Four quantities describe a simple load: voltage, current, resistance and power. Two formulas link them, so any two give the other two. This page sets out the formulas on d.c., then shows what changes on single‑phase and three‑phase a.c.
The two formulas
Ohm's law. The voltage across a resistance is the current through it times the resistance.
V = I × R
The power formula. The power is the voltage times the current.
P = V × I
Put one into the other and you get two more forms. P = I² × R is the power from the current and the resistance. P = V² ÷ R is the power from the voltage and the resistance.
| Quantity | Symbol | Unit |
|---|---|---|
| Voltage | V | volt, V |
| Current | I | ampere, A |
| Resistance | R | ohm, Ω |
| Power | P | watt, W |
Any two give the other two
There are six pairs you can know. Each pair gives the other two quantities.
| You know | You find | Formulas |
|---|---|---|
| V and I | R and P | R = V ÷ I, P = V × I |
| V and R | I and P | I = V ÷ R, P = V² ÷ R |
| V and P | I and R | I = P ÷ V, R = V² ÷ P |
| I and R | V and P | V = I × R, P = I² × R |
| I and P | V and R | V = P ÷ I, R = P ÷ I² |
| R and P | V and I | V = √(P × R), I = √(P ÷ R) |
Two runs on d.c.:
- A 50 W lamp on 12 V draws 4.17 A through 2.88 Ω. Open the 12 V lamp.
- A 470 Ω resistor carrying 20 mA has 9.4 V across it and gives off 0.188 W. Open the resistor.
Single‑phase a.c.
On a.c. the voltage and the current swing back and forth fifty times a second. The figures you read on a meter are the r.m.s. values, and the formulas above use those.
A heater is a pure resistance. Its current is in step with the voltage, and the d.c. formulas hold as they are. A 2 kW heater on 230 V draws 8.7 A and works at 26.45 Ω. Open the 2 kW heater.
A motor or a transformer also has a coil. The coil makes the current lag the voltage. Two things change.
- Impedance takes the place of resistance. The opposition to current is the impedance, Z. Ohm's law becomes V = I × Z. Z has two parts at right angles: the resistance R and the reactance X.
- The power factor enters the power. The voltage times the current is the apparent power, S, in VA. Only part of it is real power, P, in W. The power factor, pf, is that part.
| Quantity | Formula |
|---|---|
| Impedance | Z = V ÷ I |
| Apparent power | S = V × I |
| Real power | P = V × I × pf |
| Resistance part | R = Z × pf |
| Reactance part | X = Z × sin φ |
| Real power from current | P = I² × R |
Take a load on 230 V that draws 10 A at a power factor of 0.8. Open the load at 0.8 power factor.
| Result | Value |
|---|---|
| Impedance | 23 Ω |
| Resistance part | 18.4 Ω |
| Reactance part | 13.8 Ω |
| Apparent power | 2.3 kVA |
| Real power | 1,840 W |
| Reactive power | 1.38 kvar |
The real power is the current squared times the resistance part, not the impedance. On a motor the resistance part is a working figure: it stands for the power the shaft delivers as well as the heat in the windings. A meter across a motor winding reads only the copper, far less than this. At a power factor of 1 the reactance is zero and Z equals R.
Three‑phase a.c.
A three‑phase supply has three actives. Two kinds of value matter.
- The line voltage is between two phases: 400 V in most NZ and Australian installations.
- The phase voltage is from a phase to the neutral or star point. It is the line voltage over √3.
- The line current is the current in each phase conductor, the figure a clamp meter reads.
For a balanced load the real power is:
P = √3 × V × I × pf
with V the line voltage and I the line current. The apparent power is √3 × V × I.
The impedance depends on how the load is connected.
- Star. Each part of the load sits between a phase and the star point. It sees the phase voltage and carries the line current. Its impedance is (V ÷ √3) ÷ I.
- Delta. Each part sits across two phases. It sees the full line voltage and carries the line current over √3. Its impedance is three times the star value for the same line current.
The tool gives the star value per phase. A motor on 400 V drawing 32 A at a power factor of 0.85 looks like 7.22 Ω per phase and takes 18,844.71 W. Open the motor.
A star heater bank of 20 Ω per phase on 400 V draws 11.55 A in each line and gives 8,000 W. Open the heater bank.
Where the formulas stop
- Resistance that changes with heat. A lamp filament or a heater element has a higher resistance hot than cold. The formulas use the working value. A meter on a cold element reads lower.
- Unbalanced three‑phase loads. The √3 formulas hold only when the three phases carry the same current at the same power factor. An unbalanced load has to be worked phase by phase.
- Harmonics. Drives, LED drivers and chargers draw current that is not a clean sine wave. The simple power factor no longer tells the whole story.
For a worked water heater element and a three‑phase heater bank, read Ohm's law on a 3 kW water heater element. For the voltage a cable drops, which is Ohm's law on the cable itself, read Voltage drop basics and d.c. voltage drop and conductor resistance. For loads with a low power factor, read Power factor and reactive power. To turn kW into amps for a motor or a board, use the full load current calculator or the three phase power calculator.