Ohm's law calculator: U, I, R and power P
The Ohm's law relates the voltage U, the current I and the resistance R by U = R · I. This online calculator derives the missing quantity from the other two, calculates the dissipated power P and visually represents the U/R/I triangle. DC and resistive AC cases plus common LED examples.
Ohm's law: statement and detailed formulas
Formulated by Georg Simon Ohm in 1827, Ohm's law is the founding equation of linear electricity. It states that, in a metallic conductor at constant temperature, the current flowing through it is proportional to the voltage across its terminals, the proportionality coefficient being the inverse of the resistance.
The three equivalent forms are:
U = R · I— voltage from resistance and current.I = U / R— current from voltage and resistance.R = U / I— resistance from voltage and current (voltmeter/ammeter method).
The dissipated power by Joule effect in the resistance is obtained through three equivalent forms:
P = U · I— direct formulation in watts.P = R · I²— useful when R and I are known (shunts, LV lines).P = U² / R— useful when U and R are known (load resistance).
All these formulas assume consistent SI units: volts, amperes, ohms, watts. Multiplying mA by kΩ gives volts directly: it is the king of mnemonic tricks in signal electronics.
Table of common values
| Application | U (V) | I | R (Ω) | P |
|---|---|---|---|---|
| Red LED on 5 V | 3 V (drop) | 20 mA | 150 | 60 mW |
| White LED on 12 V | 9 V (drop) | 20 mA | 470 | 0,18 W |
| 1 W power LED | 3,3 V | 350 mA | 9,4 (across 6,6 V) | 1,15 W |
| 12 V 50 W halogen lamp | 12 | 4,17 A | 2,88 | 50 W |
| Heating element 230 V 2 kW | 230 | 8,7 A | 26,4 | 2000 W |
| T13 socket outlet 16 A 230 V | 230 | 16 A max | 14,4 min. | 3680 VA |
| Three-phase motor 400 V 4 kW (cos φ=0.85) | 400 (line-to-line) | 6,8 A | — (reactive) | 4 kW |
| I²C pull-up 3.3 V | 3,3 | 0,75 mA | 4 700 | 2,5 mW |
| Shunt 0.1 Ω 5 A (current measurement) | 0,5 | 5 A | 0,1 | 2,5 W |
| Cu 2.5 mm² cable, 50 m at 16 A | 7,2 V drop | 16 A | 0,45 (loop) | 115 W losses |
Ohm's law in AC vs DC: the nuance that changes everything
The most frequent confusion in the field: applying U = R · I in AC on any load. In direct current (DC), the resistance R is the only parameter opposing the current: the formula applies as is, with voltage and current as constant values.
In alternating current (AC), two cases arise. On a purely resistive load (heating resistor, incandescent lamp, conventional ceramic hob), Ohm's law remains valid in RMS values: U[V rms] = R · I[A rms]. On a reactive load (asynchronous motor, transformer, electromagnetic ballast, capacitor), the pure resistance R is replaced by theimpedance Z = √(R² + X²), with XL = 2πfL or XC = 1/(2πfC).
The major consequence concerns power. In reactive AC, three quantities are distinguished: the apparent power S = U · I [VA] which sizes cables and transformers, the active power P = U · I · cos φ [W] which corresponds to the energy actually converted, and the reactive power Q = U · I · sin φ [VAr] which circulates without useful work. NIBT 2020 requires a cos φ ≥ 0.9 on industrial feeders, failing which a compensation using capacitor banks must be provided for.
LED examples: series resistor and combinations
Single LED on a 12 V supply
Typical case of an LED signalling strip or a distribution board indicator. Take a white LED with forward voltage Vf = 3 V and rated current If = 20 mA on a 12 V DC supply. The limiting series resistor is:
R = (Vsupply − Vf) / If = (12 − 3) / 0,020 = 450 Ω
Choose the next higher standard value in the E12 series, i.e. 470 Ω (which slightly reduces the current to 19.1 mA, with no visible impact on brightness). The dissipated power is P = (Vsupply − Vf) · If = 9 · 0,020 = 0,18 W, i.e. a 1/4 W (0.25 W) package with a ×1.4 margin.
LEDs in series on 12 V
Three white LEDs at Vf = 3 V each in series drop 9 V, leaving 3 V for the resistor. With If = 20 mA: R = 3 / 0,020 = 150 Ω, standardized 150 Ω E12, dissipated power 60 mW (1/8 W is enough). This series arrangement triples the efficiency compared with three separate LEDs: 60 mW of losses instead of 540 mW, for the same total flux.
LEDs in parallel: avoid
Putting LEDs directly in parallel without a dedicated resistor for each creates a thermal imbalance: the hottest LED sees its Vf drop, draws more current, heats up further, until runaway (current hogging). The rule is to allocate one resistor per LED, calculated individually. For 4 parallel LEDs on 12 V: 4 × 470 Ω, total current 80 mA, total power 0.72 W spread across 4 resistors of 1/4 W.
FAQ
What is the formula of Ohm's law?
U = R · I with U in volts, R in ohms and I in amperes. Permutations: I = U/R and R = U/I. Strictly valid in DC and in AC on a resistive load. In reactive AC, R is replaced by the impedance Z: U = Z · I. Power is P = U · I = R · I² = U² / R in watts; in AC with phase shift φ, active power becomes P = U · I · cos φ.
How do you calculate a series resistor for an LED?
R = (Valim − Vf) / If. Example white LED Vf=3 V, If=20 mA on 12 V: R = 9 / 0.020 = 450 Ω, standard value 470 Ω in E12. Dissipated power P = 0.18 W, 1/4 W package. At 5 V, the same LED requires 100 Ω 1/8 W. For a red LED Vf=2 V on 12 V, R = 500 Ω, standardised 510 Ω.
Does Ohm's law apply to AC?
Yes, on a purely resistive load in RMS values. On a reactive load (motor, transformer, capacitor, ballast), R is replaced by Z = √(R² + X²). Apparent power S = U · I [VA], active P = U · I · cos φ [W], reactive Q = U · I · sin φ [VAr]. Cos φ ≈ 0.85 for an asynchronous motor, 1 for a resistor, 0.5-0.7 for a non-PFC switch-mode power supply.
How do you convert from mA to A and from kΩ to Ω?
1 A = 1000 mA so 20 mA = 0,020 A. 1 kΩ = 1000 Ω, 1 MΩ = 1 000 000 Ω. 1 kV = 1000 V, 1 kW = 1000 W. Tip: multiplying mA by kΩ gives volts directly. Example 20 mA × 0,47 kΩ = 9,4 V. The calculator accepts V, A, Ω in SI units; convert beforehand if necessary.
What power for what resistance?
Safety factor ≥ 2 on P = R · I² = U² / R. Packages: 1/8 W SMD 0805/1206, 1/4 W axial, 1/2 W, 1 W, 5 W cement, 25-200 W wirewound with heatsink. Examples: 100 Ω at 50 mA → 0.25 W, choose 1/2 W; 0.1 Ω shunt at 5 A → 2.5 W, choose 5 W; 4 Ω load on 12 V → 36 W, allow 100 W with heatsink.
What is the difference between active power P and apparent power S?
In DC, P = U · I [W]. In AC: apparent power S = U · I [VA] sizes the cable and transformer, active power P = U · I · cos φ [W] corresponds to the energy converted, reactive power Q = U · I · sin φ [VAr] circulates without useful work. Power factor cos φ = P/S. NIBT requires cos φ ≥ 0.9 on industrial feeders; capacitor compensation below that.
To go further
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- ElectroBlocs IEC 60617 library — 1187 IEC 60617 symbols for AutoCAD.
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