# Why is resistance across a load the V per I rather than the V per Q?

There is a load that resists the flow of electrons. This causes a voltage drop, meaning it creates a potential difference between the 2 sides of itself. This property, a load creating a voltage drop, due to electrons bumping into stuff, is called resistance. Right? My question is this:

Why is resistance NOT calculated simply by looking at how much voltage drop is created by a certain amount of charge passing through, as in R=V/Q. Instead, it is supposed to be an indication of how much voltage drop is created by a certain amount of charge, in 1 second, as in R=V/Q/t=V/I.

Why is the time factor necessary? Is it arbitrary? A whole physics is based on this equation. Or is it simply inconsequential whether time is involved? Or it makes a difference that I am missing right now?

• More correctly, it's $I=dQ/dt$, no? Jul 7 '15 at 0:50
• If $V/Q$ is a constant, then the device that you are looking at would be called a capacitor. If $V/I$ is constant, then the device is called a resistor. Jul 7 '15 at 1:04
• Resistors need not necessarily have $V/I$ constant. Jul 7 '15 at 1:19
• Yes. (Note that, in general, a statement does not necessarily imply its logical converse.) Jul 7 '15 at 2:00

Why is resistance NOT calculated simply by looking at how much voltage drop is created by a certain amount of charge passing through, as in R=V/Q.

Your statement in bold is one way to define resistance. But your words do not match the expression $V/Q$.

We're interested in the voltage drop per charge passing through, right? Well, how do you measure how much charge passes through? Or rather, by what standard do we measure a change? By timing it of course. It's then a bit more accurate to think of resistance as the constant of proportionality between a voltage drop and the amount of charge passing by per unit time, $R=V/ \frac{\Delta Q}{\Delta t}$.

Take the limit, and you get current, $R=V/I$.

After twisting my own brain for a LONG time about this, I have come to this conclusion:

Resistance is a measure of how FAST a load is able to absorb ALL the potential/kinetic energy of a given number of electrons passing through it. (I say all, because with 1 load the V drop will be equal to the V of the source. Which tells me that all energy will be gone and the "dead" electrons will be just pushed from behind as the water of a river flows on a plain, no gravity acting, just being pushed by the water "dropping" further up.)

Is this right?

The only thing I have left to grapple with is understanding the nature of voltage. Let's say it's 1 electron moving in a wire of a thickness of a few atoms. In this case current will describe how fast it's circling around, if it could, as in a particle accelerator. Still, what would its voltage describe? A sideways oscillation? What property is it that will make it suitable to do work. I would think its speed alone, but that's current.

If you look at a normal circuit with loads of electrons, it's supposed to be some kind of pressure. So is voltage how densely packed they are inside the wire? Which would be dependent on how many are pushed out of the battery per second. But that is called current, so it can't be. Still confused, what is voltage? What is an electron's potential energy? Or it's said V is DIFFERENCE in potential. Potential of what, where?

• Voltage is electric potential. Imagine a steep cliff with a whole bunch of boulders about to tumble down due to their gravitational potential. If you were to keep the boulders from falling, you'd need to exert work on the system to combat the potential. Same thing in a circuit. The voltage is "how steep the cliff is" and the electrons are the boulders. The greater the voltage, the more potential the electrons have to be thrown down the cliff. The more work you'd do to keep them from flowing. We call it a potential difference, because we focus on the relative height of the cliff. Jul 7 '15 at 22:16
• The units may help here. $1 V = 1 J/C$. Each volt corresponds to an amount of energy per unit charge. A high voltage implies a greater energy per charge. This does not imply a high current, however. If you have a few hundred electrons, you barely have a current at all, though it could be at a high voltage. This would be like a sink with high water pressure but a thin stream of water coming out. Jul 7 '15 at 22:21

Resistance is defined as $$R(i) = \frac{dV}{di}$$ and, as all definitions, is a matter of terminology. Also, as you stated, since it has to measure the drop against the flow of electrons it makes sense to take the derivative with respect to the current, which is exactly what the flow of electrons is.

Since the difference of potential must be calculated in two different points $V(x+dx)-V(x)$ you need to calculate the amount of charges moving between these two points; the same amount of charges may take different times to move between them according to the velocity: that is why you need to normalise the overall charge to the time it takes to move from $x\to x+dx$.

• That's the differential resistance
– Rol
Jul 7 '15 at 4:21