Two equations do most of the work in everyday electrical work: V = I × R and P = V × I. They are short enough to memorise, which is exactly why they are so often used in situations where they do not apply. The useful thing is not the memorised form but knowing what each symbol is a measure of, so you can tell when a question is hiding a fourth quantity you have not accounted for.
What Ohm's law actually says
Ohm's law states that the voltage across a conductor is proportional to the current through it, and the constant of proportionality is the resistance:
V = I × R · I = V ÷ R · R = V ÷ I
Three forms, one relationship, and you pick whichever solves for the unknown. The ohm's law calculator returns all three rearrangements at once, which is a cheap way to stop picking the wrong one under time pressure.
The important caveat is that R is a property of the component and the geometry, not of the voltage. A 100 Ω resistor is 100 Ω whether you put 1 V or 1,000 V across it — until it heats up, cooks itself, or changes value with temperature. That is why the linear relation holds so well for metals and resistors, and why it fails completely for a diode, an LED, a discharge lamp or a filament lamp.
Where V = IR actually comes from
Ohm's law is not a law of nature; it is a consequence of what happens to free electrons in a metal when you apply a field. The derivation is short and worth doing once.
Apply an electric field E (volts per metre) across a conductor. The field pushes the free electrons, which do not accelerate freely because they collide constantly with the lattice of stationary atoms. They reach a steady average drift velocity vd that is proportional to the field:
vd = μ × E
where μ (mobility) is a number describing how freely the carriers slide in this particular material. The current density — current per unit area — is then the number of carriers per cubic metre (n), times the charge each one carries (q), times the area (A), times the drift velocity:
J = n × q × A × vd = n × q × A × μ × E
Write the field as E = V ÷ L, and the current as I = J × A. The area cancels, and you are left with:
I = n × q × μ × A × V ÷ L
Group everything except the voltage into a single constant and call it 1 ÷ R. That gives I = V ÷ R, which is Ohm's law. The reciprocal constant ρ = 1 ÷ (n × q × μ) is the material's resistivity, and it depends only on the substance and its temperature — not on the length or the thickness.
So the resistance of a uniform conductor is:
R = ρ × L ÷ A
Read that as a physical statement: resistance grows with length and shrinks with cross-section. A 100 m cable has ten times the resistance of a 10 m one of the same type; a cable with twice the cross-section has half the resistance. This is the same equation the voltage drop guide uses, and it is worth being completely comfortable with it, because almost every cable sizing decision follows from it directly.
Why current is inversely proportional to resistance
The algebra says I = V ÷ R. The intuition is worth having, because "halve the resistance, double the current" is a statement about what is happening to the field, not a rule to memorise.
The field-length view. For a fixed supply voltage, adding resistance means adding material. The same voltage now has to be dropped across more of the chain, so the electric field per metre — and therefore the drift velocity of the electrons — falls. Fewer electrons per second cross any given cross-section. That is the whole mechanism.
The series-chain view. Put two identical resistors in series and the supply voltage splits between them: each gets half, so each carries half the current. Add a third identical resistor and each now gets a third of the voltage, and the current drops to a third of the single-resistor value. Three resistors in series is three times the resistance and one third of the current, exactly as I = V ÷ R requires. This is the easiest way to feel the relationship, and it is why cable resistance calculations add lengths together.
The area view. Widen the conductor and you have placed more lanes in parallel for the electrons to travel down. A cable with four times the cross-section carries roughly four times the current at the same voltage and length. Nothing about the electron speed has to change — there are simply four times as many of them available per second.
One consequence deserves emphasis: resistance and current are coupled, so you cannot change one without changing the other. Doubling the supply voltage doubles the current and quadruples the power. Halving the resistance doubles the current and the power. Keeping that coupling in mind is what keeps the three formulas consistent with each other.
The three power formulas and when each applies
Power is the rate of energy flow, in watts. Three forms come from Ohm's law, and each is useful precisely when it omits the quantity you do not know:
P = V × I · P = I² × R · P = V² ÷ R
- P = V × I — use when you have volts and amps. This is the field-meter case: a clamp meter on the conductor and a voltmeter across the terminals give you everything needed directly. It is also the only one of the three that is purely electrical, with no resistance anywhere, so it is the safest when the resistance is unknown or uncertain.
- P = I² × R — use when you know the current and the load resistance. This is the motor and heater case, where a nameplate gives the running current and the element resistance is known or measurable. Note the square on the current: losses in a conductor are quadratic in current, which is the entire reason oversized conductors exist.
- P = V² ÷ R — use when you have the supply voltage and the load resistance. This is the fixed-supply case, and it is the most convenient form for comparing loads: halve the resistance and the power quadruples, because the voltage stays put and the current rises to match.
Two warnings about switching between them. First, they are only interchangeable if the circuit is resistive and in steady state. On an AC circuit with inductive or capacitive loads, the voltage and the current are not in step, so the simple product V × I is not the real power. Second, in any of the three forms, all quantities must refer to the same operating point — a nameplate resistance measured cold is not the resistance of a filament once it is hot.
The power calculator and the watt calculator handle the real-versus-apparent distinction, which is where most of the confusion actually lives.
Series and parallel resistance
Real circuits are rarely a single resistor, so the two combining rules get used constantly.
Series resistors add:
Rs = R₁ + R₂ + R₃ + …
The same current flows through all of them, and each drops a share of the voltage. This is why the chain intuition above works so well.
Parallel resistors combine as reciprocals:
1 ÷ Rp = 1 ÷ R₁ + 1 ÷ R₂ + … or, for two: Rp = (R₁ × R₂) ÷ (R₁ + R₂)
Adding a resistor in parallel always reduces the total resistance below the smallest branch, because every path offers a new way for current to flow. Parallel combinations always share the same voltage; series combinations always share the same current. Those two facts are usually enough to work out an unfamiliar network by inspection.
Apparent power versus real power
Everything above assumes a resistive load, where voltage and current rise and fall together. Real loads rarely do. The quantity a supply has to deliver is the apparent power:
S = V × I, in voltamperes (VA)
The quantity that does useful work is the real power P, measured in watts. For a purely resistive load the two are the same number — a 1,800 W resistive heater at 120 V draws 15 A and 1,800 VA, so its power factor is 1. The power factor guide develops where they part company and what the triangle between them means.
The reason to know the difference now: it determines what size of cable, breaker and transformer you need. A load drawing 30 kW at a power factor of 0.7 needs 42.9 kVA of supply capacity, and the cable has to carry the larger current that comes with it. Sizing a cable on real power alone is a common and expensive mistake.
kW versus kWh: the mistake worth naming
This is the single most common unit error in the whole subject, and it is not really an electrical error — it is a rate-versus-quantity error.
kilowatt (kW) = power = energy per second = a rate. A 2 kW heater is delivering 2,000 joules every second, and it will do that for as long as it is plugged in. The number says nothing about duration or cost.
kilowatt-hour (kWh) = energy = a quantity. It is the integration of power over time, and it is what you are actually billed for.
Energy (kWh) = Power (kW) × Time (h)
Two heaters of very different power can cost completely different amounts over the same period, and two identical heaters can cost completely different amounts over different periods. Neither fact is visible in the wattage alone, which is exactly why the two must never be substituted for one another. A 3,000 W heater running for 10 minutes uses 0.5 kWh, not 3,000 kWh and not 0.5 kW.
Worked example 1: a 120 V circuit with an 8 Ω load
A resistive load of 8 Ω connected to a 120 V supply. Find the current and the power.
Step 1 — current from Ohm's law. I = V ÷ R = 120 ÷ 8 = 15 A.
Step 2 — power, choosing the form that avoids the unknown. We have V and I, so use P = V × I = 120 × 15 = 1,800 W.
Step 3 — check with a different form. P = I² × R = 225 × 8 = 1,800 W. Identical, so the arithmetic is consistent. This check takes ten seconds and catches most errors, because the two routes to the answer share almost nothing except the load resistance.
Step 4 — what the supply has to deliver. The load is resistive, so current and voltage are in step, apparent power S = 120 × 15 = 1,800 VA. Power factor = P ÷ S = 1,800 ÷ 1,800 = 1.0.
Step 5 — turn it into an energy figure. At 1.8 kW, running continuously for 30 days:
Time = 30 days × 24 h = 720 h
Energy = 1.8 kW × 720 h = 1,296 kWh
That is 1.296 MWh, and at a typical industrial tariff of 0.15 currency per kWh it comes to roughly 194.40 in cost. The heater dissipates 1,800 W continuously and its filament gets that hot, which is the second part of the story — see the voltage drop and wire sizing guide, which covers the other limit that a correctly-rated circuit has to respect.
Worked example 2: three resistors, series and parallel
Two resistors, 4 Ω and 6 Ω, are connected in parallel. That combination is in series with a 12 Ω resistor. The whole network sits on a 24 V supply. Find the total resistance, the total current and the total power.
Step 1 — the parallel pair. Rp = (R₁ × R₂) ÷ (R₁ + R₂) = (4 × 6) ÷ (4 + 6) = 24 ÷ 10 = 2.4 Ω.
Step 2 — add the series resistor. Rtotal = 2.4 + 12 = 14.4 Ω. Note that this is larger than the parallel part but smaller than the 12 Ω branch — exactly the behaviour expected, since the pair is in the conducting path of the 12 Ω resistor.
Step 3 — total current. I = V ÷ R = 24 ÷ 14.4 = 1.667 A. All of it flows through the 12 Ω resistor.
Step 4 — total power. P = V × I = 24 × 1.667 = 40.0 W.
Step 5 — verify the split, because this is where mistakes hide. Voltage across the 12 Ω resistor: 1.667 × 12 = 20.0 V. So the parallel pair has 24 − 20 = 4.0 V across it. Branch currents: 4.0 ÷ 4 = 1.000 A and 4.0 ÷ 6 = 0.667 A, and those add to 1.667 A. ✓
Step 6 — verify the power distribution. The 12 Ω resistor dissipates 1.667² × 12 = 33.3 W. The 4 Ω branch dissipates 4.0² ÷ 4 = 4.0 W. The 6 Ω branch dissipates 4.0² ÷ 6 = 2.67 W. Total: 33.3 + 4.0 + 2.67 = 40.0 W. ✓ The physically sensible check is that the largest resistor dissipates the most power, because it carries the most current and has the most resistance in series with it.
Step 7 — the current is not the same in every branch. This is the single fact to take away. In series everything carries 1.667 A; in parallel the voltage is common and the current divides. Mixing the two up is the usual cause of a network calculation that produces a number nobody can reconcile with a meter.
Common mistakes
Substituting kW where kWh belongs. Already covered, and it is the most expensive one. A 2 kW heater for 8 hours is 16 kWh.
Using V × I for a low-power-factor load. It gives apparent power in VA, not real power in watts. On a motor at 0.75 power factor, the same calculation overstates the useful power by a third.
Ignoring temperature. Copper's resistance rises about 0.39% per degree Celsius above 20 °C, so a conductor that has been carrying full load for an hour has roughly 20% more resistance than its catalogue figure. Small effect on a healthy design, large effect on a marginal one.
Forgetting the series rule and averaging instead. Two resistors in series do not take the average resistance; they take the sum. Averaging works for two identical resistors in parallel, and for nothing else in particular.
Frequently asked questions
What is the difference between kW and kWh?
A kilowatt is a rate of energy flow: 1 kW means 1,000 joules of energy every second. A kilowatt-hour is an amount of energy, and 1 kWh is what a 1 kW load consumes in one hour. Power tells you how fast equipment draws energy; energy tells you how much it drew over a period. Bills are priced in kWh, so multiply kW by hours to get cost.
Why does adding resistance reduce the current?
Because Ohm's law ties the two together directly. Adding resistors in series makes the total resistance larger, so for an unchanged supply voltage the current falls in proportion. Doubling the resistance halves the current. Think of resistance as a restriction in a pipe: a narrower, longer pipe restricts flow more than a short, wide one.
Which power formula should I use when I do not know the resistance?
Use P = V × I when you have a voltage and a current, which is what a multimeter and a clamp meter give you directly. Use P = I² × R when you know the current and the load resistance, common in motor and heater work. Use P = V² ÷ R when you have the supply voltage and the load resistance, the usual case for a fixed load.
Does a lower resistance always mean more power?
At a fixed supply voltage, yes: P = V² ÷ R, so a smaller resistance dissipates more. But the two are linked, so a low resistance does not create energy. A short thick cable has low resistance and carries large current, and the power is dissipated in whatever that resistance actually is. Near zero resistance gives a short circuit, not useful heating.