Ygthkb

This is a statement that I read mainly on books about electronics.

Could someone explain why after the resistance current is what makes things work and not voltage?

Where did you read that? That's not really a correct statement if its all by itself.

In context, I'm assuming the book is referring to $frac{V}{R} = I$, which is a correct statement- but then again, so are a lot of statements if you put them in context.

I think you should try to make your question more specific as to what specifically is confusing you, but I'll try my best to explain, with as few equations as possible:

First off, it's not voltage (directly) that causes charges to move through a circuit. It's an electric field that forms within the circuit and pushes charges (electrons) through the circuit.

(And if you want to get really specific, it's an electric field due to rings of charge that build on the circumference of the components of the circuit - but that's for another question.)

However, what causes this electric field in the first place?

(What causes the rings of charge that create an electric field throughout the entire circuit)?

Usually, its a battery inserted somewhere in the circuit. Without getting into the specifics of how a battery works, let's just say there's an overabundance of electrons on one side of it and a lack of electrons of the other, and since electrons repel each other, they want to move from the side with more electrons to the side with less of them. This cramming of electrons creates an electric field opposite to the direction the electrons want to move (since the electric field points from positive to negative, while electrons move from negative to positive).

When we connect insert the battery in a circuit, the battery causes an electric field throughout the circuit that will give the electrons some net velocity (drift velocity) opposite to the direction of the electric field (since, once again, electrons move opposite to the direction of the electric field).

The electrons' velocity is approximately proportional to the strength of the electric field created by the battery. Again, we could go into the details of why that is, but it's also for another question.

The current through a circuit, by definition, is the amount of charge that flows through a cross-sectional area of the circuit per second. Obviously then, the current will be proportional to both the velocity of the electrons in the circuit and the cross-sectional area of the circuit components.

Now, let's get back to voltage, and your original question. I said we weren't going to get into the specifics of how a battery works, but there is one thing you have to accept: for some (electrochemical) reason or another, when you purchase a battery, it doesn't come with an "electric-field rating" that tells you how big of an electric field it'll create when connected to a circuit. No! Batteries come with their VOLTAGE specified! They come with the potential difference between the positive and negative terminals specified!!!

(From the wording of your question, I sugggest you REALLY try and understand voltage before trying to understand OHM's law, and stop reading my answer here. But I'll continue anyways)

Let's look at what the fact that batteries come with a set potential difference between the positive and negative terminals implies. Remember, voltage is the product of the dot product of the electric field pushing the electrons and the distance the electrons move ($V = int [vec{E}cdot dr]$):

1. Let's say we get some wire and connect the positive and negative terminals of the battery together with it. This wire isn't perfectly conductive, so that we don't short our circuit.

(At this point you're probably angry about the fact that I said "perfectly conductive" to avoid saying "this wire has some resistance" Relax-you can understand the term "not perfectly conductive" without "resistance". That the wire isn't completely conductive means that there is something of the material that composes the wire which resists the movement of electrons a bit. Because of this, in order for the wire to conduct electricity, the battery needs to create an electric field through it. If the wire was perfectly conductive ("super-conductive" is the correct term actually) then even without an electric field pushing them forwards, electrons could move through the wire at REALLY FAST SPEEDS.)

Right, so we connect the positive and negative terminals of the battery with this wire, and the battery creates an electric field within the wire that causes electrons to start moving from the negative terminal of the battery to the positive terminal. The electric field points in the direction of the wire, meaning that the voltage will be the product of the electric field and the length of the wire.

But remember: the electric field wasn't what the battery declared to be constant - for some reason or another, the battery comes with a fixed voltage difference between the positive and negative terminals.

So, let's say we were to double the length of the wire. In order for the voltage to stay the same, the electric field in the wire would get divided by 2. And since the electric field was approximately proportional to the velocity of the electrons in the wire, the velocity of the electrons would also get divided by 2. Meaning that the current would get divided by 2 as well.

We've just discovered our first relationship: for a certain voltage (which I can't emphasize enough is what is FIXED in the circuit due to the way in which a battery works) caused by some battery throughout the circuit, the current is inversely proportional to the length of the wire, but directly proportional to the voltage, since the voltage was directly proportional to the electric field caused within the wire.

$I propto frac{V}{L}$

2. Now, let's say we were to double the cross-sectional area of this wire. Once again, remember what's fixed is the voltage of the battery, which is given by the electric field the battery generates within the wire multiplied by the length of the wire.

Since the length of the wire didn't change, the electric field in this wire won't change either. Meaning that the velocity of the electrons moving in this wire won't change either.

However, going back to the definition of current, we saw it was both proportional to the velocity of the electrons and the cross-sectional area those electrons were traveling through!

That means that if we double the cross-sectional area of the wire, the current (amount of charge flowing through that area) must've double as well. Aaaand, we've discovered our second relationship: the current is proportional to the cross-sectional area of the wire.

$I propto AV$ (where $A$ is the cross-sectional area, and once again the current is proportional to the voltage $V$ because the voltage is proportional to the electric field created within the wire, which is proportional to the velocity of the electrons. I know I'm repeating myself a lot, but I feel like it'll help. If the repetition is annoying, tell me and I'll edit this post).

3. We've got one more relationship left.

Remember when I said the material making the wire wasn't "perfectly conductive", so that it required an electric field of some significant magnitude to be created within the wire in order for electrons to move through it? Well, there's actually a way to measure how "perfectly conductive" this wire is - how much it resists the flow of electrons through it. It's called the "resistivity" of the material. We won't go into what causes materials to have more or less resistivity, but the best way to think of it is that the higher the resistivity of the material, the more tightly its atoms hold on to their electrons.

What that means is that the higher the resistivity, the greater of an electric field will be needed in order to move the electrons at a certain velocity through the wire. If two wires are identical in shape, but the material composing the second wire has twice the resistivity as the material composing the first wire, then in order to cause the same current through the second wire as through the first wire, we'll need an electric field of twice the magnitude through the second wire.

The resistivity of a material is usually denoted by $ρ$

And thus, we've discovered our third relationship:

$I propto frac{V}{ρ}$.

Let's now put the three relationships together. We have:

$I propto frac{V}{L}$

$I propto AV$

$I propto frac{V}{ρ}$

Putting them all into one...

$I = frac{AV}{ρl}$

Now, all Ohms did (or whoever invented the concept of resistance) was give the term $frac{ρl}{A}$ a name. He called it the RESISTANCE of the wire.

$R = frac{ρl}{A}$

ANd thus, the current through the wire is given by:

$I = frac{V}{R}$

Which agrees with the wording of your original question. YAY!!!!

Hope that helped :). Please suggest edits and comment if something's confusing!!!! I'll edit this answer as much as need be - after like 40 minutes of writing, I'm really attached to it!!!

Best of luck with your electrical endeavors!

"Resistance converts voltage to current" is not an accurate statement.

The electric potential difference maintained by the two ends of a battery creates an electric field through the circuit. Charges flow along this electric field in an attempt to minimize their energy.

But materials have large numbers of densely-packed atoms, so the electrons have to bounce around, colliding with atoms in an attempt to get to the bottom of the potential well. They actually move quite slowly for this reason. This is a basic model of resistance.

I don't know if this helps you---perhaps I'm misunderstanding the statement you say you read in books, but I think it is a false statement.

I could imagine the resistance being thought of as a "conversion factor" or sorts, giving the corresponding voltage for a given current via Ohm's law, which is taken as faith in some electronics books. But I think that's not particularly physically helpful.