Ohm's Law Wheel — All 12 Voltage, Current, Resistance & Power Formulas

The Ohm's law and power wheel: all twelve V, I, R, and P formulas with worked examples, computed for DC and unity-power-factor AC.


Updated August 20, 2026

Two laws generate the whole wheel: Ohm’s law, V = I × R, and Joule’s power law, P = V × I. Substitute one into the other and every one of the four quantities — voltage, current, resistance, power — can be written three ways, one for each pair of knowns. That is the twelve-formula “Ohm’s law wheel” printed inside meter lids and on shop walls, tabulated below as four tables: solve for V, solve for I, solve for R, solve for P. The formulas hold exactly for DC circuits and for AC circuits with purely resistive loads (unity power factor); with reactive loads, P = V·I becomes apparent power and the power-factor forms take over.

Every worked example uses one consistent circuit — 120 V across 24 Ω drawing 5 A and dissipating 600 W — and each result cell is computed through the same solver the Ohm's law calculator uses, from exactly the two quantities that row treats as known. Any two known quantities pin the other two: read the row for what you know, and the example shows the arithmetic once through.

Solve for Voltage (V)

Known pairFormulaWorked example
Current & resistanceV = I × R5 A × 24 Ω = 120 V
Power & currentV = P ÷ I600 W ÷ 5 A = 120 V
Power & resistanceV = √(P × R)√(600 W × 24 Ω) = 120 V

Solve for Current (I)

Known pairFormulaWorked example
Voltage & resistanceI = V ÷ R120 V ÷ 24 Ω = 5 A
Power & voltageI = P ÷ V600 W ÷ 120 V = 5 A
Power & resistanceI = √(P ÷ R)√(600 W ÷ 24 Ω) = 5 A

Solve for Resistance (R)

Known pairFormulaWorked example
Voltage & currentR = V ÷ I120 V ÷ 5 A = 24 Ω
Voltage & powerR = V² ÷ P120² ÷ 600 W = 24 Ω
Power & currentR = P ÷ I²600 W ÷ 5² = 24 Ω

Solve for Power (P)

Known pairFormulaWorked example
Voltage & currentP = V × I120 V × 5 A = 600 W
Current & resistanceP = I² × R5² × 24 Ω = 600 W
Voltage & resistanceP = V² ÷ R120² ÷ 24 Ω = 600 W

DC / unity-power-factor AC. With reactive AC loads, V·I is apparent power (VA); true power needs the power factor.

Sources & Further Reading

  • Ohm’s law (V = I·R) and Joule’s power law (P = V·I) — definitional identities; the twelve wheel formulas are their algebraic rearrangements (no empirical constants)
  • Worked examples computed through the Ohm’s law calculator’s solver, which encodes and tests all six known-quantity pairs

Frequently Asked Questions

How do I use the Ohm’s law wheel?

Identify the two quantities you know and the one you want, then read the matching row: each of the four tables solves for one quantity, and its three rows cover the three possible known pairs. Knowing any two of V, I, R, and P fixes the other two — the wheel is just all twelve rearrangements written out.

Do these formulas work for AC circuits?

For purely resistive AC loads — heaters, incandescent lamps — yes, using RMS voltage and current, because the power factor is 1. For motors, magnetic ballasts, and electronics, V × I gives apparent power in volt-amperes, and true watts require multiplying by the power factor; the wheel’s resistance forms also stop describing a simple resistance once reactance enters.

What current does a 600 W, 120 V load draw?

Use I = P ÷ V: 600 W ÷ 120 V = 5 A. The same row backwards sizes loads: a circuit that can supply 15 A at 120 V can carry up to P = V × I = 1,800 W of resistive load — before applying the 80% continuous-load rule used in circuit design.

Why are there two power formulas besides P = V × I?

They are substitutions that let you skip a step. P = I² × R drops V (useful for conductor heating — the “I-squared-R loss” in a wire of known resistance), and P = V² ÷ R drops I (useful for a fixed-resistance element on a known supply). All three agree exactly on any consistent circuit, as this chart’s examples show.

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