# How does resistance convert voltage to current?

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, it's 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 the battery and a lack of electrons of the other side, 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). Again, this isn't REALLY how a battery works, but it's just meant to be a super handwavy explanation of what a battery does

When we 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. We could go into the details of why that is, but it's really 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 moving through the circuit and the cross-sectional area of the wire in the circuit.

$$I \propto AV_d$$

(the reason for the $$_d$$ in $$V_d$$ is because it's the drift velocity of the electrons, which you can reasearch more on - but if that doesn't make sense, you can ignore it for now and simply read: $$I \propto AV_e$$, where $$A$$ is the cross-sectional area of wire in the circuit and $$V_e$$ is the velocity of the electrons)

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 used to power 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 try and understand voltage in the context of electro-statics and electric fields before trying to understand Ohm's law, and stop reading my answer here. But, for after you've understood voltage, 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 "...isn't perfectly conductive..." to avoid saying "...wire has some resistance..."

Relax-you can understand the term "not perfectly conductive" without understanding "resistance".

Me saying that the wire isn't perfectly conductive means that there is something about the material composing the wire which resists the movement of electrons through it - like the way sand in a pipe would resist the flow of water through the pipe.

Because of this, in order for the wire to conduct electricity, the battery needs to create an electric field throughout that wire. 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 QUICKLY AND FOREVER. Trying to create super-conductive electrical components is one of the things our current generation of physicists is obsessed with doing.)*

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 (for why this is, why the electric field in a wire points along the direction of the wire, search it up on Stack-Exchange. I remember seeing some really good answers a while back), meaning that the voltage will simply be the product of the magnitude of the electric field through the wire, and the wire's length.

But remember: the electric field wasn't what the battery came set with as a 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 wire's length. In order for the voltage to stay the same, the electric field through 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 some 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 $$\rho$$

And thus, we've discovered our third relationship:

$$I \propto \frac{V}{\rho}$$.

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

$$I \propto \frac{V}{L}$$

$$I \propto AV$$

$$I \propto \frac{V}{\rho l}$$

Putting them all into one...

$$I = \frac{AV}{\rho l}$$

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

$$R = \frac{\rho l}{A}$$

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!

A resistor does NOT convert voltage to current. A voltage difference across a resistor causes a current proportional to the voltage difference to flow in it.

A resistor can be used to convert a voltage SIGNAL into a current SIGNAL.

An example of this is in the typical application of an operational amplifier to form an inverting amplifier. The operational amplifier tries to maintain its two inputs at the same voltage. In this case, the non-inverting input is at zero volts, so the inverting input can also be assumed to be at zero volts (unless the op amp is operating outside the ideal range.)

Therefore a voltage applied to V in causes a current to flow in R inof value I in = V in / Rin

Thus the voltage SIGNAL at V in is converted into a current SIGNAL, namely the current flowing in R in (which I have named I in.)

The operational amplifier responds by adjusting the output voltage V out to maintain its two inputs at the same voltage. V out is sent to the load, an also causes a current to flow in R f. This creates a feedback current SIGNAL I fand the op amp works to keep the current in both resistors equal so that there is no net current at the input.

Thus

I in - I f = 0

And the circuit amplifies the signal according to the equation

(V in / R in) - (V out / R f) = 0

image from https://en.wikipedia.org/wiki/Operational_amplifier_applications

There are many other circuits where a resistor is used in this way. It relies on there being a point in the circuit where the voltage is constant, but a variation in current can achieve some effect in the circuit. In such cases, a current signal is required, but if all we have is a voltage signal, we can use the fact that the flow of current in a resistor is proportional to the voltage across it to convert a voltage SIGNAL into a current SIGNAL.

Another example is a bipolar transistor used as a common emitter amplifier. A brief oversimplification: the bipolar transistor is fundamentally a current-to-current signal amplifier, but it is common to add a resistor on the input for voltage-to-current signal conversion, and a resistor on the output for current-to-voltage signal conversion.

This answer has already crossed the line from physics to electronics engineering. Any further details would be best asked on Electronics stack exchange.