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Head-to-head

Ohm's Law vs Voltage Drop

Ohm's law treats a circuit as one resistor: divide the voltage by the current and you have the total resistance in the path. Voltage drop treats the same circuit as two resistors in series — the cable and the load — and asks how much of the supply the cable takes before the load sees any of it.

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The difference is not academic. A 12 V circuit drawing 10 A has 1.2 Ω somewhere in it by Ohm's law, and if the cable is short and thick that resistance is essentially all in the load. Stretch the same circuit over 30 metres of 2.5 mm² copper and the cable contributes 0.4138 Ω on its own, a third of the total, and the load gets 7.86 V instead of 12.

Cable resistance comes from resistivity, length and cross-section: ρ × 2L ÷ A, with the factor of two because the current goes out and comes back. That is why low-voltage systems are so sensitive to distance — the same cable at 230 V would drop the same 4.14 V, which is 1.8% instead of 34.5%.

So the two calculations are complementary rather than competing. Ohm's law tells you what the circuit must total; voltage drop tells you how much of that total is wasted in the wiring, and therefore whether the gauge is adequate. Sizing a run by Ohm's law alone is how a 12 V installation ends up with lights that dim when the pump starts.

How They Differ

Ohm's law compared with Voltage drop
AttributeOhm's lawVoltage drop
What it computesTotal resistance in the circuitVolts lost along the conductor
InputsSupply voltage and currentCurrent, length, cross-section, resistivity
Treats the cable asPart of one lumped resistorA resistor in series with the load
FormulaR = V ÷ IV = I × ρ × 2L ÷ A
Result here1.200 Ω in total0.4138 Ω and 4.14 V
Depends on distanceNoYes, linearly
Depends on supply voltageYes, directlyNo — the drop is set by current and cable
What it is forFinding an unknown in a circuitChoosing a wire gauge

The Same Numbers, Both Ways

A 12 V circuit drawing 10 A through 30 m of 2.5 mm² copper, checked once by Ohm's law and once for voltage drop.

Supply voltage
12 V
Load current
10 A
Cable length
30 m each way
Cross-section
2.5 mm²
Copper resistivity
1.724 × 10⁻⁸ Ω·m
Worked example: Ohm's law versus Voltage drop
LineOhm's lawVoltage drop
Resistance in the path1.200 Ω total0.4138 Ω in the cable
Share of the supply100.0%34.5%
Volts reaching the load12.00 V7.86 V
Power lost in the conductor0.00 W41.38 W
Over a 10 m run instead1.200 Ω total1.379 V, 11.5%
The same cable at 230 V1.200 Ω total4.138 V, 1.8%
Depends on distanceNoYes, linearly

Ohm's law puts 1.200 Ω somewhere in a 12 V, 10 A circuit. Running 10 A through 30 m of 2.5 mm² copper makes the cable itself 0.4138 Ω — 34.5% of the supply, 4.14 V gone and 41.38 W dissipated as heat, leaving 7.86 V for the load.

When To Use Each

Ohm's law

  • You know two of voltage, current and resistance and need the third.
  • You are checking whether a stated current is plausible for a given supply.
  • The conductor is short enough that its resistance is negligible against the load.

Voltage drop

  • You are choosing a wire gauge for a run of any significant length, especially at low voltage.
  • You need to know whether a load will see enough volts to start or run properly.
  • You are checking a code requirement, most of which are expressed as a percentage drop.

Common Questions

Why does the cable count twice the length?

Because the circuit is a loop: the current travels out along one conductor and back along another, so the resistance is that of twice the run. A 30 m run has 60 m of copper in the path, which is why long runs surprise people.

What counts as acceptable voltage drop?

Common guidance is 3% for a branch circuit and 5% overall from the point of supply. Those are conventions rather than physics — the real limit is whatever the load needs, and 12 V equipment is far less tolerant than a heater.

Does voltage drop waste power?

Yes. The drop multiplied by the current is power dissipated in the conductor as heat, which in the example is 41.38 W — more than many of the loads such a circuit would run. That is why long low-voltage runs get thick cables rather than long thin ones.

Can I just use a bigger cable?

Yes, and resistance falls in proportion to cross-section, so doubling the area halves the drop. It is usually the cheapest fix, but check the terminals and the breaker rating before assuming the circuit can take a larger conductor.

Why is the drop the same at 230 V but the percentage so different?

Because the drop depends on current and cable resistance, not on supply voltage. The absolute loss is unchanged; what changes is how much of the supply it represents, which is why the same cable is fine at mains voltage and hopeless at 12 V.

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All comparisons →

Sources

Both methods sit in Science.