**When talking about resistance, we cannot avoid ohm’s law, which shows the connection between current, voltage and resistance.**

George Simon Ohm was a German physicist who lived from 1788 to 1854. He was the first one who was able to coherently describe the relation between current, voltage and resistance. To honor him the formula he discovered is known as ohm's law till today. What Ohm did was basically a row of measurements to show the mathematical correlation between the applied voltage and the corresponding current that flows through a piece of wire. Ohm's law is not a mathematical construction, it is based on George Simon Ohm's empirical findings.

Today we can easily recreate his experiments. This is because we have all the required tools at hand. At the times Ohm lived, there were no good tools to measure current and voltage. His findings were a real break through, although the relevance of his findings was not quickly accepted.

Let's have a look at what Ohm found out. For this we need a simple circuit, in which we measure current and voltage. The only other components we need are an adjustable voltage source and a resistor. The circuit is shown below. I used a 51 Ω resistor with a power rating of 1 W for my measurements. If you only have standard resistors you might want to use a 100 Ω resistor instead, to not exceed its maximum power rating.

I now increase the voltage step by step. I start at 0 V and then increase the voltage to 5 V in 1 V increments. For each voltage I measured the current. The results are shown in the table below.

Voltage | Current |
---|---|

0 V | 0 mA |

1 V | 19 mA |

2 V | 38 mA |

3 V | 59 mA |

4 V | 78 mA |

5 V | 98 mA |

If we plot this data in a graph, we get an almost perfectly straight line starting at (0 | 0). This is essentially what Ohm discovered. The current rises proportionally to the applied voltage.

Instead of talking about a straight line, we can put it in other words and say that we have got a function with a fixed gradient. We can get this gradient using a gradient triangle.
Let's try for 5 V and the current of 98 mA:

\({5 V \over 0.098 A} = 51.02 {V \over A}\)

We can also pick another value like 1 V and the corresponding current of 19 mA:

\({1 V \over 0.019 A} = 52.63 {V \over A}\)

As you can see the quotient of voltage and current (the gradient of our function) is roughly constant, albeit there is a minor variation between both values due to the inaccuracy our measurements.
This is exactly what ohm's law states: The quotient of voltage and current is a constant value.

\({ U \over I } = const\)

Your might have already noticed something odd: This quotient is roughly \(51 {V \over A}\) which matches the value of the resistor we used. This is no coincidence, but the definition of resistance. This quotient is what mathematically is resistance. All the other values for the conductivity and resistivity of materials are calculated based on this definition.

\(R = { U \over I }\)

The unit ohm is equivalent to volts per ampere: \(1 Ω = 1 {V \over A}\)

The voltage \(U\) in this formula is the voltage drop over the resistor. As we only have one resistor in our circuit, this voltage drop is equal to the configured voltage. The whole voltage drops over this one component. If we have multiple components in our circuit, however, we have to be a bit more careful and only measure the voltage drop over the component whose resistance we want to calculate.

Now that we have a formula that describes the correlation between resistance, voltage and current we can make use of this when designing our circuits.
Here is an example. If we use a 9 V battery and want to achieve a current of 20 mA, we can now calculate the necessary resistor:

\(U = 9 V\)

\(I = 20 mA = 0.02 A\)

\(R = {U \over I} = {{9 V} \over {0.2 A}} = 450 Ω\)

As this is not a standard resistor value, we probably need use a 470 Ω resistor instead. Feel free to build the circuit and check if the result is correct.

We can also transform the formula to for instance calculate the current we get with a 9 V battery and a resistor of 1 Ω.

\(U = 9 V\)

\(R= 1 Ω\)

\(I = {U \over R} = {{9 V} \over {1 Ω}} = 9 A\)

That much? Yes indeed, the calculation is correct and this is why we have to be careful with small resistors. For small resistance values we quickly get high currents. In the extreme case of a short circuit we only have the tiny resistance of our wire, in consequence we get a very high current - at least if our power source is able to deliver such high currents.

As we get very high currents with small resistors we have to be careful with them. In the last tutorial we learned that resistors convert some electrical energy into heat. If too much heat is produced and the power rating of the resistor is exceeded it might catch fire.
Let's take a look at our example and calculate the amount of energy per second that is converted into heat:

\(P_{loss} = I^2 * R = (9 A)^2 * 1 Ω = 81 W\)

If we use a resistor with a power rating of 0.25 W it will certainly die. We can derive from both formulas, ohm's law and this one, that when we half our resistor value at a fixed voltage, we also double the amount of energy that is converted to heat in each second. I say it once more: Be careful with high currents and small resistor values.

In reality not all electric components follow ohm's law. Don't get me wrong, the resistance is still calculated according to the formula you learned today. The point is, that not all components have fixed resistance. For these components the quotient \(U \over I\) is not a constant. Prominent examples for this are all semiconductor components, including LEDs. If you want to be really exact ohm's law does not even fully apply to resistors. As we know they can heat up when a higher current is flowing trough them. The higher temperature will affect their resistance value. In most use cases, this effect is irrelevant, however.