Ohm's law and power formulas
Voltage, current and resistance are tied by one relationship, and power adds a second. Know the four forms and you can solve any pair:
- V = I × R, so I = V ÷ R and R = V ÷ I.
- P = V × I, power in watts from volts and amps.
- P = I² × R, when you know the current and the resistance.
- P = V² ÷ R, when you know the voltage and the resistance.
The calculator picks the right pair of formulas for whichever two boxes you fill, and shows which ones it used.
Three-phase power with power factor
A balanced three-phase load draws real power from all three lines, and the line-to-line voltage multiplies by the square root of 3, about 1.732:
- Real power (W) = √3 × V × I × PF, with line-to-line voltage and line current.
- Apparent power (VA) = √3 × V × I, the same product with no power factor.
- Solve for current with I = P ÷ (√3 × V × PF), and for voltage with V = P ÷ (√3 × I × PF).
Power factor is the share of the apparent power that does useful work. It is a number from 0 to 1, and a purely resistive load has a power factor of 1.
Worked example 1: a 240 volt heating element
A 240 V single-phase element is rated 4,500 W. Current = P ÷ V = 4,500 ÷ 240 = 18.75 A. Resistance = V² ÷ P = 57,600 ÷ 4,500 = 12.8 Ω. The current you just found is the number the next steps start from. The conductor and overcurrent steps, by NEC article and table, are in the Calculations portion guide.
Worked example 2: current and resistance known
A load has 8 Ω of resistance and carries 15 A. Voltage = I × R = 15 × 8 = 120 V. Power = I² × R = 225 × 8 = 1,800 W. Check it the other way: 120 × 15 = 1,800 W.
Worked example 3: three-phase load with a power factor
A balanced load on 208 V, three-phase draws 30 A at a power factor of 0.9. Apparent power = 1.732 × 208 × 30 = about 10,808 VA. Real power = 10,808 × 0.9 = about 9,727 W, or 9.73 kW. Reporting kVA when the question asks for kW is the classic miss.
Worked example 4: current from kW
A balanced 15 kW load runs on 480 V, three-phase at a power factor of 0.85. Current = 15,000 ÷ (1.732 × 480 × 0.85) = 15,000 ÷ 706.7 = about 21.23 A. Convert kW to watts first, then divide.
These are original arithmetic examples, not exam questions.
Where Ohm's law shows up in voltage drop
A conductor is a resistor too. A 20 A load through 0.5 Ω of total wire resistance drops 20 × 0.5 = 10 V in the wire and wastes 20² × 0.5 = 200 W as heat. The NEC voltage drop guide turns that into the conductor-size method with your Chapter 9 table.
Ohm's law questions
When do I use P = I² × R instead of P = V × I?
Use whichever one uses the two values you were given. If the problem gives current and resistance, I² × R saves you from finding the voltage first.
Why is the three-phase formula 1.732 and not 2?
The 1.732 comes from the 120 degree spacing between the three phases. Single-phase two-wire voltage drop uses 2 for the out-and-back path, which is a different idea. Do not swap them.
What is the difference between kW and kVA?
kVA is apparent power, V × I × √3 for three-phase. kW is real power, which is kVA times the power factor. A transformer is sized in kVA, and a heater is rated in kW.
To reduce resistors in series and parallel and see every voltage drop, use the series and parallel resistance calculator.
To turn a transformer's kVA rating into current on each side, use the transformer calculator.
Ohm's law needs no NEC lookup, so it is the fastest arithmetic you can bank. Practice it alongside the free practice guide.