Ohms law was found by Georg Ohm and it goes a little something like this:

**V = RI**

V is voltage, R is resistance and I is current.

The law states that a voltage potential equals the current multiplied by the resistance.

It is used 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.

### How to remember Ohms law

A simple way of remembering things is to make a stupid association with it 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 your driving a your car really fast, then suddenly you hit the breaks really hard. What sound do you hear?

“VRIIIIIIIIIIII!”

And this way you can remember V=RI 😉

### A practical example

The best way to teach how to use it is by example.

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 Ohm. How much current flows through the circuit?

To find the amount of current, we rearrange Mr. VRI and get I = V/R. Now we can calculate the current by using the voltage and the resistance.

I = 12 V/600 OhmI = 0.02 A = 20 mA (milli Ampere)

So the current in the circuit is 20 mA.

If you don’t like calculating things yourself, check out this calculator for Ohm’s law.

### Another example

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 (milli Ampere).

The resistance of the resistor is 600 Ohm. What is the voltage of the battery? By using our friend VRI we get

V = RI

V = 600 Ohm * 3 mA

V = 1.8 V

So the voltage of the battery must be 1.8 V.

Dipo says

Why didn’t you convert the 3mA to Ampere before evaluating in the second examlpe?

Dipo says

Kindly help remove my comments. Gosh! I suck at math.

admin says

I’m glad you figured it out =)

Oyvind

vamsi says

because if we covert 3mA to Amphere we have do a long process

jake says

Hey im new to the whole electronics thing was wondering if anyone could mentor me?

jake says

If so email me, [email protected]

Abdul Latif says

Hi sir I like electronics kindly help me how to make a circuit on pcb is there any software for it?

Thanks.

admin says

Yes,

check out http://www.build-electronic-circuits.com/pcb-design/

Cheers!

Oyvind

SnowBerry says

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

admin says

Thanks!

SnowBerry says

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

admin says

Join the robotics club! It sounds awesome =)

Oyvind

Seair says

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

Mike says

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.

Nick says

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?

admin says

Hi Nick,

3 mA is 0.003 A (so you were lacking a zero 😉

Best,

Oyvind

Bunky says

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…

admin says

@Bunky;

Oooops! Sorry, that was a mistake by me. I’ve edited my comment now.

3 mA is 0.003A.

30 mA = 0.03A

louis says

Great little website ….really happy i found it

Bunky says

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

🙂

admin says

Exactly =)

ismaili seleman jafo says

its good and educated

vamsi says

because if we convert it it will be a long process

kiprono says

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

Chris says

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)

admin says

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

Hudson Mesritz says

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

admin says

Hey, let me know what you don’t understand and I’ll try to explain.

Best,

Oyvind

Amit says

Give me an example of electrical component which obeys ohm’s law

gayathri says

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??

irshad says

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