Practical Electricity: Circuits, Power, and Machines
The core relationships
Nearly every computational item reduces to three tools: Ohm's law (V = IR), the power family (P = VI = I²R = V²/R), and Kirchhoff's laws. Energy is power × time, so a 1,500 W load for 4 hours consumes 6 kWh. In series circuits the current is common and resistances add; in parallel circuits the voltage is common and conductances (reciprocals) add. For AC, distinguish RMS from peak: V_peak = √2 × V_rms, which is why 230 V RMS has a peak near 325 V. Impedance Z replaces R, and the phase angle gives power factor: single-phase P = V·I·cos θ, three-phase P = √3·V_L·I_L·cos θ, with apparent power S in VA or kVA.
Traps that show up in questions
Mixing peak and RMS values; reporting kW where the item asks for kVA; adding parallel resistances directly; omitting √3 on a three-phase item; and ignoring power factor when converting kW to amperes. Watch unit prefixes (W vs kW, mm² vs m²) and remember conductor resistance rises with temperature, so a hot conductor has more resistance and more voltage drop.
Worked example
A 10 kW, 230 V single-phase load at 0.8 power factor: I = P / (V·cos θ) = 10,000 / (230 × 0.8) = 54.3 A. If the same 10 kW were three-phase at 400 V, I = 10,000 / (√3 × 400 × 0.8) = 18.0 A - the √3 factor is the usual discriminator between the two answers offered.
Rule of thumb
kVA = kW ÷ power factor. Size feeders and switches on kVA/amperes, but size energy bills and heating on kW.