Ohm's Law and Power Calculator
Enter any two of voltage, current, resistance and power to get the other two.
Formulas
V = I·R and P = V·I = I²R = (V²)/(R). Knowing the power a resistor dissipates tells you which rating to buy (¼ W, ½ W and so on). A common rule is to allow at least twice the calculated power.
Worked example: 470 Ω across 12 V
- Current: I = (V)/(R) = (12 V)/(470 Ω) = 25.5 mA
- Power: P = (V2)/(R) = ((12 V)2)/(470 Ω) = 306 mW
- A common ¼ W resistor would run at 123% of its rating and burn. With 2× margin you need a 1 W part.
Current and power by resistor (5 V and 12 V)
| Resistor | I (5 V) | P (5 V) | Rating (5 V) | I (12 V) | P (12 V) | Rating (12 V) |
|---|---|---|---|---|---|---|
| Resistor100 Ω | I (5 V)50 mA | P (5 V)250 mW | Rating (5 V)½ W | I (12 V)120 mA | P (12 V)1.44 W | Rating (12 V)5 W |
| Resistor220 Ω | I (5 V)22.7 mA | P (5 V)114 mW | Rating (5 V)¼ W | I (12 V)54.5 mA | P (12 V)655 mW | Rating (12 V)2 W |
| Resistor330 Ω | I (5 V)15.2 mA | P (5 V)75.8 mW | Rating (5 V)¼ W | I (12 V)36.4 mA | P (12 V)436 mW | Rating (12 V)1 W |
| Resistor470 Ω | I (5 V)10.6 mA | P (5 V)53.2 mW | Rating (5 V)⅛ W | I (12 V)25.5 mA | P (12 V)306 mW | Rating (12 V)1 W |
| Resistor1 kΩ | I (5 V)5 mA | P (5 V)25 mW | Rating (5 V)⅛ W | I (12 V)12 mA | P (12 V)144 mW | Rating (12 V)½ W |
| Resistor2.2 kΩ | I (5 V)2.27 mA | P (5 V)11.4 mW | Rating (5 V)⅛ W | I (12 V)5.45 mA | P (12 V)65.5 mW | Rating (12 V)¼ W |
| Resistor4.7 kΩ | I (5 V)1.06 mA | P (5 V)5.32 mW | Rating (5 V)⅛ W | I (12 V)2.55 mA | P (12 V)30.6 mW | Rating (12 V)⅛ W |
| Resistor10 kΩ | I (5 V)500 µA | P (5 V)2.5 mW | Rating (5 V)⅛ W | I (12 V)1.2 mA | P (12 V)14.4 mW | Rating (12 V)⅛ W |
Ratings are the smallest standard rating at least twice the power. If your ammeter reads below the calculation, see the guide on why an ammeter reads low.
Checked on our bench: one current, three readings
The numbers below are computed from values the site owner recorded in lab classes. Measured values are left exactly as recorded.
A 1 kΩ and a 3 kΩ resistor in series, with the supply set to 10 V and then 16 V. The same current was worked out three ways: the supply voltage put through this calculator, the voltage across each resistor divided by its nominal value, and the ammeter reading.
| Supply | Calculator (Vdc ÷ 4 kΩ) | VR1 ÷ 1 kΩ | VR2 ÷ 3 kΩ | Ammeter |
|---|---|---|---|---|
| Supply10 V | Calculator (Vdc ÷ 4 kΩ)2.506 mA | VR1 ÷ 1 kΩ2.48 mA | VR2 ÷ 3 kΩ2.501 mA | Ammeter2.438 mA (−2.7%) |
| Supply16 V | Calculator (Vdc ÷ 4 kΩ)4.004 mA | VR1 ÷ 1 kΩ3.963 mA | VR2 ÷ 3 kΩ3.952 mA | Ammeter3.912 mA (−2.3%) |
Both voltage-based currents land within about 1% of the calculator, while the ammeter reads 2.3 to 2.7% low. Ohm's law isn't wrong here. The ammeter and its leads added resistance to the loop. The full analysis is in why an ammeter reads low.
What this means when you use the calculator: when you need a small current accurately, measure the voltage across a resistor you know and let the calculator do I = V ÷ R. A voltmeter barely disturbs the circuit.
Working with Ohm's law
Sizing a resistor from two known values
- Type the two values you know. Prefixes work, so 4.7k, 20mA or 0.5W are fine and you never have to convert units first.
- Read the other two values in the result card. The power line tells you how hot the part will run.
- Pick a resistor rating at least twice the power shown, and check the supply can deliver the current.
Where Ohm's law sums go wrong
- Mixing milliamps and amps. 20 typed into the current field means 20 A, not 20 mA. Add the m.
- Using it on a diode, LED or transistor. These parts are not resistors; their current changes sharply with voltage. Use the LED or transistor calculators instead.
- Ignoring the rest of the loop. Wires, a meter's shunt and the supply's own resistance all add to R. With small resistors (under about 10 Ω) they can change the current noticeably.
Reading current without an ammeter
Measure the resistor out of the circuit first, because a 5% part can sit anywhere in its band. Then measure the voltage across it while powered and divide by the measured resistance; that current is usually more trustworthy than an ammeter reading on a low range. A resistor that runs too hot to touch is close to its rating even if the numbers look fine, because the rating assumes free air at 70 °C or below.
Frequently asked questions
How do I choose a resistor's power rating?
Pick a rating at least twice the power the resistor dissipates. If it dissipates 0.1 W, a ¼ W (0.25 W) resistor is a good choice.
Are there parts that don't follow Ohm's law?
Yes. Diodes, LEDs and transistors are non-linear, so current is not proportional to voltage and V = IR does not apply directly. Use the voltage-current curve in the datasheet instead.