Ohm’s law an Electronics Master Key
Part 1 | Part 2 | Part 3 | Part 4
I love swimming, albeit in the last few months I haven’t swam as much as I would love to, you see at the time of this writing its approaching winter in the southern hemisphere.
A few years ago I was attending swimming lessons and the first few weeks were very tough on me, whenever I would practice my free stroke I would have to pause every now and then to catch my breath.
My instructor noticed this and she started encouraging me to slow down and asked me to trust the water.
Pfft! Yeah right, just trust the water … my thoughts exactly! However after a few days of struggling I decided to give it a try, and every time I got in the water I would tell myself ‘trust the water and slow down’
Trust the water and slow down
The more I did this, the easier it became to cover longer distances faster!
In electronics we find this kind of mantra in Ohm’s law. The relationship between voltage, current and resistance in a circuit is one of the most important concepts in electronics.
Understanding it will not only enable you to design your own circuits successfully but also to analyse circuits that others have designed with confidence.
Before diving into Ohm’s law lets explore how current is driven through a resistance. When you power a component that has a measurable resistance for example a LED. The movement of electrons through the component is what makes up an electrical current.
When you apply a greater voltage you are pushing said electrons harder and you build a stronger flow which translates to a larger current through the component hence it goes the stronger the force the stronger the current.
Ever~changing Relationships
In the early 1800s a scientist by the name Georg Ohm discovered a relationship between Voltage V and Current I in a component with a Resistance R.
Ohm figured that for components with a fixed resistance voltage and current vary in the same way i.e double the voltage and the current is doubled, halve the voltage and the current is halved.
Ohm’s law states that voltage equals current multiplied by resistance
V = I * R
Consider the example below where a 12volt battery is applied across a 2200Ω resistor this would produce a current of 0.005 amps and can be calculated as
V = 0.005 * 2200
V = 12V
Notice!!
It’s important to note that when using Ohm’s law, we need to be very careful with the units in question. Convert any kilos and millis before calculating.
It helps to think of Ohm’s law as
volts = amps * Ohms
That way a circuit with a 100Ω resistor and 50 milliamps of current has a voltage of
V = I * R
V = 0.05 * 100
V = 5V
as opposed to 50 * 100 which is a 5000, which would simply be catastrophic if applied to the circuit
One law three equations
Using basic algebra, we can rearrange ohm’s law into 3 equations that would enable us derive all 3 elements.
Voltage
To calculate an unknown voltage multiply the current by the resistance
V = I * R
Current
To calculate an unknown current take the voltage and divide it by the resistance
I = V/R
Resistance
To calculate an unknown resistance take the voltage and divide it by the current
R = V/I
And that is Ohm’s law! Practice this with as much as you can as you delve into electronics and you are well on you way to designing and analysing circuits confidently.
Using Ohms Law
Now that you have a firm grasp on Ohm’s law, you must be wondering where and how it is used in everyday electronics work.
You can use Ohms law in many ways however I would like to introduce you to two of the most common ways of using it
Analysing Circuits
Imagine that you have a circuit with three resistors, two in parallel formation and one in series formation with aforementioned two as in the diagram below
To calculate the equivalent resistance of the circuit above we would do the following
R = 220 + (1000 * 2200)/(1000+2200)
R ≈ 908 Ω
To calculate the total current supplied by the battery
I = V/R
I = 6v/908Ω
I = 0.0066amps
To calculate the voltage drop across the first resistor
Vr = 0.0066amps * 220Ω
Vr = 1.45V
To calculate the voltage drop across the parallel resistors
Vp = I * R
Vp = 0.0066 * (1000 * 2200)/(1000 +2200)
Vp = 4.5V
Try and calculate the voltage drop across each parallel, resistor It’s simpler than you think …. leave the answer in the comments
Designing Circuits
You can use Ohms law to determine what components to use in a circuit design.
Lets consider the diagram below you would really like to power and LED with a 9 Volt source, You know the LED needs 2 Volts and can handle a maximum current of 0.025 amps. What amount of resistance would you need to put in series with the LED so that the current never exceeds 0.025amps
To do this you would first need to determine the voltage drop you need from the resistor, since you know the supply voltage is 9V and the LED needs 2V we can see that the resistor needs to use up the difference voltage.
Vr = 9V - 2V
Vr = 7V
Now that we know what amount of voltage drop we need from the resistor, if we want to limit the current that gets to the LED to 0.025amps we can calculate the Resistance needed as
R = V/I
R = 7V/0.025amps
R = 280Ω
From this you can see that Ohms law comes in very handy even when we would want to tweak an existing circuit, We can use it to make small changes to our circuits with big impacts on the results we get from them!
Did you calculate the voltage drop across the parallel resistors?
Let us know in the comments or better yet join our Discord and tell us the answer
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