Electrical fundamentals

Electrical calculator

Solve DC resistive-load quantities, convert AC current and power, or relate balanced wye line and phase voltages.

DC resistive load

Calculation inputs

Inputs

V
A

Example values are loaded from the worked method example: 24 V and 12 Ω. Replace them with job values; results update when both entries are valid.

Electrical quantities

Enter both known values to calculate.

Ideal DC resistive-load arithmetic only. This result does not select conductors, breakers or equipment.

AC power and current

Calculation inputs

AC load details

V RMS

For single-phase use circuit voltage; for three-phase use line-to-line voltage, not line-to-neutral voltage.

0–1

Use the load's power factor, not motor/inverter efficiency. This calculation assumes sinusoidal supply and a balanced three-phase load.

A

Example values are loaded from the worked method example: 600 V line-to-line, 0.8 power factor and 10 A on a balanced three-phase load. Replace them with nameplate or measured values.

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AC calculation result

Enter values and calculate.

Use power on the same AC side as the current. Enter an inverter's rated AC output directly; do not apply DC-to-AC efficiency again. Motor shaft output needs an efficiency conversion before it represents electrical input. No conductor or breaker selection is provided.

Balanced wye line and phase voltage

Calculation inputs

Known wye-system voltage

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Balanced wye voltages

Enter one known RMS voltage.

Balanced sinusoidal star-connected voltage relationship only. No nominal supply or installation conclusion is inferred.

What it calculates

The calculator solves Ohm's law and the electrical power relationship for a single ideal resistive DC load. All six pairs of known values share this page. Values are magnitudes, with positive resistance and power consumption.

Gather any two quantities describing the same load and operating point. Convert milliamperes to amperes, kilohms to ohms and kilowatts to watts before entering them. Inputs and results are not stored or submitted.

Method and example

V = I × R and P = V × I. Substitution gives P = I² × R and P = V² ÷ R. For power and resistance, the positive current is √(P ÷ R). This tool reports the positive solution.

For example, 24 V across 12 Ω gives 2 A and 48 W. Specifying any other pair from those four values produces the same result. Calculations retain numeric precision; the display uses three decimal places, with scientific notation for very small or large quantities.

AC method and limits

Single-phase apparent power is S = V × I ÷ 1000 in kVA. Balanced three-phase apparent power is S = √3 × V × I ÷ 1000, using line-to-line RMS voltage and line RMS current. Real AC power is P = S × power factor in kW. The inverse modes solve for current or apparent power using the same relationships.

For a balanced sinusoidal star-connected three-phase system, line-to-line RMS voltage is √3 times line-to-neutral (phase) RMS voltage. This conversion is not valid as a general rule for delta-connected systems or unbalanced phase voltages.

Example: a balanced three-phase load at 600 V line-to-line and 10 A, with power factor 0.8, draws approximately 10.392 kVA and 8.314 kW. A 4.8 kW single-phase resistive heater at 240 V and power factor 1 draws 20 A. A balanced star system with 120 V line-to-neutral has approximately 207.846 V line-to-line. These are arithmetic examples, not installation recommendations.

The power-factor current multiplier mode reports ratios only. For unchanged real power, voltage and phase arrangement, current is inversely proportional to PF: the multiplier versus unity PF is 1 ÷ PF. No energy savings, bill impact or equipment capacity is inferred.

Use real power from the same AC side as the current. For inverter AC output current, enter rated AC output power directly; do not apply DC-to-AC efficiency again. For motor current, shaft output is not electrical input, so account for efficiency before using shaft power. Supply the actual load power factor; the tool does not infer it. The AC model assumes sinusoidal operation and balanced three-phase loads. It does not model harmonics, variable drive input behaviour or neutral current.

Zero load is supported. Power factor zero can be used when current or kVA is known, but cannot determine current from real power. No code demand factor, continuous-load rule, conductor sizing or breaker choice is calculated.

DC interpretation and limits

The result describes an ideal resistive operating point. It is not an AC power-factor, three-phase, impedance or nonlinear-device model. Do not apply the displayed resistance as a motor, inverter or LED driver selection rule.

Zero resistance, open circuits, zero power and negative power flow need different or sometimes indeterminate models. They are excluded here, rather than silently assigning a result. No Canadian code acceptance decision, conductor sizing or overcurrent-protection recommendation is made.

Common questions

Where do I enter AC voltage and current?

Use the AC power and current section, which requires phase, RMS voltage and power factor. Do not use the DC resistance result to describe a motor, inverter or other reactive/nonlinear load.

Why do my inputs disappear when I change the known pair?

The quantity and unit of each field may change. Clearing both prevents the old number being reinterpreted as a different physical quantity.

Does the watt value specify a safe resistor rating?

No. A real component rating also depends on temperature, duty, manufacturer limits and other conditions. The output is the calculated power at the supplied operating point.

Source

Aaron Lee and Chad Flinn, Basic Motor Control, “Ohm's Law and Watt's Law”, BCcampus, 2020. Source, formulas and unit meanings checked November 9, 2026. The book states the two foundational relationships; the six modes here are algebraic rearrangements. This source is electrical education, not a provincial code approval.

Natural Resources Canada, Variable Frequency Drives reference guide, “Power in AC Circuits”. The guide supplies single- and three-phase kVA, kW and power-factor relationships. This implementation uses √3 instead of the guide's rounded 1.73 three-phase factor. Checked November 9, 2026; no equipment-selection rules are imported.

BC Trades Training, Construction Electrician Apprenticeship Program, Level 3, D-5: Electric Motors, version 1 (2016), lesson on three-phase systems, identifies line-to-line voltage as √3 times line-to-neutral voltage for the illustrated wye connection. This is used only for the balanced sinusoidal wye conversion, not as a jurisdictional installation rule.

Report a problem or suggest a source. Calculation values are not included automatically.