Power factor hides amps
Cut pf from 1.0 to 0.5 and the current doubles for the same watts. Same load, twice the copper heating (\(I^2R\) losses).
Formulas use RMS values. For three-phase (balanced), \(k=\sqrt{3}\) with line-to-line \(V\) and line \(I\).
In AC circuits, voltage and current can be out of phase by an angle \(\varphi\). This creates three related kinds of power:
These form the power triangle: \(\; S^2 = P^2 + Q^2 \;\) and the power factor \(\mathrm{pf} = P/S = \cos\varphi\).
For line-to-line voltage \(V\) and line current \(I\):
Tip: Use RMS quantities. If you only know power factor, \(\varphi = \arccos(\mathrm{pf})\).
Inductive loads are lagging (current lags voltage, positive \(Q\)); capacitive loads are leading (current leads voltage, negative \(Q\)).
\(S\) is in VA, \(P\) in W, and \(Q\) in VAR—symbols help separate total apparent power, useful real power, and energy-swapping reactive power.
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Given \(V = 230\,\text{V}\), \(I = 3.5\,\text{A}\), \(\mathrm{pf} = 0.8\) (lagging):
Checkpoint: \( S^2 \approx P^2 + Q^2 \) should hold (allow rounding).
Cut pf from 1.0 to 0.5 and the current doubles for the same watts. Same load, twice the copper heating (\(I^2R\) losses).
Power factor is just \(\cos\varphi\). A pf of 0.8 means voltage and current are ~37° apart; 0.95 is only ~18°.
Inductive motors “borrow” reactive power; capacitor banks “lend” it back. Utilities install massive caps to pull pf toward 1 and free grid capacity.
Balanced three‑phase magic: line voltage = phase voltage × √3, and S = √3 × VLL × IL. That constant drops out of the geometry of 120° phases.
Inductive loads make positive Q (lagging). Capacitive correction pushes Q negative, sometimes even past zero—data centers intentionally run slightly leading to offset cable inductance.