Ohm's Law and Power Formulas for Electricians

Ohm's law and the power formulas are the base layer under almost every other calculation on the exam. Voltage drop, motor currents, heater output and transformer loads all come back to the same four quantities. This guide lists the formulas you need, shows when to use each one, and works through the problems that come up most often in electrical work.

The four quantities

Symbol	Quantity	Unit
E (or V)	Voltage	volts (V)
I	Current	amperes (A)
R	Resistance	ohms (Ω)
P	Power	watts (W)

The formulas

Ohm's law: E = I × R I = E / R R = E / I Power: P = E × I P = I² × R P = E² / R Rearranged: I = P / E I = √(P / R) E = √(P × R) R = E² / P R = P / I²

You do not need to memorize all twelve. Learn E = I × R and P = E × I, and you can build the rest by substitution. If you know any two of the four quantities, you can find the other two.

Problem 1: current and resistance of a heater

A 1,500 W portable heater runs on 120 V. What current does it draw, and what is its resistance?

I = P / E = 1,500 / 120 = 12.5 A R = E / I = 120 / 12.5 = 9.6 Ω Check: P = E² / R = 120² / 9.6 = 14,400 / 9.6 = 1,500 W

The check line is worth the five seconds it takes. If the check does not return the starting value, you made an arithmetic slip.

Problem 2: a 240 V element on 208 V

This one appears in many forms. A water heater element is rated 4,500 W at 240 V. It is installed on a 208 V system. What is the actual power?

The element is a resistance. Its resistance does not change with the supply voltage (ignoring temperature effects), so find R first, then use the new voltage.

R = E² / P = 240² / 4,500 = 57,600 / 4,500 = 12.8 Ω P at 208 V = 208² / 12.8 = 43,264 / 12.8 = 3,380 W

The element puts out about 75% of its rated power, because power in a resistance changes with the square of the voltage: (208 / 240)² = 0.751. A common wrong answer is to scale the power by 208 / 240 (about 87%), which treats power as proportional to voltage instead of voltage squared.

Problem 3: series and parallel resistance

Series: R(total) = R1 + R2 + R3 ... Parallel: 1 / R(total) = 1 / R1 + 1 / R2 + 1 / R3 ... Two in parallel: R(total) = (R1 × R2) / (R1 + R2) Equal resistors: R(total) = R / number of resistors

Three resistors of 10 Ω, 20 Ω and 20 Ω are connected in parallel across 120 V. Find the total resistance and total current.

1 / R = 1/10 + 1/20 + 1/20 = 0.1 + 0.05 + 0.05 = 0.2 R(total) = 1 / 0.2 = 5 Ω I(total) = 120 / 5 = 24 A Branch currents: 120/10 = 12 A, 120/20 = 6 A, 120/20 = 6 A → 12 + 6 + 6 = 24 A

In parallel, the total resistance is always smaller than the smallest branch. In series, it is always larger than the largest. If your answer breaks that rule, something went wrong.

Watts, volt-amperes and power factor

In AC circuits with motors, transformers or electronic loads, the current and voltage are not always in step. Then volts × amps gives apparent power in volt-amperes (VA), and the true power in watts is smaller.

Apparent power: VA = E × I True power: W = E × I × PF Power factor: PF = W / VA

For a resistive load like a heater, PF = 1 and watts equal volt-amperes. Exam questions will tell you the power factor when it matters. If none is given for a heating load, treat it as 1.

Three-phase power

Three-phase: P = 1.732 × E(line) × I(line) × PF I = P / (1.732 × E × PF)

A 480 V three-phase load draws 50 A at a power factor of 0.85.

P = 1.732 × 480 × 50 × 0.85 = 35,333 W ≈ 35.3 kW Apparent power = 1.732 × 480 × 50 = 41,568 VA ≈ 41.6 kVA

Use 1.732 only for three-phase. For single-phase 240 V, current is simply P / 240. Mixing these up is one of the most common calculation mistakes.

Horsepower and efficiency

1 hp = 746 W (output) Input W = output W / efficiency

A 10 hp motor is 90% efficient. Its input power is 10 × 746 / 0.90 = 8,289 W. For sizing motor circuits, though, the NEC does not want you to calculate current from horsepower like this. You use the full-load current tables in Article 430. See motor circuit calculations.

Putting it to work: a water heater circuit

Ohm's law questions often lead into a code question. A 4,500 W, 240 V storage water heater with a capacity of 120 gallons or less is treated as a continuous load under 422.13, so the branch circuit must be rated at least 125% of the heater's current.

I = 4,500 / 240 = 18.75 A 18.75 × 1.25 = 23.4 A minimum branch-circuit rating Next standard overcurrent device: 25 A Conductor: 10 AWG copper (12 AWG copper is limited to 20 A by 240.4(D))

The 125% factor for continuous loads also shows up in 210.19(A) and 210.20(A) for branch circuits in general. Any load expected to run for three hours or more counts as continuous.

Quick reference

Know two values, find the other two.
Resistance stays the same when the voltage changes; power does not.
Power in a resistance changes with the square of voltage and the square of current.
Parallel total is smaller than the smallest branch.
Three-phase uses 1.732; single-phase does not.

Voltage drop is Ohm's law applied to the wire itself. To see that connection step by step, try the voltage drop calculator.

Frequently asked questions

Why does a 240 V heater put out less heat on 208 V?

Its resistance stays the same, and power in a resistance changes with the square of the voltage. A 4,500 W, 240 V element on 208 V produces about 3,380 W, roughly 75% of its rating.

Are watts and volt-amperes the same?

Only when the power factor is 1, as with resistive loads like heaters. Otherwise watts equal volt-amperes times the power factor.

When do I use 1.732 in a power formula?

Only for three-phase circuits: P = 1.732 x E x I x PF, using line-to-line voltage and line current.

What is the total resistance of resistors in parallel?

Add the reciprocals and take the reciprocal of the sum. The result is always smaller than the smallest branch resistance.

FREE TOOL

Voltage drop calculator with every step shown

Single-phase or three-phase, copper or aluminum, 14 AWG to 500 kcmil.