Contact resistance turns amps into heat and a small voltage drop at the connector. Calculate connector heat and voltage drop (I²R): the tool reports loss watts from I²R; multiply amps by R (mΩ ÷ 1000) for drop volts on the same preset.
Benefits
- Heat (W) = amps² × R—same I²R the tool outputs as connector loss.
- Voltage drop (V) = amps × R from the preset mΩ—explains pack sag at the pins.
- Compare XT30 vs XT60 at identical amps—lower mΩ cuts both heat and drop.
How it works
- Enter current draw (A) and pick XT30, XT60, or XT90 resistance preset.
- Read connector loss (W)—that is I²R heat at the contact pair.
- Voltage drop (V) ≈ amps × (mΩ ÷ 1000)—add session minutes for Wh waste.
FAQ
How do I calculate connector heat and voltage drop (I²R)?
Convert mΩ to Ω (÷ 1000). Heat watts = I² × R. Voltage drop volts = I × R. The calculator outputs I²R as connector loss (W). Example: 12 A, XT60 (0.8 mΩ = 0.0008 Ω): heat ≈ 0.12 W, drop ≈ 0.0096 V (~10 mV).
Why is heat I²R but drop only I×R?
Both use the same contact resistance. Power dissipated as heat is I²R; the voltage lost across the connector is I×R (Ohm's law). Doubling amps doubles drop but quadruples heat—peak launches heat pins faster than they sag voltage.
Example: 18 A on XT30 vs XT60?
XT30 ~1.5 mΩ: heat ≈ 0.49 W, drop ≈ 27 mV. XT60 ~0.8 mΩ: heat ≈ 0.26 W, drop ≈ 14 mV. Same amps through a better pair halves heat and roughly halves drop—loose pins raise effective mΩ beyond presets.
Does connector drop affect range?
Small per-pin drops add to pack sag under peak amps—the controller sees slightly lower voltage. Pair I²R heat with peak-amps and range tools so burst current, connector loss, and SOC sag are modeled on the same commute.
Technical specifications
- R (Ω) = connector mΩ ÷ 1000.
- Heat / loss (W) = amps² × R (I²R).
- Voltage drop (V) = amps × R.
- 12 A, XT60 (0.8 mΩ): ~0.12 W heat, ~10 mV drop.
- Related: e-scooter-connector-power-loss-calculator, escooter-peak-amps.