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How to Use Ohms Law

By 35 Comments

I usually don’t use a lot of math when doing electronics, but Ohms law is extremely useful!

The law was found by Georg Ohm and is based on how voltage, current and resistance are related:

Ohms law cartoon

Look at the drawing above and see if it makes sense to you that:

  • If you increase the voltage in a circuit while the resistance is the same, you get more current.
  • If you increase the resistance 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 = RI

  • 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.

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 the remaining letters are over each other, it means divide the top one with 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 divide V by R:

I = V / R

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 brakes 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?

Demo of ohms law on circuit with battery and resistor

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:

I = 12 V/600 Ohm
I = 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).

Demo of ohms law on circuit with battery and resistor

The resistance of the resistor is 600 Ohm. What is the voltage of the battery?

By remembering the “VRIIII!” rule, you get:

V = RI
V = 600 Ohm * 3 mA
V = 1.8 V

So the voltage of the battery must be 1.8 V.

Return from Ohms Law to Electronic Schematics

Reader Interactions

Comments

  5. 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

    Reply

  6. 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

    Reply

  7. 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

    Reply

    • 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.

      Reply

  8. 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?

    Reply

      • 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…

        Reply

        • @Bunky;

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

          3 mA is 0.003A.
          30 mA = 0.03A

          Reply

  10. 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

    :)

    Reply

  13. 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

    Reply

  14. 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)

    Reply

    • 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

      Reply

  15. 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

    Reply

  17. 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??

    Reply

  18. 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

    Reply

  20. 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. ..

    Reply

  21. Thank you, that explanation was simple yet very complete…before this I had no clue about ohms law, now I understand

    Reply

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