⚡ Line Losses & Voltage Drop

Power delivery begins with a simple fact that hides in plain sight: if you hold power constant, raising voltage lowers current. Lower current matters because the heat in the wire goes as current squared. That squared term is why power grids use high voltage, why USB-C raised its voltage for high-power charging, and why a robot hand's wrist cable can be a thermal design problem rather than a wiring detail.

By the end, you can

1. Compute current from load power and bus voltage using $I = P/V$
2. Estimate cable voltage drop from loop resistance and current
3. Explain why copper loss follows $I^2R$ and why doubling voltage cuts wire loss by roughly four
4. Use voltage-drop percentage to decide whether a conductor, run length, or bus voltage is plausible

Intuition first

Think of power as the useful work arriving at the load. Current is how much charge has to march through the cable each second to deliver that work. If the voltage is low, each coulomb carries only a little energy, so many coulombs must move. If the voltage is high, each coulomb carries more energy, so fewer coulombs do the same job.

The cable resists that march. It wastes energy as heat according to

$$P_\text{loss} = I^2 R_\text{loop}$$

The square is the punchline. If you double the bus voltage for the same load power, current halves. Half the current squared is one quarter the heat. You did not change the copper. You changed how hard the copper had to work.

The loop resistance is the resistance of the full outbound-and-return path, not just the one-way run. A two-wire DC cable that is 10 m one way has about 20 m of copper in the current loop.

Drive it

Set a load, bus voltage, cable length, wire gauge and temperature. Watch current, drop, heat and efficiency move together. The default starts near a low-voltage robot or battery system: 1200 W at 48 V over a 12 m run.

Try three moves:

1. Raise the bus from 48 V to 96 V. Current roughly halves and copper heat roughly quarters.
2. Drop the wire from 12 AWG to 18 AWG. The resistance climbs and the cable becomes the load.
3. Push the conductor temperature up. Hot copper has higher resistance, so a marginal cable gets worse after it warms.

?A 1000 W load runs from a 50 V bus. What current flows before considering cable loss?

• A 20 A, because current is power divided by voltage
• B 50 A, because bus voltage sets the current directly
• C 1000 A, because the load is 1000 W
• D 0.05 A, because voltage is power divided by current

?At fixed load power and cable resistance, what happens to cable heat if you double the bus voltage?

• A It stays the same
• B It halves
• C It becomes one quarter
• D It doubles

↘ Why AC distribution uses the same current-squared logic

The same loss term appears in AC distribution. RMS current through a conductor still heats it by $I_\text{RMS}^2R$. Single-phase and three-phase systems differ in how much power they move for a given line voltage and current, and power factor changes the current needed for a given real power, but the conductor loss still follows the square of the current in each conductor. The details change; the design instinct does not.

Key takeaways

• ▸Wire loss is $I^2R$, so current is the expensive quantity.
• ▸For fixed load power, raising bus voltage lowers current and cuts copper heat quickly.
• ▸Use the full loop resistance: outbound plus return conductor.
• ▸Voltage-drop percentage is a practical accept/reject check for cable sizing.
• ▸Temperature matters because hot copper has higher resistance.

Practice 1 core

A 500 W load runs from 24 V through a cable whose loop resistance is 0.08 Ω. Estimate the current, cable loss, and voltage drop.

↘ Show worked solution

Current is $I = P/V = 500/24 \approx 20.8$ A.

Cable heat is $P_\text{loss} = I^2R \approx 20.8^2 \times 0.08 \approx 34.7$ W.

Voltage drop is $\Delta V = IR \approx 20.8 \times 0.08 \approx 1.7$ V, about $1.7/24 \approx 7$ percent. That is high for many electronics loads, so the fix is larger copper, a shorter run, parallel conductors, or a higher bus voltage.