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Ohm's law calculator

Ohm's law calculator: enter any two of voltage, current, resistance and power to get the other two, shown in the units you pick (mA, kΩ…).

Drafted with AI writing toolsUpdated Checked examples: 6

Accepts 4.7k or 2.2M
Try
Current
A
Current: 2 A
Significant figures: 6; Nearest, ties away from zero
Voltage
12V
Resistance
6Ω
Power
24W
Conductance
0.166667S

12 V across 6 Ω drives 2 A and turns 24 W into heat.

Ohm's-law triangle (solved values highlighted)

V12 VI2 AR6 ΩP = V × I24 WCover the one you want: V = I × R, I = V ÷ R, R = V ÷ I
How it's calculated S

  1. Given, in base units

    V=12 V,R=6 ΩV = 12\ \mathrm{V},\quad R = 6\ \mathrm{\Omega}

  2. Find I

    I=VR=126=2 AI = \frac{V}{R} = \frac{12}{6} = 2\ \mathrm{A}

  3. Find P

    P=V2R=(12)26=24 WP = \frac{V^2}{R} = \frac{(12)^2}{6} = 24\ \mathrm{W}

  4. Check

    IR=12 V,VI=24 WIR = 12\ \mathrm{V},\quad VI = 24\ \mathrm{W}

    Ohm's law holds for ohmic conductors (resistance independent of current); diodes, lamps and LEDs are not ohmic.

About the Ohm's law calculator

Ohm's law says the current through a resistor equals the voltage across it divided by its resistance, I = V/R. Combined with the power law P = VI, any two of voltage, current, resistance and power fix the other two: P = I²R and P = V²/R follow by substitution. Solved values are shown in the unit you choose, so a milliamp current does not appear as 0.002 A.

The default, 12 V across 6 Ω, gives 2 A and 24 W. The same arithmetic sizes fuses and cables (a 2,300 W kettle on 230 V draws 10 A) and checks resistor ratings (a 100 Ω, ¼ W resistor carries at most 50 mA).

The law holds for ohmic conductors, whose resistance stays constant as the current changes. Diodes, LEDs and filament lamps are not ohmic, so a single resistance value describes them only at one operating point.

Ohm's law formula: V = IR

Voltage equals current times resistance: V = IR, so I = V/R and R = V/I, in volts, amperes and ohms. With 5 V across a 1 kΩ (1,000 Ω) resistor, I = 5 ÷ 1,000 = 0.005 A, or 5 mA. The law holds for resistors and other conductors whose resistance stays constant.

The Ohm's law triangle is a memory aid: V sits on top, with I and R below. Cover the quantity you want, and the two that remain show the operation: I × R, V ÷ R or V ÷ I.

OpenStax's Ohm's law lesson derives the relationship and shows where it breaks down.

Power formula: P = VI, I²R or V²/R

Electrical power is voltage times current, P = VI, in watts. Substituting V = IR gives P = I²R, and substituting I = V/R gives P = V²/R. For a resistor all three give the same answer, so use the one that matches the two values you know.

A kettle on 230 V drawing 10 A uses 230 × 10 = 2,300 W, and its element is 230 ÷ 10 = 23 Ω. A 100 Ω resistor with 5 V across it dissipates 5² ÷ 100 = 0.25 W.

Power grows with the square of voltage or current. Doubling the voltage across that 100 Ω resistor to 10 V raises its dissipation fourfold, to 1 W.

All twelve Ohm's law and power formulas

Any two of V, I, R and P fix the other two. The full set, often drawn as a power wheel:

  • Voltage: V = IR = P/I = √(PR)
  • Current: I = V/R = P/V = √(P/R)
  • Resistance: R = V/I = V²/P = P/I²
  • Power: P = VI = I²R = V²/R

Worked examples

Each row is one operating point. Any two of its values reproduce the other two.

Voltage (V)Current (A)Resistance (ohms)Power (W)
122624
50.00510000.025
23010232300
50.051000.25
  • 5 V across 1,000 Ω: I = 0.005 A, and P = 5² ÷ 1,000 = 0.025 W, or 25 mW.
  • 100 Ω dissipating 0.25 W: I = √(0.25 ÷ 100) = 0.05 A, and V = √(0.25 × 100) = 5 V.

Milliamps, kilohms and other prefixes

Convert prefixes before using the base formulas: 1 mA = 0.001 A, 1 kΩ = 1,000 Ω, 1 MΩ = 1,000,000 Ω, 1 mW = 0.001 W and 1 kW = 1,000 W.

Milliamps times kilohms gives volts directly, because the two prefixes cancel. A 2 mA current through 4.7 kΩ drops 2 × 4.7 = 9.4 V and dissipates 0.002² × 4,700 = 0.0188 W, or 18.8 mW.

Conductance is the reciprocal of resistance, G = 1/R, measured in siemens. A 1 kΩ resistor has a conductance of 0.001 S, or 1 mS.

LEDs, lamps and AC loads

Diodes and LEDs are non-ohmic. Almost no current flows below their forward voltage, then current rises steeply, so a single V/I ratio describes them at one operating point only. That is why an LED needs a series resistor.

A filament lamp's resistance rises as it heats, and ordinary resistors drift with temperature. V/I still gives the resistance at each operating point, but that value changes with temperature.

For AC heaters and kettles, P = VI works with RMS voltage and current.

Motors and many power supplies have a power factor (PF) below 1, and real power is P = VI × PF. A 1,000 W load on 230 V at a power factor of 0.8 draws 5.43 A, not the 4.35 A that 1,000 ÷ 230 suggests.

The OpenStax lesson on electrical energy and power covers the power formulas and the difference between power and energy.

Common mistakes

  • Dropping a prefix. 5 V across 4.7 kΩ gives 5 ÷ 4,700 ≈ 1.06 mA. Dividing by 4.7 instead gives 1.06 A, a thousand times too much.
  • Using the supply voltage for one resistor in a series chain. Each resistor sees only its share of the voltage, so P = V²/R needs the voltage across that resistor.
  • Running a resistor at its full rating. A 100 Ω resistor at 5 V dissipates the whole 0.25 W a ¼ W part is rated for; a 0.5 W resistor leaves headroom.
  • Confusing watts with watt-hours. Power is a rate. A 24 W load running for one hour uses 24 Wh, which is 86,400 J.

Input formats and units

  • Pick the two quantities you know; the other two are solved. Every entry must be positive.
  • Resistance accepts suffixes such as 4.7k or 2.2M. Typing 4.7k with the unit set to kΩ applies both prefixes and means 4.7 MΩ.
  • Solved values are shown in the units you choose, such as mA or kΩ. Conductance follows the resistance unit, so kΩ gives mS.
  • Power can carry its own unit: 5 V and 500 mW give 0.1 A and 50 Ω.

For energy use and running cost over hours, use the electricity cost calculator. To read a resistor's value from its bands, use the resistor color code calculator.

For networks of resistors, use the series and parallel calculator or the voltage divider and LED resistor calculator.

Worked examples

12 V across 6 Ω

I know
Voltage and resistance
Voltage
12
Voltage unit
V
Resistance
6
Resistance unit
Ω
Show current in
A
Show power in
W
Current
2 A
Power
24 W

Checked against: I = 12/6, P = 12²/6

Kettle: 230 V drawing 10 A

I know
Voltage and current
Voltage
230
Voltage unit
V
Current
10
Current unit
A
Show resistance in
Ω
Show power in
W
Resistance
23 Ω
Power
2,300 W

Checked against: R = 230/10, P = 230 × 10

Quarter-watt 100 Ω resistor at full rating

I know
Resistance and power
Resistance
100
Resistance unit
Ω
Power
0.25 W
Show voltage in
V
Show current in
A
Current
0.05 A
Voltage
5 V

Checked against: I = √(0.25/100), V = √(0.25 × 100)

5 V, 500 mW

I know
Voltage and power
Voltage
5
Voltage unit
V
Power
500 mW
Show current in
A
Show resistance in
Ω
Current
0.1 A
Resistance
50 Ω

Checked against: I = 0.5/5, R = 25/0.5

Questions

How do you calculate current from voltage and resistance?

Divide the voltage by the resistance: I = V/R. With 12 V across 6 Ω the current is 2 A; with 5 V across 1 kΩ it is 0.005 A, or 5 mA. Convert kilohms to ohms (×1,000) before dividing, or pick kΩ and mA in the unit selectors and let the calculator convert.

How do you convert watts to amps?

Divide the power by the voltage: I = P/V. A 2,300 W kettle on a 230 V supply draws 10 A, and a 60 W load on 12 V draws 5 A. For AC this holds for resistive loads such as heaters and kettles; motors and power supplies with a power factor below 1 draw more current than P/V.

What is the formula for electrical power?

P = VI, which becomes P = I²R or P = V²/R once Ohm's law is substituted. A 100 Ω resistor with 5 V across it dissipates 5²/100 = 0.25 W, exactly the limit of a ¼ W part, so a 0.5 W resistor is the safer choice there.

Does Ohm's law apply to LEDs and light bulbs?

Not with a single fixed resistance. An LED conducts almost nothing below its forward voltage (about 2 V for red, about 3 V for blue and white) and then current rises steeply, so it needs a series resistor. A filament lamp's resistance rises as the tungsten heats, so the V²/P value, 882 Ω for a 60 W, 230 V bulb, applies only when it is lit.

What is conductance?

Conductance is the reciprocal of resistance, G = 1/R, measured in siemens (S). A 6 Ω resistor has a conductance of 1/6 ≈ 0.167 S. Conductances in parallel add directly, which is why it is convenient for parallel networks: 1/R_total = G₁ + G₂ + …

About this calculator

V=IR,P=VI=I2R=V2RV = IR,\qquad P = VI = I^2R = \frac{V^2}{R}

Sources

  1. OpenStax University Physics Volume 2, §9.4 Ohm's law and §9.5 Electrical energy and power
  2. HyperPhysics — Ohm's law, power

Checked against references

This calculator includes 6 worked examples with independently sourced answers. They run in the test suite, and you can run them here too.

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