Electrical calculators

Ohm’s law calculator

Enter any two of voltage, current, resistance and power. The other two appear instantly, with the working shown step by step.

Calculated with electrician.wiki · https://electrician.wiki/calculators/ohms-law/

Result

I = 2.55 mA · P = 30.6 mW

With 12 V across 4.7 kΩ, 2.55 mA flows and 30.6 mW is dissipated.

Voltage (V): 12 V
Current (I): 2.553 mA
Resistance (R): 4.7 kΩ
Power (P): 30.64 mW

If this is a resistor, use one rated at least 0.125 W (about twice the 30.6 mW it dissipates) so it runs cool.

Show the work

1. Current from Ohm’s law
$I = \frac{V}{R} = \frac{12 V}{4.7 kΩ} = 2.553 mA$
2. Power
$P = \frac{V^{2}}{R} = \frac{{(12 V)}^{2}}{4.7 kΩ} = 30.64 mW$

DC, or AC with a purely resistive load (power factor 1). For motors, transformers and other reactive loads use the power and impedance calculators.

New to this? Ohm’s law explained

Report a problem with this result

How to use this calculator

Pick the two quantities you already know, type them in, and the calculator fills in the other two. It accepts the shorthand printed on parts and schematics: 4k7 means 4.7 kΩ, 2M2 means 2.2 MΩ, 4R7 means 4.7 Ω, and 250m means 0.25. Results update as you type, every step is shown under the answer, and the address bar always holds a link that reproduces exactly what you see.

The twelve Ohm’s law and power formulas

Ohm’s law says the current through a resistance is proportional to the voltage across it: V = I × R. Combine it with the power relation P = V × I and any two of voltage, current, resistance and power give you the other two. These are the twelve rearrangements, sometimes drawn as the “power wheel”:

Voltage V (volts)

V = I \cdot R

V = \frac{P}{I}

V = \sqrt{P \cdot R}

Current I (amperes)

I = \frac{V}{R}

I = \frac{P}{V}

I = \sqrt{\frac{P}{R}}

Resistance R (ohms)

R = \frac{V}{I}

R = \frac{V^{2}}{P}

R = \frac{P}{I^{2}}

Power P (watts)

P = V \cdot I

P = I^{2} \cdot R

P = \frac{V^{2}}{R}

Worked examples

A pull-up resistor on a 5 V logic line

A 10 kΩ resistor between a 5 V rail and an input pin. How much current flows when the pin is pulled low, and how much heat does the resistor make?

1. Current from Ohm’s law
$I = \frac{V}{R} = \frac{5 V}{10 kΩ} = 500 µA$
2. Power
$P = \frac{V^{2}}{R} = \frac{{(5 V)}^{2}}{10 kΩ} = 2.5 mW$

With 5 V across 10 kΩ, 500 µA flows and 2.5 mW is dissipated.

A 1,500 W space heater on a 120 V circuit

The nameplate says 1,500 W at 120 V. What current does it draw, and what is its hot resistance?

1. Current from the power formula
$I = \frac{P}{V} = \frac{1.5 kW}{120 V} = 12.5 A$
2. Resistance
$R = \frac{V^{2}}{P} = \frac{{(120 V)}^{2}}{1.5 kW} = 9.6 Ω$

1.5 kW at 120 V draws 12.5 A, equivalent to 9.6 Ω.

A 3 kW immersion heater element on 230 V

A typical European or UK water-heater element. The current tells you the circuit and fuse size to look at next.

1. Current from the power formula
$I = \frac{P}{V} = \frac{3 kW}{230 V} = 13.04 A$
2. Resistance
$R = \frac{V^{2}}{P} = \frac{{(230 V)}^{2}}{3 kW} = 17.63 Ω$

3 kW at 230 V draws 13 A, equivalent to 17.6 Ω.

Choosing a resistor’s power rating

Power is where small circuits go wrong. A resistor that dissipates close to its rated wattage runs hot, drifts in value and shortens the life of everything near it. A good rule is to pick a part rated for at least twice the power it will dissipate, which is what the calculator suggests under each result. Through-hole resistors are commonly rated ⅛ W, ¼ W, ½ W, 1 W and 2 W; surface-mount sizes run from about 0.05 W (0402) to 1 W (2512). Above a few watts you are into wirewound or chassis-mount power resistors that need airflow or a heatsink.

Watch the square in $P = I^{2} \cdot R$ : doubling the current through a resistor quadruples its heat. The same is true of cable: a conductor carrying twice the current produces four times the heat per foot, which is why the electrical codes limit current by conductor size.

Where Ohm’s law stops being enough

AC loads with reactance. Motors, transformers, fluorescent ballasts and LED drivers draw current that is out of phase with the voltage, so V × I gives volt-amperes, not watts. Use the amps to watts calculator with the power factor, or the three-phase power calculator.
Resistance that changes with temperature. A tungsten lamp filament has roughly a tenth of its hot resistance when cold, so the inrush current is many times the running current. Copper’s resistance rises about 0.4 % per °C, which is why cable resistance tables state a temperature (75 °C in the NEC tables).
Non-linear parts. Diodes and LEDs do not have a fixed resistance; their current rises steeply above a threshold voltage. For an LED you size the series resistor from the voltage left over: LED resistor calculator.
Long cable runs. The wire itself is a resistance in series with the load. The voltage drop calculator works out how many volts it costs you.

Method and verification

The calculator solves the two equations $V = I \cdot R$ and $P = V \cdot I$ exactly in double precision, then displays four significant figures with SI prefixes. All six input pairs are tested to give the same circuit back (for example, 12 V across 4.7 kΩ returns 2.553 mA and 30.64 mW from every direction). Values are only rounded for display; nothing is rounded between steps.

Assumes DC, or AC with a purely resistive load. Updated 2026-10-14. Found a problem? Report it and we log every fix publicly.

Questions people ask

What is Ohm’s law?

Ohm’s law states that the voltage across a resistance equals the current through it times the resistance: V = I × R. It holds for resistive parts like heaters and resistors at a steady temperature.

How do I find amps from watts and volts?

Divide the power by the voltage: I = P ÷ V. A 1,500 W heater on 120 V draws 1,500 ÷ 120 = 12.5 A. For motors and other AC loads with a power factor below 1, divide by the power factor too.

How do I find resistance from voltage and current?

Divide the voltage by the current: R = V ÷ I. If 12 V drives 0.5 A through a load, its resistance is 24 Ω.

What wattage resistor do I need?

Work out the power it dissipates with P = I² × R (or V² ÷ R), then choose a resistor rated at least twice that. The calculator suggests the next standard rating for you.

Does Ohm’s law work for AC?

Yes for purely resistive loads such as heaters and incandescent lamps (use RMS values). For motors, transformers and electronic loads, current and voltage are out of phase, so you need impedance and power factor rather than plain resistance.