Electrical Power Calculator (P, V, I, R)
Enter any two known electrical quantities and the calculator instantly solves the other two using Ohm's Law (V = I x R) and Watt's Law (P = V x I). Works for DC circuits and single-phase AC circuits with a power factor. Choose your preferred unit prefix (micro, milli, base, kilo, mega) for each quantity independently.
How to use this calculator
Select "Solve for" to choose the unknown quantity: Power, Voltage, Current, or Resistance. Then enter any two of the remaining three known values. The calculator immediately applies the correct formula from Ohm's Law (V = I x R) and Watt's Law (P = V x I) and fills in all four quantities. For AC circuits, switch the circuit type to "AC (single-phase)" and enter your power factor (a number between 0 and 1). Enable "Calculate energy cost" to add a daily, monthly, and annual cost estimate based on your electricity rate.
Ohm's Law and Watt's Law explained
Georg Ohm showed in 1827 that the voltage across a resistor equals the current through it multiplied by its resistance: V = I x R. James Watt's Law links electrical power to those same quantities: P = V x I. Combining the two gives the complete set of twelve formulas shown in the table below: three ways to find each of P, V, I, and R from any other pair. Because the relationships are purely algebraic, every one of the twelve paths gives the same answer as all others when applied consistently.
DC vs AC power and the power factor
In a direct current (DC) circuit, power is simply P = V x I and all the energy delivered by the source is consumed as heat or mechanical work. In alternating current (AC) circuits, reactive components such as capacitors and inductors store and release energy each cycle. The power factor (pf) is the cosine of the phase angle between voltage and current, ranging from 0 (purely reactive, no real power consumed) to 1 (purely resistive, all apparent power is real). Real power P = S x pf, where S is the apparent power in volt-amperes (VA). The reactive power Q = sqrt(S^2 - P^2) represents energy cycling back and forth but doing no useful work. A pf below about 0.85 is generally considered poor and attracts penalty charges from many utilities.
Common reference values
A 60-watt incandescent bulb at 120 V draws 0.5 A and has a resistance of 240 ohms. A 1500 W electric kettle at 230 V draws about 6.5 A. A 100 W LED equivalent at 120 V uses less than 15 W, drawing only 0.125 A. Household circuits in North America are typically 15 A at 120 V (maximum 1800 W) or 20 A at 120 V (maximum 2400 W). European circuits are usually 16 A at 230 V (maximum 3680 W). These reference points help you quickly sanity-check a calculated result.
All twelve Ohm's Law and Watt's Law formulas
| Known values | Find Power (P) | Find Voltage (V) | Find Current (I) | Find Resistance (R) |
|---|---|---|---|---|
| V and I | P = V x I | - | - | R = V / I |
| V and R | P = V^2 / R | - | I = V / R | - |
| I and R | P = I^2 x R | V = I x R | - | - |
| P and V | - | - | I = P / V | R = V^2 / P |
| P and I | - | V = P / I | - | R = P / I^2 |
| P and R | - | V = sqrt(P x R) | I = sqrt(P / R) | - |
Given any two of P, V, I, R, use the matching row to find the others.
Frequently asked questions
Which two values do I need to enter?
You need any two of the four: Power (P), Voltage (V), Current (I), or Resistance (R). The calculator can derive the other two from any pair. However, not every pair is independent in all situations: for example, if you know P and the ratio of V to I already encodes R, you are effectively giving the same constraint twice. In practice, the most common pairs are V and I (you measured both on a circuit), V and R (you know the supply voltage and the load spec), and P and V (you know the rated power and the supply).
What is the power factor and when does it matter?
Power factor (pf) is the ratio of real power (watts) to apparent power (volt-amperes) in an AC circuit. It matters whenever inductors or capacitors are present, including motors, transformers, fluorescent ballasts, and many switch-mode power supplies. A pf of 1.0 means all drawn power is converted to useful work; a pf of 0.7 means 30% of the apparent current drawn from the supply is reactive and does no work. For DC circuits, pf = 1 always, so there is no distinction.
How do I calculate electricity cost from power?
Multiply the power in kilowatts by the hours of use to get kilowatt-hours (kWh), then multiply by your electricity rate in dollars per kWh. For example, a 100 W (0.1 kW) appliance running 8 hours uses 0.8 kWh. At $0.15/kWh that is $0.12 per day, $3.60 per month, and $43.80 per year. The energy cost section of this calculator does all that arithmetic automatically once you enable it and enter your hours and rate.
What is the difference between watts and volt-amperes?
Watts (W) measure real power: the energy per second actually consumed or converted to heat, light, or mechanical work. Volt-amperes (VA) measure apparent power: the product of the RMS voltage and RMS current. For DC circuits and AC circuits with purely resistive loads, they are equal. For AC circuits with reactive loads, VA is always greater than or equal to W. The ratio W/VA is the power factor.
Can this calculator handle microelectronics (milliwatts, microamps)?
Yes. Use the unit prefix selectors next to each input to choose the right scale: microamps (µA), milliamps (mA), milliwatts (mW), kilohms (kΩ), and so on. The calculator converts everything to base SI units internally, so a 10 mA signal through a 4.7 kΩ resistor correctly produces 470 mW of power and 47 V across the resistor.
Why does a higher power factor matter for my electricity bill?
Most residential meters measure only real power (kWh), so households rarely face a direct power-factor surcharge. However, industrial and commercial consumers are often charged for reactive power or penalized when pf falls below a threshold (commonly 0.85 or 0.9). Even at home, a low power factor means the wiring and breakers carry more current than the watts alone would imply, increasing resistive losses in the conductors. Power-factor-corrected power supplies and variable-frequency drives are the common engineering responses.