# Ohms Law

**Understanding and Applying Ohms Law**

In our post Voltage, Current and Resistance we covered what these are and how they are related. But how do we calculate these and use them mathematically to help us analyse and create circuits?

Ohms Law is probably one of the most useful things you’ll learn as an engineer, if not thee most. It is used by engineers on a daily basis. It shows us the relationship between Voltage, Current and Resistance.

**OHMS LAW!!**

**OHMS LAW!!**

The equation: V = I * R

Voltage (Volts) = Current (Amps) * Resistance (Ohms)

It is more commonly written as V = IR, as the algebraic mathematics states that two variables beside each other are to be multiplied together.

Thus V = IR is the same as V = I * R. This can be manipulated to find I or R.

As we progress we gain an intuitive understanding of voltage Current and Resistance and can rewrite the equation as:

I = V / R

R = V / I

Take a look at the Ohms Law triangle, it can be quite useful for people starting out to understand ohms law.

*Okay let’s show an example:*

*Okay let’s show an example:*Take this simple LED circuit for example, this is a good example as you have a Voltage source, a current limiting resistor and a component that will have a voltage drop across it.

Let’s use what we’ve learnt to calculate the total current (i), the voltage drop over the resistor (V1), and the resistance of the LED, assuming that we have a 5V power supply (VCC).

*Finding the Voltages*

*Finding the Voltages*Due to the characteristics of an LED it will typically have a voltage drop of 1.8V to 3V, in this case we’ll assume it is 1.8V.

Thus V2 = 1.8V. To find the voltage across the resistor we can simply subtract 1.8V from 5V, which will give us a 3.2V drop, which will be over the resistor. We know we this is the case as the 3.2V cannot simply disappear, and we only have one other component that it could be dropped across.

*Finding th*e Current

*Finding th*e CurrentTo find the total current we start with: 5 = I * R as we know that our power source VCC is 5V. Now to calculate the total circuit current, we need to rearrange the formula to find I.

I = 5 / R

*Note: As it is a series circuit the current will be the same the whole way around the circuit, so the current through the resistor will be the same as the current through the LED.*

As we’re looking to find the current, we can use the resistance value of the resistor as it’s given to us. In this case, we have a resistor of 330 Ohms. Now we can substitute in 330 to our equation and find the current (i).

I = 3.2 / 330

Thus I = 0.009696 Amps, however in engineering we would write this as 9.69mA (9.69 milli Amps).

*Finding the Total Resistance*

*Finding the Total Resistance*Now we can find the resistance of the LED, so R = V / I

R = 1.8V / 9.96mA

R = 180 Ohms

Add the known resistance of the resistor to the calculated resistance of the LED to get our total circuit resistance.

Rt = (330+180)

Rt = 510 Ohms

*Note: Rt here denotes the total resistance*

The handy thing about ohms law is you can use it to check your own calculations (assuming your logic and methods are correct). So, we know the current is 9.69mA and the resistor is 330 Ohms and the resistance of the LED is 180 Ohms. If we’re looking to check if this is correct we can try to calculate the voltage which we know should be 5V, we’ll rearrange the equation to V = I R and sub in our calculated values and see if we get the 5V we’re expecting.

V = 9.69mA * 510

V = 4.94V

This is still correct as we did some rounding with our calculations; nearly 5V is still considered to be correct.

*Congratulations you’ve just finished your first ohms law calculations!*

*Congratulations you’ve just finished your first ohms law calculations!***Written by Joel Gray**

16/03/2020