Electricity Calculator

V = I · R and P = V · I

V

Voltage across the component (V)

I

Current through it (A)

R

Resistance (Ω)

P

Power it dissipates (W)

The first identity is Ohm's law and ties together voltage, current and resistance. The second brings power in. Substituting one into the other and rearranging gives the six combinations in the table on this page — each one solves for the two quantities you did not supply.

1. A 230 V circuit drawing 10 A

   1. Write down what is known: V = 230 V, I = 10 A.
   2. Resistance from Ohm's law: R = V / I = 230 / 10 = 23 Ω.
   3. Power from the watt identity: P = V · I = 230 × 10 = 2300 W.
   4. Check with the other form of the same identity: P = I² · R = 100 × 23 = 2300 W. ✓

   2300 W is a high-load domestic appliance — an oven or a kettle — which is why it draws 10 A rather than the 0.5 A of a laptop charger.

2. 12 V across a 100 Ω resistor

   1. Known: V = 12 V, R = 100 Ω.
   2. Current from Ohm's law: I = V / R = 12 / 100 = 0.12 A (120 mA).
   3. Power: P = V · I = 12 × 0.12 = 1.44 W.
   4. Check with the resistance form: P = V² / R = 144 / 100 = 1.44 W. ✓

   A quarter-watt resistor would survive 1.44 W only briefly — this is the calculation that decides whether you need a 2 W part.

3. 100 W into 4 Ω

   1. Known: P = 100 W, R = 4 Ω.
   2. This pair has no way in through V = I · R directly, so use the rearrangements: V = √(P · R) = √(100 × 4) = √400 = 20 V.
   3. Then I = √(P / R) = √(100 / 4) = √25 = 5 A.
   4. Check: P = V · I = 20 × 5 = 100 W. ✓

   Speaker and amplifier ratings are quoted this way. Note that both derived values come from a square root, so doubling the power does not double the voltage.

4. 5 V at 20 mA

   1. Known: V = 5 V, I = 0.02 A.
   2. Enter 0.02 with the unit set to A, or switch the unit to mA and enter 20.
   3. R = V / I = 5 / 0.02 = 250 Ω.
   4. P = V · I = 5 × 0.02 = 0.1 W.

   Small-signal work: the numbers are far from the amp range, which is what the mA unit option is for.

This calculator assumes a steady direct current and a load that actually obeys Ohm's law — a resistor, a heater element, a filament at its working temperature. It does not cover alternating current, where voltage and current can drift out of phase, nor reactive loads such as motors, transformers and most mains power supplies, whose power factor makes P = V · I an overestimate of real power. The four values returned are magnitudes only: the page does not track polarity, so read the signs off your circuit diagram rather than from here. Resistance is treated as constant, which fails for filaments and semiconductors as they heat.

How do I use the electricity calculator?

Fill in exactly two of the four boxes — voltage, current, resistance, power — and leave the rest empty. The calculator solves for the two you left blank, using the pair that has a direct rearrangement. The first screen comes pre-filled with 230 V and 10 A, so clear one of those two before typing a different pair.

How many amps does a 2300 W appliance draw at 230 V?

10 amps. Current is power divided by voltage: 2300 / 230 = 10 A. That is a 2.3 kW load on a 230 V supply, and it is the same relationship the other way round — 10 A at 230 V dissipates 2300 W.

Why does the calculator complain when I fill in three boxes?

Because three values are an over-specification: any two already determine the other two, so the third has to agree with them. When it does not, the page raises an error instead of silently overwriting what you typed. If you are seeing that, the two values you meant to use are the ones that agree — check for a unit slip, such as entering mA as A.

Which units does it accept?

Volts (V, mV, kV) for voltage, amps (A, mA) for current, ohms (Ω, kΩ, MΩ) for resistance and watts (W, kW) for power. Every box has its own unit selector, and the displayed results always come back in V, A, Ω and W. Switching a unit is also the easiest way to get more precision — 0.02 A and 20 mA are the same current, but only one of them reads cleanly at two decimal places.

Does Ohm's law work for AC and for motors?

Only partly. Ohm's law in the form V = I · R holds for resistive loads, and for AC it holds instant by instant, giving impedance rather than plain resistance. Motors, transformers and switch-mode supplies draw a current that is out of phase with the voltage, so V · I gives apparent power in volt-amperes rather than the real power in watts that you are billed for. For those loads you need the power factor as well.

What is the difference between power and energy?

Power in watts is the rate at which energy is used right now; energy in kilowatt-hours is that rate multiplied by how long it runs. A 2300 W kettle boils for three minutes and uses about 0.1 kWh, while a 10 W router left on all month uses around 7 kWh. To turn watts into a bill, use the electricity cost calculator.

• Ohm's Law and the V-I-R Relationship — The Physics Classroom
• The International System of Units (SI Brochure) — BIPM