Electrical logic, short of BS 7671

Start from resistance, then voltage, then power. Stop before protection and cable selection.

Logic

Ohm's law is V = IR. 230 V across 46 Ω draws 5 A and dissipates 1,150 W. Two resistors add in series and combine as the product over the sum in parallel: 10 Ω and 15 Ω are 25 Ω and 6 Ω. An unloaded divider is Vin × R2 / (R1+R2). 12 V across 1 kΩ then 2 kΩ is 8 V, until something is connected. Use Ohm's law, resistors and the divider.

The copper voltage-drop estimate is resistivity times go-and-return length. 16 A, 25 m, 2.5 mm² is about 5.5 V. It ignores reactance and installation method. Use voltage drop.

Single-phase power is VI times power factor. 230 V, 10 A, pf 0.9 is 2.07 kW. Three-phase power inserts √3: 400 V, 32 A, pf 0.85 is 18.85 kW. Capacitor correction is P times the change in tan φ. That 18.85 kW from 0.85 to 0.95 needs about 5.49 kVAr. Use single-phase power, three-phase power and power factor.

The electrical chain

Start with the load you can point at. A single-phase nameplate at 230 V, 10 A and power factor 0.9 is 2.07 kW. A balanced three-phase load at 400 V, 32 A and 0.85 is 18.85 kW. The second uses √3 once. A divider and a resistor pair are only the voltages and resistances you typed, and only while nothing else draws current.

The voltage-drop estimate is deliberately the 20°C copper screen. Quoting it as a BS 7671 drop is the misread this guide is here to stop.

Correction from 0.85 to 0.95 on that 18.85 kW load is about 5.49 kVAr. That is a difference of tangents, not a capacitor catalogue.

Last checked 3 October 2026. The logic is the textbook relationship named above. It is not a code check. See standards these tools do not replace.