# Circuit Basics - Ohm's and Joule's Laws

Now that we have introduced some electrical properties like voltage, current, and resistance we need to understand how they relate to each other. Amazingly, most electrical properties relate to each other in ways that are beautifully simple.

## Ohm’s Law

All of the electronics begins with one simple equation called *Ohm’s Law*. Georg Simon Ohm was the physicist and mathematician who first proved that the amount of electrical current (amps) that flows through a circuit is proportionally related to the voltage difference in the circuit. He called the property that governs this relationship *resistance*. Meaning, that the ratio of the voltage applied to a circuit and the current that flowed through the circuit was equal to the resistance of the circuit. This relationship between voltage, current, and resistance came to be known as Ohm’s law and the unit of resistance was named Ohm in his honor.

$latex Voltage = Current \cdot Resistance \quad or \quad V = I \cdot R&s=2$

This law is the foundation of all other relationships in electronics. With this law, you can calculate voltage, current, or resistance, if you know the other two properties.

$latex V = I \cdot R, \quad I = \frac{V}{R}, \quad R = \frac{V}{I}&s=2$

In circuit design, we use components known as resistors to add very controlled amounts of resistance to circuits.

Calculating Voltage from the Current Flowing Through a Resistor

This time, we have a 470Ω resistor and we know that there is 200mA (0.200A) of current flowing through this resistor. Knowing this, we can calculate the voltage across the resistor using Ohm’s law.

$latex V = I \cdot R = 0.200A \cdot 470\Omega = 94V&s=2$

Using Ohm’s law, we found that the voltage across the resistor is 94V.

Calculating A Resistor Value to Achieve a Desired Current

Let’s say that you have a 12V battery and you want to make a circuit with the battery and a resistor that has 100mA (0.100A) or current flowing through it. In this case, we can use Ohm’s law to solve for the needed resistor value.

$latex R = \frac{V}{I} = \frac{12V}{0.100A} = 120\Omega&s=2$

Using Ohm’s law (solved for R) we calculate that a 120 Ohm resistor wired in series with the 12V battery will result in a circuit with 10mA of current flowing.

## Joule’s Law

Our second foundational equation was discovered by a physicist by the name of James Prescott Joule. Joule discovered the relationship between energy (Q) and the flow of electricity. He said that amount of electrical energy expended in a circuit is equal to the square of the current, times the resistance of the circuit, times the amount of time it was on.

$latex Q = I^2 \cdot R \cdot t&s=2$

However, when we work with electricity, we’re rarely interested in the total amount of power used in a period of time (unless you work for the power company). We are much more interested in the amount of power we are using *per* time. Energy per time is the definition of *power*. We can rearrange Joule’s law so that the power consumed is equal to the square of current times resistance.

$latex P = \frac{Q}{t} = I^2 \cdot R&s=2$

Finally, we can combine Joule’s law with Ohm’s law and get a couple of other useful equations for power.

$latex I = \frac{V}{R}&s=2$

$latex P = I^2 \cdot R = I \cdot I \cdot R = (\frac{V}{R}) \cdot I \cdot R = V \cdot I&s=2$

$latex P = I^2 \cdot R = (\frac{V}{R})^2 \cdot R = \frac{V^2}{R}&s=2$

Now, we have derived three different equations for electrical power:

$latex P = V \cdot I, \quad P = I^2 \cdot R, \quad P = \frac{V^2}{R}&s=2$

Determining the Required Power Rating for Resistor

Let’s suppose that we have a circuit with a 220Ω resistor and there are 12V across that resistor. We need to know what power rating of capacitor we need to specify. We can easily calculate the power that will be dissapated as heat in the resistor using Joule’s law.

$latex P = \frac{V^2}{R} = \frac{(12V)^2}{220\Omega} = 0.65W&s=2$

In this circuit the resistor will be generating

0.65W of heat.This means that 0.65W is ourminimumpower rating for this resistor. The power ratings of common resistors are 1/16W, 1/10W, 1/8W, 1/4W, 1/2W, 1W, 2W, and so on. This means that we would need to choose a resistor rated for 1W of power dissipation to be sure that it could handle the 0.65W of heat that will be generated in it.

## Try It Yourself!

Let’s go ahead and try both of these examples in the simulator!