Ohms law is a simple formula that makes it easy to calculate voltage, current, and resistance. You can use it to find what resistor value you need for an LED. Or to find out how much power your circuit uses. And much more.
This is one of the few formulas in electronics that you’ll use on a regular basis. In this guide, you’ll learn how it works and how to use it. And I’ll also show you an easy way to remember it.
The law was discovered by Georg Ohm (hence the name) and shows how voltage, current, and resistance are related. Check out this Ohm’s law cartoon below to see how they relate:
Look at the drawing above and see if it makes sense to you that:
- If you increase the voltage (Volt) in a circuit while the resistance is the same, you get more current (Amp).
- If you increase the resistance (Ohm) in a circuit while the voltage stays the same, you get less current.
Ohm’s law is a way of describing the relationship between the voltage, resistance, and current using math:
V = R * I
- V is the symbol for voltage.
- I is the symbol for current.
- R is the symbol for resistance.
I use it VERY often. It is THE formula in electronics.
You can switch it around and get R = V/I or I = V/R. As long as you have two of the variables, you can calculate the last.
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Ohm’s Law Triangle
You can use this triangle to remember Ohm’s law:
How to use it:
Use your hand to cover the letter you want to find. If one of the remaining letters is above the other, it means dividing the top one by the bottom one. If they are next to each other, it means multiply one with the other.
Example: Voltage
Let’s find the formula for voltage:
Place your hand over the V in the triangle, then look at the R and the I. I and R are next to each other, so you need to multiply. That means you get:
V = I * R
Example: Resistance
Let’s find the formula for resistance:
Place your hand over the R. Then you’ll see that the V is over the I. That means you have to divide V by I:
R = V / I
Example: Current
Let’s find the formula for current:
Place your hand over the I. Then you’ll see the V over the R, which means dividing V by R:
I = V / R
Quick Tip: How to Remember Without the Triangle?
A simple way of remembering things is to make a stupid association so that you remember it because it’s so stupid.
So to help you remember Ohm’s law let me introduce the VRIIIIIIII! rule.
Pretend that you’re driving a car really fast, then suddenly you hit the brakes really hard. What sound do you hear?
“VRIIIIIIIIIIII!”
And this way you can remember V=RI ;)
Example: Using Ohm’s Law to Calculate Current in a Circuit
The best way to learn how to use Ohm’s law is by looking at some examples.
Below is a very simple circuit with a battery and a resistor. The battery is a 12-volt battery, and the resistance of the resistor is 600 Ohms. How much current flows through the circuit?
To find the amount of current, you can use the triangle above to the formula for current: I = V/R.
Now you can calculate the current by using the voltage and the resistance. Just type it into your calculator to get the result:
I = 12 V / 600 Ω
I = 0.02 A = 20 mA
So the current in the circuit is 20 mA.
Example: Choosing a Resistor for an LED
To safely power a Light-Emitting Diode (LED), you should always have a resistor in series with it to limit the current that can flow. But what value should you choose?
This is one of those practical situations where Ohm’s law becomes really useful.
Below you can see a typical LED circuit with a resistor in series. The LED grabs 2V from the battery, so the rest of the voltage (9V minus 2V = 7V) drops across the resistor. Let’s imagine that this LED can only handle up to 10 mA.
Since there is only one path for the current to take, the current through the resistor is the same as the current through the LED. So if you find the resistor value needed to get 10 mA through the resistor, then that’s what you’ll get through the LED as well.
The battery voltage is 9V. The voltage across the LED is 2V. So the rest of the battery voltage has to drop across the resistor. That means the voltage across the resistor is 7V.
And you want 10 mA (0.01A) through the resistor.
Plug this into the formula for finding resistance (see above), and you’ll get the needed resistor value:
R = V / I
R = 7 V / 0.01 A
R = 700 Ω
This means you need a resistor of 700 Ω to set the current to 10 mA.
Example: Figuring Out the Battery Voltage
Let us try another example.
Below we have a circuit with a resistor and a battery again. But this time we don’t know the voltage of the battery. Instead, we imagine that we have measured the current in the circuit and found it to be 3 mA (same as 0.003 A).
The resistance of the resistor is 600 Ω. What is the voltage of the battery?
By remembering the “VRIIII!” rule, you get:
V = R * I
V = 600 Ω * 0.003A
V = 1.8 V
So the voltage of the battery must be 1.8 V.
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Why didn’t you convert the 3mA to Ampere before evaluating in the second examlpe?
Kindly help remove my comments. Gosh! I suck at math.
I’m glad you figured it out =)
Oyvind
because if we covert 3mA to Amphere we have do a long process
Hey im new to the whole electronics thing was wondering if anyone could mentor me?
If so email me, [email protected]
i am interested in electronics
Hi sir I like electronics kindly help me how to make a circuit on pcb is there any software for it?
Thanks.
Yes,
check out https://www.build-electronic-circuits.com/pcb-design/
Cheers!
Oyvind
Hi. I am not so new to electronics but I didn’t have enough recourses to play around with the various components. Keep being awesome and make more interesting articles about electronics:D
Thanks!
Hello Øyvind! I am new to all of these concepts and I really need some help. I have considered joining a robotics club. Please keep on making awesome article about electronics!:D
Join the robotics club! It sounds awesome =)
Oyvind
Hi. Hoping you can help.
I’m using three(3) 1.5V bulbs(not leds) running off of one(1) 1.5V AAA battery.
When I start with a brand new battery, the bulbs are nice and bright. However, after about 2 minutes, I can notice the bulbs all starting to get dimmer. They’re just about completely out after 30 minutes.
Would using resistors keep the bulbs burning at a more equal brightness from start to ending of the battery?
I believe they are wired in parallel. Would it matter if in series or parallel?
Thanks,
Seair
The rate of discharge you are describing sounds more like a 3W 1.5V LED bulb. AAA batteries are rated 1.5V, with about 1200mAh. So a single 3W 1.5V bulb is going to draw a 2A current. Technically that would give you a 36 minute run time, but since 2A is such a high current for these batteries, the actual run time would be significantly less. Now add 2 more of these bulbs in the equation, for a total of three, and you would see them dimming almost immediately, a lot like how you described. As far as the circuit type, normally if three bulbs we’re to be wired in series they would be 1/3rd the brightness they normally are. However in this case, they probably wouldn’t work at all, because now your at 4.5 volts between the three of them. I would double check your bulb type. If they actually are regular incadseccent, then they are probably around 20-30ma bulb, which should run for at least an hour before noticeably dimming. If your wiring job is good, that leaves the batteries as the only other possible problem. For the most part you get what you pay for, when buying these type batteries.
Stupid question, in the second example
we had a current, I, or .03 and a resistance of 600. Multiplied they give us 18. How do you know to make it 1.8 volts?
Hi Nick,
3 mA is 0.003 A (so you were lacking a zero ;)
Best,
Oyvind
Hi,
I am nerd in electronics, but I guess my maths is good.
If 1000mA = 1A
then 100mA = 0.1A
10mA = 0.01A
implicates
30mA = 0.03A
so howcome your statement, it is 30 mA is 0.003 A (so you were lacking a zero ????
Am so confused…
@Bunky;
Oooops! Sorry, that was a mistake by me. I’ve edited my comment now.
3 mA is 0.003A.
30 mA = 0.03A
600 x 3 = 1800
0r, 3 mA doit être convertit en Ampère donc, on aura 0,003 A
d’où le calcul est: 600 x 0,003 = 1.8v
ou 600 x3 = 1800
1800 divisé par 100 = 1.8v
Great little website ….really happy i found it
V = RI where V = 1 volt, R = 1 Ohms, I = 1 Ampere
========================================
Given example, R = 600 Ohms, I = 3 mili Ampere
1000mA = 1A, then 3mA = 0.003A
On putting the values
V = 600 (R) x 0.003 (I)
V = 1.8
:)
Exactly =)
its good and educated
because if we convert it it will be a long process
I am really new for electronics but i have a little ideas over this. my questions is under programming how can understand it? Am a first year pursuing computer science. my email is [email protected]
regards
Kenneth
Nice site.
In your example, 12v batt and 600ohm resister, you determine current is .02 (20ma)
What if batt was a 3.7v battery (with say 3000mah)
Hey,
The 3000mah just tells you how much energy your battery can store. It doesn’t affect this calculation. So just switch 12V with 3.7V:
3.7V / 600 ohm = 0.006A = 6mA
Best,
Oyvind
Can u please just write me a summary or something on this because i still don’t fully understand all of this. Actually can anyone describe ohms law better for me, thanks
Hey, let me know what you don’t understand and I’ll try to explain.
Best,
Oyvind
Ohms law states that the current (charges)moving through a conductor(wire) is directly proportional to the applied potential difference(volts) and inversely proportional to the resistance in the wire. This means as the voltage increases, the current increases hence the resistance is kept constant.
I once had it explained to me in terms of water and pipes. Voltage is water pressure, current is how much water is actually flowing (it is possible to have very high pressure, but little actual water flow – think of a pressure washer for a car. You can also have a LOT of water flowing at low pressure – think of a large storm drain pipe during a heavy rain), and resistance is a restriction on the water flow – think of a valve or a narrowing of a water pipe.
Give me an example of electrical component which obeys ohm’s law
It’s not about components. It’s about how current, voltage and resistance relate to each other.
Hii! Ohms law explanation is very nice…
But I have a doubt that in Ohms law there is a direct relation between V & I (V is directly proportional to I).
P=VI
In this relation, its look like V is inverse to I
So, can you explain??
you not convert mA into A
and resistor denoted by v
and in drawing v denoted by u
why?
I think second example is wrong
Very Good Information & You Made it easy to learn….
Thanks.
It very useful than my physics class.
Now I can study electric after a long time of my school days.
Thanks , you are my teacher. ..
Thank you, that explanation was simple yet very complete…before this I had no clue about ohms law, now I understand
What if resistance is 0? (If you dont have a resistor, for example). Does the voltage will be 0 too? Why?
I guess you are referring to the V=R*I form?
This gives you the voltage drop across the resistance you put into the formula. If the resistance is zero (like a wire), there will be no voltage drop across the wire, so yes, V becomes 0 in this case.
Best,
Oyvind
AM NEW AT LEARNING ELECTRONIC. Am from scratch
is it possible to to find out about resistance,
if u have V (volts) and I (current)
how to find amount of RESISTANCE
It’s voltage divided by the current:
R=V/I
Best,
Oyvind
I did not know anything before reading the material. I can see how voltage works and how you can use math is a asset.
Its been cool and by time ill be perfect.
Mr. Øyvind, you are truly creative and amazing in teaching electronics. My journey towards electronics has become so easy and interesting because of you! Thank you!
Give me an example of Kirchoff law
I’m still a little confused about voltage. Does it influence the speed at which electrons flow?
More voltage means more current if the resistance stays the same. More current means more electrons flowing.
In Italy we use to remember this law with a sentence: “Vittorio Re d’Italia” (V=RI) which translated in English means: “Vittorio king of Italy”
Vittorio was in fact king of Italy in the past ahah.
I know this comment wasn’t necessary, but it’s fun
Thanks for sharing!
thanks!
its very easy to understand.
Hi there, thanks for providing some examples for people to practice Ohm’s law. The first example is incorrect. milliamps is 10^-3 and centiamps is 10^-2. So technically 0.02 is 2cA right?
Hi Ben,
You’re right that 0.02A is 2cA – although “centi” is usually not used.
But 2 cA is the same as 20 mA.
Hello, i am making presentation about electronic components and i want to use first image that explains it. Some research bring me to this website, but i am not sure if this image is actually yours. If it is, can i use it? If not, do you remember source of it?
Hi, all images belong to me unless it’s noted otherwise, so yes. Send a message through https://www.build-electronic-circuits.com/contact/ with how you want to use it (is it for web? is the presentation going to be available for download?) and we’ll figure something out.
Hello,
I´m using your image to teach Ohms law in school. Actually I changed it a bit. I copied the yellow person and pasted it on the right side. I also changed the size to only 75%. So now it looks like two persons are pushing an the current is going the way were there is less “force”.
Thanks a lot for the drawing.
I liked the way you explained Ohm’s law with the help of a triangle. Your way of explaining is very good.
I suck at math that is why is i did bad