report distance in miles. Note that I’m multiplying different units (hours) and (miles/hours).
Well, the “correct” units for I = V/R to work is if current is in amps, and potential is in volts, then resistance will be in ohms.
An amp in SI base units is simply A
A volt in SI base units is 1 kg m^2 / (A s^3)
And an ohm in SI base units is 1 kg m^2 / (A^2 s^3)
So if you look at the units,you’ll see the math works.
You could measure current, potential, and resistance in other units (see ESU and EMU units for older electrical units). If you did, you’d need to convert before doing math between systems.