12 volts across a 6 Ω resistor drives 2 A of current and dissipates 24 W. Pick a voltage and a resistance to see all twelve formulas worked out.
| Find | Formula | Known values | Result | Unit |
|---|---|---|---|---|
| Voltage | V = I × R | I = 2 A, R = 6 Ω | 12.000 | V |
| Voltage | V = P / I | P = 24 W, I = 2 A | 12.000 | V |
| Voltage | V = √(P × R) | P = 24 W, R = 6 Ω | 12.000 | V |
| Current | I = V / R | V = 12 V, R = 6 Ω | 2.000 | A |
| Current | I = P / V | P = 24 W, V = 12 V | 2.000 | A |
| Current | I = √(P / R) | P = 24 W, R = 6 Ω | 2.000 | A |
| Resistance | R = V / I | V = 12 V, I = 2 A | 6.000 | Ω |
| Resistance | R = V² / P | V = 12 V, P = 24 W | 6.000 | Ω |
| Resistance | R = P / I² | P = 24 W, I = 2 A | 6.000 | Ω |
| Power | P = V × I | V = 12 V, I = 2 A | 24.000 | W |
| Power | P = I² × R | I = 2 A, R = 6 Ω | 24.000 | W |
| Power | P = V² / R | V = 12 V, R = 6 Ω | 24.000 | W |
How it is computed
Ohm's law, V = I × R, combined with the power law, P = V × I, gives three ways to find each of V, I, R and P from any two others. The table starts from the voltage and resistance you choose, derives I and P, then checks every formula against them.
FAQ
How do I remember the formulas?+
Use the wheel: cover the quantity you want and the remaining two show the operation. V over I × R is the core triangle; power adds P = V × I.
Does Ohm's law work for AC circuits?+
For purely resistive loads, yes, with RMS values. With motors or capacitors, use impedance instead of resistance, and real power also depends on the power factor.