## Level 2

**Level 2**requires school mathematics. Suitable for pupils.

# Ohm's law: Formula, Graph and 3 Easy Examples

Ohm's law, named after the German physicist Georg Simon Ohm, describes how electric current \(I\) and electric voltage \(U\) are related.

So if you want to understand and build up electric circuits, you won't get very far without Ohm's Law. So it's worth to understand this law. Especially because it is the simplest physical law you can imagine.

In previous lessons, you learned what electric current and electric voltage physically mean. Here is a quick recap!

## The 1st ingredient: Voltage

If you separate positive and negative electric charges and put them in two boxes, the bunch of negative charges forms a **minus pole** and the bunch of positive charges forms a **plus pole**.

An **electrical voltage** is created between the negative and positive poles. The voltage is abbreviated by the letter \(U\) and is sketched like this, for example. The voltage tells how much kinetic energy a positive charge would gain if it traveled from the positive to the negative pole. Voltage is measured in volts, abbreviated with a \(\text{V}\). Let's assume that there is a voltage of \( 10 \, \text{V} \) between the poles.

## The 2nd ingredient: Electrical conductor

We can't do much with voltage alone. Next, we need an **electrical conductor**. This could be a wire made of copper or aluminum, for example. We need this conductor to connect the two poles with each other. Of course, it makes a difference whether we use copper or aluminum for the connection, because different materials conduct the charges differently. But more about that later.

## The 3rd ingredient: Electric current

Since the two poles are now conductively connected, the positive charges can move to the negative pole along the conductor. They are eventually attracted by the negative charges. Thus, there is an **electric current** \(I\) through this conductor. The negative charges, of course, would also travel to the positive pole. However, we have fixed them in the box so that they cannot move, in order not to make our thought experiment unnecessarily complicated.

The resulting electric current \(I\) is measured in amperes (or short amps), abbreviated with the letter \(\text{A}\). Let us assume that a current of \( 1 \, \text{A}\) flows through the conductor.

Over time, the number of positive charges at the positive pole will decrease because the charges are moving to the negative pole all the time. Thus, the voltage and current would also decrease over time because fewer charges are separated. To prevent this, we will constantly supply the charges to maintain the charge separation. This resupply of charges is the purpose of a *voltage source*.

With a voltage source, we make sure that the voltage and current stay constant and don't get lower.

## Ohm's law as a graph

Our voltage between the poles is set to \( 10 \, \text{V}\) and a current of \( 1 \, \text{A}\) flows. Now let's change the voltage and see how this affects the electric current. We can change voltage by changing the number of separated charges.

- If we increase the voltage from \( 10 \, \text{V}\) to \( 20 \, \text{V}\), then the current increases from \( 1 \, \text{A}\) to \( 2 \, \text{A}\). We have
*doubled*the voltage \(U\) and thus the current \(I\) has also*doubled*. - If we increase the voltage from \( 20 \, \text{V}\) to \( 60 \, \text{V}\), then the current increases from \( 2 \, \text{A}\) to \( 6 \, \text{A}\). We have
*tripled*the voltage \(U\) and thus the current \(I\) has also*tripled*. - If we decrease the voltage from \( 60 \, \text{V}\) to \( 40 \, \text{V}\), then the current decreases from \( 6 \, \text{A}\) to \( 4 \, \text{A}\). We have decreased the voltage \(U\)
*by the factor*\(\frac{2}{3}\) and thus the current \(I\) has also decreased*by the factor*\(\frac{2}{3}\).

No matter by which factor we change the voltage, the current also changes by the same factor.

We can illustrate the measured values in a diagram. On the \(y\) axis, that is the vertical axis, we plot the voltage. And on the \(x\) axis, that is the horizontal axis, we plot the current. This graph is called **voltage-current graph** because it illustrates the relationship between voltage and current.

If we now connect the measured data points with each other, we get a *straight line*. Whenever a straight line comes out on a plot, then we say that the plotted quantities are related *linearly* to each other. So we have found a law that voltage and current are linearly related. And this is exactly the statement of Ohm's law!

What if we plot our voltage and current measurements and get such a graph like in illustration (4)?

Does this conductor, which connects the poles and through which a current flows, fulfill Ohm's law? No, it does not! Because it is not a straight line. A conductor fulfills Ohm's law only if the voltage-current graph results in a straight line. An electric current through such conductors, which give a straight line, are called **ohmic conductors**, because they fulfill exactly Ohm's law. For example, a conductor made of copper is an ohmic conductor. This is because if we apply a voltage to the copper conductor and a current flows as a result, then the current and voltage are linearly related.

The straight current-voltage line is represented by its *slope*. We denote the slope as **electrical resistance** and abbreviate it with the letter \(R\): `\[ \text{Slope of the straight line = Resistance } R \]`

We can therefore say:

- A
*steep*straight line has a large slope and thus represents a*large*electrical resistance. - A
*shallow*straight line has a small slope and thus represents a*small*electrical resistance.

What effect does the slope of the straight line have on the voltage and current?

- A shallow straight line, that is a small resistance, means: If you increase the voltage \(U\) only very slightly, then the current \(I\) increases very much.
- On the other hand, a steep straight line, that is a large resistance, means: If you increase the voltage \(U\) just a little bit, then the current \(I\) also increases only very slightly.

## Ohm's law as a formula

How can we now unite the three quantities, voltage \(U\), current \(I\) and resistance \(R\) in one formula? Let's use the language of physics, that is mathematics, to translate this straight line into a formula.

From mathematics we know that a straight line passing through the origin of the coordinate system is described by the equation of a straight line:`1\[ y ~=~ m \, x \]`

Here \(m\) is the slope of the straight line. In our case the slope of the resistance is \(R\). So let's replace \(m\) with \(R\): \( y = R \, x \). The \(y\) is at the \(y\) axis and is the voltage \(U\). So we replace the \(y\) with \(U\): \( U = R \, x \). The \(x\) is on the \(x\) axis and in our case represents the current \(I\). Let's replace the \(x\) with \(I\). And we have a straight line in the diagram translated into a formula. So Ohm's law as a formula is:

**Ohm's law as formula**

`2\[ U ~=~ R \, I \]`

## Electrical resistance

You can find out the unit of resistance from the formula for Ohm's law. You only have to solve the formula for the resistance \(R\). Bring \(I\) to the other side and you get:`3\[ R ~=~ \frac{U}{I} \]`

The voltage has the unit Volt and the current has the unit Ampere. The resistance must therefore have the unit *Volt per Ampere*:`\[ [R] ~=~ \frac{\text{Volt}}{\text{Ampere}} ~=~ \frac{\text{V}}{\text{A}}\]`

We abbreviate *Volt per Ampere* briefly with the unit *Ohm*:`\[ [R] ~=~ \Omega \]`By the way, this character \(\Omega\) is a Greek letter 'Omega'.

The value of this electrical resistance \(R\) depends on the conductor used to connect the positive and negative poles.

- A conductor made of iron has a lower resistance \(R\) than a conductor made of copper. For the iron conductor we expect a shallower straight line than for the copper conductor.
- And a conductor made of copper has a smaller resistance \(R\) than a conductor made of aluminum. So an aluminum conductor has a steeper straight line than a copper conductor.

## 3 examples of how the formula is applied

Let's take a look at some concrete examples of how you can apply Ohm's Law formula.

**Example: Resistance unknown**The voltage between the poles is \(U ~=~ 10 \, \text{V}\) and the current flowing through the conductor is \(I ~=~ 1 \, \text{A}\). What is the resistance of the conductor?

We solve Ohm's law for the resistance \(R\):`4\[ R ~=~ \frac{U}{I} \]`

Then we insert given values:`4.1\[ R ~=~ \frac{10 \, \text{V}}{1 \, \text{A}} ~=~ 10 \, \Omega \]`

**Example: Voltage unknown**The current flowing through the conductor is \(I = 2 \, \text{A} \) and the resistance of the conductor is \(R = 100 \, \Omega \). What is the voltage between the ends of the conductor, that is, between the two poles?

There we can use URI-formula directly:`5\[ U ~=~ R\, I \]`

Insert given values:`5.1\[ U ~=~ 100 \, \Omega ~\cdot~ 2 \, \text{A} ~=~ 200 \, \text{V} \]`

**Example: Current unknown**The voltage between the ends of the conductor is \(U ~=~ 6 \, \text{V}\) and the resistance of the conductor is \(R ~=~ 2 \, \Omega \). What is the electric current through the conductor?

To determine the current \(I\), we solve '\(U ~=~ R\,I\)' for the current by bringing the resistance \(R\) to the other side. Then we get:`6\[ I ~=~ \frac{U}{R}\]`

Insert given values:`6.1\[ I ~=~ \frac{6 \, \text{V}}{2 \, \Omega} ~=~ 3 \, \text{A} \]`