Kalkulix

Ohm's law explained: U = R x I, with worked examples

Ohm's law is one equation – U = R × I – but it answers three different questions depending on which of the three values you already know. Here's how to use it, and how it connects to power.

7 min read · Updated: 7 September 2026

Key points

  • Ohm's law states voltage U (volts, V) equals resistance R (ohms, Ω) multiplied by current I (amps, A): U = R × I.
  • The same formula rearranges to find any of the three: I = U ÷ R (current), or R = U ÷ I (resistance), depending on which two values are already known.
  • A 100 Ω resistor with 2 A flowing through it has 200 V across it: U = 100 × 2 = 200 V.
  • Electrical power P (watts, W) is a separate but linked formula: P = U × I. Combined with Ohm's law, this gives two more useful forms: P = I²R and P = U²/R.
  • A UK kettle rated 3,000 W on a 230 V supply draws 13.04 A and has an effective resistance of 17.63 Ω while running.

The formula and what each letter means

Ohm's law describes the relationship between three quantities in an electrical circuit: voltage (U, measured in volts, V), current (I, measured in amps, A), and resistance (R, measured in ohms, Ω). The law states that voltage equals resistance multiplied by current: U = R × I. Voltage is the electrical “push” driving current through a circuit; resistance is how much a component or wire opposes that flow; current is the resulting rate of flow of electric charge.

Named after Georg Simon Ohm, who published the relationship in 1827, it holds for any component with a constant resistance (called an ohmic component) across the voltages and currents it's normally used at – resistors and most simple wiring behave this way. Some components, such as diodes and LEDs, don't follow a straight-line relationship between voltage and current, so Ohm's law is an approximation for those, not an exact description.

Rearranging for the unknown value

Because U = R × I involves three quantities, knowing any two lets you calculate the third. The same equation rearranges three ways depending on which value is missing.

The three forms of Ohm's law

Known valuesUnknownFormulaWorked example
Resistance (Ω) and current (A)Voltage (V)U = R × I100 Ω, 2 A → U = 100 × 2 = 200 V
Voltage (V) and resistance (Ω)Current (A)I = U ÷ R230 V, 115 Ω → I = 230 ÷ 115 = 2 A
Voltage (V) and current (A)Resistance (Ω)R = U ÷ I12 V, 3 A → R = 12 ÷ 3 = 4 Ω
Voltage U (V) = Resistance R (Ω) × Current I (A)Power P (W) = Voltage U (V) × Current I (A). Cover the value you want to find: U sits above R and I (U = R × I); covering I or R instead gives I = U ÷ R or R = U ÷ I.Voltage U (V)Current I (A)Resistance R (Ω)×=Power P (W)Cover the value you want to find: U sits above R and I (U = R × I); covering I or R instead gives I = U ÷ R or R = U ÷ I.
The Ohm's law triangle: cover the unknown quantity and the remaining two show whether to multiply or divide.

Power: the formula Ohm's law connects to

Electrical power (P, measured in watts, W) is how fast energy is being converted – into heat, light, motion or another form. It's calculated as P = U × I: voltage multiplied by current. This is a separate formula from Ohm's law, but because U and I are linked by resistance, substituting one into the other gives two more useful versions: P = I²R (power from current and resistance) and P = U²/R (power from voltage and resistance).

The three power formulas and when to use each

Known valuesFormulaWorked example
Voltage and currentP = U × I230 V, 2 A → P = 230 × 2 = 460 W
Current and resistanceP = I²R2 A, 115 Ω → P = 2² × 115 = 460 W
Voltage and resistanceP = U²/R230 V, 115 Ω → P = 230² ÷ 115 = 460 W

All three power formulas give the same answer for the same circuit

The 460 W result above comes from the same circuit (230 V, 2 A, 115 Ω) calculated three different ways – pick whichever formula uses the two values you already have, rather than calculating a third value first just to use P = U × I.

Worked example: a UK kettle

A typical UK kettle is rated 3,000 W on the standard 230 V mains supply. Using P = U × I rearranged for current, I = P ÷ U = 3,000 ÷ 230 = 13.04 A. That current draw is why kettles are wired on their own socket rather than shared with several other high-power appliances on the same circuit – a standard UK 13 A plug fuse is close to that kettle's running current already.

The kettle's effective resistance while heating follows from Ohm's law: R = U ÷ I = 230 ÷ 13.04 = 17.63 Ω. This is the resistance of the heating element at its operating temperature – element resistance in real appliances changes with temperature, so this figure is only accurate while the kettle is running at its rated power, not when cold.

Calculate any of the three values

Enter any two of voltage, current and resistance, and the calculator works out the third along with the resulting power.

Go to the Ohm's law calculator

Worked example: sizing a series resistor for an LED

A common practical use of Ohm's law is choosing a resistor to protect an LED from too much current. An LED with a forward voltage drop of 2 V, run from a 9 V supply and needing a target current of 20 mA (0.02 A), needs a resistor carrying the remaining 7 V (9 V − 2 V) at that current: R = U ÷ I = 7 ÷ 0.02 = 350 Ω.

This is Ohm's law applied to only the resistor's share of the circuit, not the whole supply voltage – the LED itself doesn't obey Ohm's law in a simple way (its voltage-current relationship isn't a straight line), so only the voltage dropped across the resistor is used in the calculation, and the LED's own forward-voltage figure comes from its datasheet, not from a formula.

Ohm's law doesn't apply cleanly to every component

Diodes, LEDs, transistors and many semiconductor devices have a non-linear relationship between voltage and current – resistance isn't constant as voltage changes. Ohm's law is exact for resistors and most simple wiring, but only an approximation, or not applicable at all, for these components.

Common mistakes

  • Mixing units: resistance in kΩ or MΩ must be converted to Ω, and current in mA must be converted to A, before applying the formula directly – a resistor marked “4.7k” is 4,700 Ω, not 4.7 Ω.
  • Using the full supply voltage instead of the voltage across just the component in question, especially in circuits with more than one resistor or component in series.
  • Applying Ohm's law to a non-ohmic component (diodes, LEDs, lamps at varying temperature) and expecting an exact answer rather than an approximation.
  • Forgetting that resistance itself can change with temperature in real components – the “cold” resistance of a heating element or incandescent filament is often much lower than its resistance while hot and operating.

Frequently asked questions

What is Ohm's law?

Ohm's law states that voltage (U, in volts) equals resistance (R, in ohms) multiplied by current (I, in amps): U = R × I. Knowing any two of the three values lets you calculate the third.

How do I calculate current from Ohm's law?

Rearrange the formula to I = U ÷ R. For example, 230 V across a 115 Ω resistance gives a current of 230 ÷ 115 = 2 A.

How do I calculate resistance from Ohm's law?

Rearrange the formula to R = U ÷ I. For example, 12 V driving a 3 A current gives a resistance of 12 ÷ 3 = 4 Ω.

How is power related to Ohm's law?

Power P = U × I is a separate formula, but combining it with Ohm's law gives two more forms: P = I²R and P = U²/R. All three give the same result for the same circuit – use whichever matches the two values you already know.

Does Ohm's law apply to LEDs?

Not directly. LEDs and other diodes have a non-linear voltage-current relationship, so Ohm's law doesn't describe the LED itself accurately. It's still used to size a series resistor that limits current to the LED, based on the voltage that resistor alone needs to drop.

What current does a 3,000 W kettle draw on a 230 V UK supply?

13.04 A, from I = P ÷ U = 3,000 ÷ 230. This is close to the 13 A rating of a standard UK plug fuse, which is why high-power appliances like kettles are typically wired on their own socket.

Why does my calculation give the wrong answer even though I used the right formula?

The most common cause is mismatched units – resistance given in kΩ or current given in mA needs converting to Ω and A before applying U = R × I directly, since the formula assumes base SI units.

Sources

Related articles

Convert your own figures

Enter the apparent power in kVA and the power factor to get the real power in kW, or the other way round.

Go to the kVA ↔ kW converter
Back to guides