Power & Energy
How to calculate voltage drop in a wire
Every wire is a resistor. A current I through a wire of resistance R loses a voltage drop of V = I × R along it, and turns I² × R into heat. Long, thin wires carrying big currents lose the most, which is why low-voltage power needs thick cable and why the grid runs at high voltage.
A wire is a resistor
Copper is a good conductor, not a perfect one. A wire's resistance grows with its length and shrinks with its cross-section:
For copper, ρ is about 0.0168 Ω for each metre of 1 mm² wire. Remember the current goes out and back: a lamp 5 m away has 10 m of wire in its loop.
A worked example
A 12 V pump draws 5 A at the end of a 5 m run of 1 mm² cable, 10 m in all.
The pump gets about 11.2 V, 7% less than it should. Switch to 2.5 mm² cable and the drop falls to 0.34 V, under 3%.
Live experiment
Send 20 W down a long cable
A battery sends 20 W down 20 m of cable to Pip's lamp. The cable is a resistor, 2 Ω thin or 0.2 Ω thick. Send the same 20 W at twice the voltage and the loss falls to a quarter: 5.6 W at 12 V, 1.4 W at 24 V, 0.35 W at 48 V through the thin one.
“Lost in the wire” is lesson 4 of 6 in Watt Harbour, Power & Energy World 1. Free with an account. Lessons unlock in order, so you’ll do the 3 before it first.
Send it at high voltage
For the same power, P = V × I, twice the voltage needs half the current. The loss is I² × R, so half the current means a quarter of the loss:
| Sending 20 W through 2 Ω at | Current | Lost in the cable |
|---|---|---|
| 12 V | 1.67 A | 5.6 W |
| 24 V | 0.83 A | 1.4 W |
| 48 V | 0.42 A | 0.35 W |
That's the whole reason long-distance power lines run at hundreds of thousands of volts and step down near your home, and why bigger solar and boat systems move from 12 V to 24 or 48 V.
But the load has to match the voltage. A 12 V lamp on 24 V draws far more than it should and burns out; raising the voltage only helps when the load is built for it.
Where you'll meet it
Long USB cables that charge slowly, LED strips that dim towards the far end, and car headlights that dim when the starter cranks are all voltage drop. On a circuit board the same thing happens in narrow copper traces, which is why high-current traces are made wide.
Free lesson
Now build it yourself
Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.
“Lost in the wire” is lesson 4 of 6 in Watt Harbour, Power & Energy World 1. Free with an account. Lessons unlock in order, so you’ll do the 3 before it first.
Or start from the very first lesson, no account neededWork it out
Practice questions
Three quick questions from the lessons. Pick an answer to check it.
Show the answer
Answer: Drops to a quarter. Twice the volts for the same watts means half the amps. Loss is I² × R, and half squared is a quarter.
- Why not “Halves”? Halving the current would halve the voltage drop, but the heat goes with current squared, so it falls to a quarter.
- Why not “Stays the same”? The cable is the same, but the current isn't. Twice the volts for the same watts means half the amps.
- Why not “Doubles”? Higher voltage means less current for the same power, and less current means less heat in the cable, not more.
Show the answer
Answer: About 1.3 W. I² × R: 0.8² × 2 is about 1.3 W. The same 20 W at 12 V would need 1.7 A and waste 5.6 W.
- Why not “About 20 W”? 20 W is the power the lamp takes, not what the cable wastes.
- Why not “About 5.6 W”? 5.6 W is the loss at 12 V, where the same power needs 1.7 A. At 0.8 A it's 0.8² × 2 ≈ 1.3 W.
- Why not “Nothing”? Any wire with resistance turns I² × R into heat. A 2 Ω cable at 0.8 A wastes about 1.3 W.
Show the answer
Answer: Smaller. Less current, and the loss goes as current squared.
- Why not “Bigger”? Higher voltage means less current for the same power, and the loss goes with current squared.
- Why not “The same”? The loss depends on the current, which falls as the voltage rises.
- Why not “Zero”? Any real wire has some resistance, so the loss never quite reaches zero.
“Lost in the wire” is lesson 4 of 6 in Watt Harbour, Power & Energy World 1. Free with an account. Lessons unlock in order, so you’ll do the 3 before it first.
Questions people ask
How do you calculate voltage drop in a wire?
Work out the wire's resistance, R = ρ × L ÷ A, with ρ = 0.0168 Ω·mm² per metre for copper, L the total length out and back, and A the cross-section in mm². Then the drop is I × R. 10 m of 1 mm² copper is 0.168 Ω, so 5 A drops 0.84 V.
Why does higher voltage reduce power loss?
For the same power, P = V × I, twice the voltage needs half the current. The wire's loss is I² × R, so half the current means a quarter of the loss. That's why the grid carries power at hundreds of thousands of volts.
How much voltage drop is acceptable?
A common rule of thumb is a few percent: around 3% for important circuits. On 12 V that's under 0.4 V, which is why low-voltage systems need much thicker cable than mains for the same power.
Does wire length matter for voltage drop?
Yes, directly. Resistance grows with length, so double the length doubles the drop. Remember the current goes out and back, so a device 5 m away has 10 m of wire in its loop.
Related topics
- Ohm's law explainedV = I × R: set any two of voltage, current and resistance and the third follows.
- Electrical power explained: P = V × IP = V × I: how fast energy turns into heat, and why a resistor has a power rating.
- PCB trace width for currentCount the squares: how a trace's width sets its ohms, its voltage drop and its heat, and how wide to draw it.
- What is resistance?The squeeze on the flow, in ohms: how resistors set the current and protect an LED.