Ohm's Law Calculator
Solve voltage, current, resistance and power from any two values
Ohm's law: V = I × R. Voltage equals current times resistance, so 12 V across 6 Ω drives 2 A and dissipates 24 W. Enter any two values to find the other two.
Ohm's law calculator
LiveEnter any two values and the other two are calculated.
How Ohm's law works
Ohm's law says that the current through a conductor is proportional to the voltage across it, and the constant that links them is the resistance: V = I × R. Voltage (V, in volts) is the electrical pressure, current (I, in amperes) is the flow, and resistance (R, in ohms, Ω) is how strongly the material opposes that flow. Combine it with the power formula P = V × I and every quantity can be found from any other two, which is exactly what the calculator does. The same two formulas sit behind every tool in the electrical calculators collection.
How to use the calculator
- Type any two known values: voltage, current, resistance or power. The two fields you edited most recently are treated as the inputs.
- The other two fields fill in automatically, rounded to six significant figures.
- Use base units: volts, amps, ohms and watts. Convert first if you have milliamps (÷ 1,000) or kilohms (× 1,000).
- Press Clear to start a new calculation.
Worked example: the default inputs are 12 V and 2 A. Resistance is 12 ÷ 2 = 6 Ω and power is 12 × 2 = 24 W. If you instead knew 24 W and 6 Ω, current would be √(24 ÷ 6) = 2 A and voltage √(24 × 6) = 12 V, the same circuit seen from a different pair of values. For household-load calculations where you already know watts and volts, the watts to amps calculator and amps to watts calculator add AC power factor and three-phase options.
The Ohm's law triangle
- V = I × R
- I = V ÷ R
- R = V ÷ I
- P = V × I
What happens to current when resistance doubles?
If the voltage stays the same, doubling the resistance halves the current: 12 V across 6 Ω drives 2 A, and across 12 Ω it drives 1 A.
Power falls by the same factor, because P = V² ÷ R at a fixed voltage: the 6 Ω load dissipates 24 W and the 12 Ω load 12 W. The pattern holds for any ratio, so tripling the resistance cuts the current to a third.
The result flips when the current is held constant instead, as with an LED driver or a bench supply in constant-current mode. Then doubling the resistance doubles the voltage across it and doubles the power, since P = I² × R: 2 A through 6 Ω needs 12 V and dissipates 24 W, while 2 A through 12 Ω needs 24 V and dissipates 48 W.
So before predicting what a change in resistance will do, ask which quantity the source holds fixed: household outlets and batteries hold voltage, while LED drivers hold current.
How do I choose a resistor's power rating?
Work out the power the resistor will dissipate with P = V² ÷ R or P = I² × R, then choose a standard rating of at least twice that, so the part runs cool.
Example: 9 V across 470 Ω dissipates 9² ÷ 470 = 0.1723 W. Twice that is 0.3447 W, so a 1/2 W resistor is the comfortable choice, even though a 1/4 W part is technically above 0.1723 W.
| Circuit | Power dissipated | With 2× margin | Rating to choose |
|---|---|---|---|
| 5 V across 1,000 Ω | 0.025 W | 0.05 W | 1/8 W |
| 9 V across 470 Ω | 0.1723 W | 0.3447 W | 1/2 W |
| 24 V across 1,000 Ω | 0.576 W | 1.152 W | 2 W |
| 12 V across 100 Ω | 1.44 W | 2.88 W | 3 W |
The margin matters because a resistor at its full rating runs very hot, datasheets reduce (derate) the allowed power at higher ambient temperatures, and heat drifts the resistance value and stresses nearby parts. Every watt a resistor dissipates becomes heat: 0.1723 W for an hour is 620.4 J, which the joules to calories converter shows as 148.3 cal.
Can I use milliamps and kilohms in Ohm's law?
Yes: milliamps × kilohms gives volts directly, because the 10⁻³ and 10³ cancel, so 2 mA through 4.7 kΩ drops 2 × 4.7 = 9.4 V.
The same shortcut works for the other rearrangements. Volts ÷ kilohms gives milliamps, so 5 V across 1 kΩ drives 5 mA. Volts ÷ milliamps gives kilohms, so dropping 3 V at 20 mA needs 0.15 kΩ, which is the 150 Ω LED resistor from the section below. Microamps × megohms also gives volts.
Power needs care: volts × milliamps gives milliwatts, so 5 V × 20 mA = 100 mW, or 0.1 W. Mixing prefixes is where mistakes creep in. Multiply milliamps by plain ohms and you get millivolts: 20 mA × 150 Ω = 3,000 mV, which is 3 V. When in doubt, convert everything to base units first, as the calculator above expects volts, amps, ohms and watts.
Volts vs amps: what is the difference?
Voltage and current are the two quantities Ohm's law ties together through resistance, and they are easy to mix up. Here is how they differ.
| Attribute | Volts (voltage) | Amps (current) |
|---|---|---|
| Symbol and unit | V (or U), measured in volts (V) | I, measured in amperes (A) |
| What it is | Potential difference: energy per unit charge, 1 V = 1 J/C | Rate of charge flow, 1 A = 1 C/s |
| Water-pipe analogy | The pressure difference that pushes | The flow rate through the pipe, with resistance as the pipe's narrowness |
| Role in Ohm's law | V = I × R: the push needed to drive a current through a resistance | I = V ÷ R: the flow a voltage drives through a resistance |
| How to measure it | Multimeter across the component, in parallel | Multimeter in series, or a clamp meter around one conductor |
| Typical values | 1.5 V AA cell, 12 V car battery, 120 V or 230 V mains | 20 mA indicator LED, 5 A headlamp, 12.5 A space heater |
5 key differences
- Voltage is applied across a component, while current flows through it.
- For a fixed resistor, current is proportional to voltage: 6 Ω passes 1 A at 6 V and 2 A at 12 V.
- A voltmeter has a very high internal resistance and connects in parallel; an ammeter has a very low one and connects in series, so swapping them can damage the meter or the circuit.
- One volt driving one ampere delivers one watt, which is why power is the product of the two.
- Since the 2019 SI revision the ampere is defined by fixing the elementary charge, and the volt follows as one watt per ampere.
All 12 Ohm's law formulas
Each quantity can be found three ways, depending on which two values you know. The examples all describe the same circuit, 12 V driving 2 A through 6 Ω and dissipating 24 W, so every route gives the same answer.
| To find | From V, I or R | Second route | Third route | Check on a 12 V, 2 A, 6 Ω circuit |
|---|---|---|---|---|
| Voltage (V) | V = I × R | V = P ÷ I | V = √(P × R) | 2 × 6 = 12 V · 24 ÷ 2 = 12 V · √(24 × 6) = 12 V |
| Current (I) | I = V ÷ R | I = P ÷ V | I = √(P ÷ R) | 12 ÷ 6 = 2 A · 24 ÷ 12 = 2 A · √(24 ÷ 6) = 2 A |
| Resistance (R) | R = V ÷ I | R = V² ÷ P | R = P ÷ I² | 12 ÷ 2 = 6 Ω · 12² ÷ 24 = 6 Ω · 24 ÷ 2² = 6 Ω |
| Power (P) | P = V × I | P = I² × R | P = V² ÷ R | 12 × 2 = 24 W · 2² × 6 = 24 W · 12² ÷ 6 = 24 W |
How do you calculate an LED resistor?
Subtract the LED's forward voltage from the supply voltage and divide by the LED current you want: the resistor has to drop that difference at that current. Choosing this current-limiting resistor is one of the most common everyday uses of Ohm's law.
- Take the supply, for example a 5 V USB or microcontroller rail.
- Subtract the LED's forward voltage from its datasheet, for example about 2 V for many red LEDs.
- Divide by the target current in amps: 20 mA is 0.02 A. Here (5 − 2) ÷ 0.02 = 150 Ω.
- Check the resistor's power: 3 V × 0.02 A = 0.06 W, so a common 1/4 W (0.25 W) resistor has plenty of margin.
On a 9 V battery the same LED needs (9 − 2) ÷ 0.02 = 350 Ω. If the exact value is not a standard size, round up to the next available resistor so the current stays at or below the target. Always use the forward voltage and maximum current from your LED's datasheet.
Ohm's law in real circuits
Current and power for a few typical voltage and resistance combinations, calculated with I = V ÷ R and P = V × I. Heater and kettle elements are sized so that their resistance produces the rated power at the supply voltage. Filament resistance is given hot; cold filaments measure much lower.
| Example | Voltage | Resistance | Current | Power |
|---|---|---|---|---|
| Heater element | 120 V | 9.6 Ω | 12.5 A | 1500 W |
| Kettle element | 230 V | 26.45 Ω | 8.69565 A | 2000 W |
| Car headlamp filament (hot) | 12 V | 2.4 Ω | 5 A | 60 W |
| Resistor on a 5 V rail | 5 V | 1,000 Ω | 0.005 A | 0.025 W |
| Resistor on a 9 V battery | 9 V | 470 Ω | 0.0191489 A | 0.17234 W |
How do you combine resistors in series and parallel?
Add series resistances together; for parallel resistances, add the reciprocals and invert the sum. Either way you end up with one equivalent resistance to use in Ohm's law.
- Series (end to end, one path): add them. R = R₁ + R₂ + … A 100 Ω and a 220 Ω resistor in series make 320 Ω, so 12 V drives 12 ÷ 320 = 0.0375 A, or 37.5 mA.
- Parallel (side by side, several paths): add the reciprocals. 1/R = 1/R₁ + 1/R₂ + … For two resistors this simplifies to R = (R₁ × R₂) ÷ (R₁ + R₂). The same 100 Ω and 220 Ω in parallel make 68.75 Ω, and 12 V drives 0.1745 A through the pair.
Two useful checks: series resistance is always larger than the largest resistor, and parallel resistance is always smaller than the smallest one. In a series string the current is the same everywhere and the voltages add up; in a parallel group the voltage is the same across each branch and the currents add up.
When does Ohm's law not apply?
Ohm's law breaks down whenever resistance changes with voltage, current or temperature, as it does in diodes, LEDs and lamp filaments, and in AC circuits with coils or capacitors it has to be extended from resistance to impedance.
The German physicist Georg Simon Ohm published the relationship in 1827, and the SI unit of resistance is named after him: one ohm is the resistance that passes one ampere when one volt is applied. The law holds well for metal wires, resistors and heating elements at a steady temperature, which is why it underpins almost every basic circuit calculation.
- Temperature changes resistance. A tungsten filament has a far lower resistance when cold than when glowing, so it draws a surge at switch-on.
- Diodes and LEDs are non-linear. Their current rises steeply once the forward voltage is reached, which is why they need a series resistor rather than a direct connection.
- AC circuits with coils and capacitors use impedance rather than simple resistance, and power also depends on the power factor.
When a result comes out in watts and you need kilowatts for an appliance rating or an energy estimate, use the watts to kW converter; the full set of electrical calculators covers the other everyday conversions. For mains wiring, the numbers are only part of the picture. Safety note: household voltages can kill. Follow your local electrical code and have a licensed electrician carry out any work on mains wiring.
Sources and further reading
The standards and references behind the numbers on this page. Links open in a new tab.
Standards and official sources
Further reading
- Electric powerWikipedia
Ohm's law FAQs
Short, exact answers to what people ask most.
What is Ohm's law?
Ohm's law states that voltage equals current times resistance, V = I × R. It means the current through a resistor rises in proportion to the voltage across it.
How do you calculate resistance from voltage and current?
Divide the voltage by the current: R = V ÷ I. For example, 12 V driving 2 A means a resistance of 6 Ω.
How do you find current with Ohm's law?
Divide voltage by resistance: I = V ÷ R. A 9 V battery across a 470 Ω resistor drives about 0.0191 A, or 19.1 mA.
How do you calculate power from current and resistance?
Use P = I² × R. A current of 2 A through 6 Ω dissipates 2² × 6 = 24 W.
What resistor do I need for an LED on 5 volts?
For a red LED with a forward voltage of about 2 V at 20 mA, use R = (5 − 2) ÷ 0.02 = 150 Ω. Check your LED's datasheet for its exact forward voltage and current.
What is the Ohm's law triangle?
It is a memory aid with V at the top and I and R at the bottom. Cover the quantity you want: V = I × R, I = V ÷ R and R = V ÷ I.
How many amps does 12 volts push through 4 ohms?
I = 12 ÷ 4 = 3 A, and the resistor dissipates 12 × 3 = 36 W.
What is the resistance of a 60 W bulb at 120 V?
When hot, R = V² ÷ P = 120² ÷ 60 = 240 Ω. A cold tungsten filament measures far less, which is why it draws a surge at switch-on.
Related conversions
Definitions of the volt, ampere, ohm and watt from the BIPM SI Brochure (9th ed.); the law as published by Georg Simon Ohm in Die galvanische Kette, mathematisch bearbeitet (1827). Last updated . How we verify our numbers.