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Ohm’s law and power: U, I, R, P

Any two of voltage, current, resistance and power give the other two.

Parameters

Note

Fill in two fields; the other two are computed.

Result

Power P (W)

24

Relations
U = R × I and P = U × I, hence P = U²/R = I²R
Given
Voltage + Current
Voltage U (V)
12
Current I (A)
2
Resistance R (Ω)
6
Note

U = 12 V, I = 2 A, R = 6 Ω, P = 24 W.

Calculation, assumptions and sources

OverviewU = RI and P = UI tie voltage, current, resistance and power: any two of them determine the other two.

U = RI and P = UI tie voltage, current, resistance and power: any two of them determine the other two. A third input is refused, since three values can contradict each other. The law holds for ohmic components, under direct current, at constant temperature.

Method & formulasU = R × I
U = R × IP = U × IP = U² / RP = I² × R

The twelve formulas of the "Ohm’s law wheel" follow from U = RI and P = UI.

U
voltage across the component, in volts.
I
current through it, in amperes.
R
resistance, in ohms: the ratio U/I, constant for an ohmic component.
P
power dissipated as heat, in watts.
Validity domainOhm’s law describes ohmic components, resistors and conductors, under direct current and at constant temperature.

Ohm’s law describes ohmic components, resistors and conductors, under direct current and at constant temperature. The filament of a 100 W bulb reads 529 Ω lit, ten times less off. An LED, a diode or a transistor does not follow the law; under AC, impedance replaces resistance as soon as a coil or a capacitor is involved. All four quantities are positive, the component being passive.

Reading the resultPower sizes the component: a 100 Ω resistor under 10 V dissipates 1 W, and a quarter-watt part would burn there.

Power sizes the component: a 100 Ω resistor under 10 V dissipates 1 W, and a quarter-watt part would burn there. At fixed voltage, P = U²/R rises as R falls; at fixed current, P = I²R rises with R.