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Complex power and sign conventions

Short answer

Complex power S = VI* = P + jQ, where P is real power (W) and Q reactive power (var). In load convention (current enters positive), positive P/Q means absorbed; in source convention (current leaves positive), positive P/Q means delivered.

Define the meaning of complex power in power system. Explain the sign conventions of power for sources and loads.

Complex power is the phasor product of voltage and the conjugate of current. Its real part is the active power and its imaginary part is the reactive power:

S=VI∗=∣V∣∣I∣∠(θv−θi)=P+jQS = VI^* = |V||I|\angle(\theta_v - \theta_i) = P + jQ
  • P=∣V∣∣I∣cos⁡ϕP = |V||I|\cos\phi (W), the average power converted to work or heat.
  • Q=∣V∣∣I∣sin⁡ϕQ = |V||I|\sin\phi (var), the power that oscillates between the source and the fields of L and C.
  • ∣S∣=P2+Q2|S| = \sqrt{P^2+Q^2} (VA), the apparent power. Equipment is rated in VA.
  • ϕ=θv−θi\phi = \theta_v - \theta_i is the power-factor angle. Q is positive for a lagging (inductive) current.

Sign conventions

Load (motor) convention: the current enters the positive terminal. S=VI∗S = VI^* is then the power absorbed.

  • P > 0: the element absorbs real power (resistor, motor).
  • Q > 0: it absorbs reactive power (inductor, lagging load).
  • Q < 0: it supplies reactive power (capacitor).

Source (generator) convention: the current leaves the positive terminal. S=VI∗S = VI^* is then the power delivered.

  • P > 0: the source delivers real power (generator action). P < 0 means it absorbs P (motor action).
  • Q > 0: it delivers reactive power (over-excited machine). Q < 0 means it absorbs Q.
 Load convention          Source convention
   I -->  +                 +  --> I
   +------o                 o------+
   | Load |  S absorbed     | Src  |  S delivered
   +------o                 o------+
          -                 -
ElementPQ (load convention)
Resistor+0
Inductor0+ (absorbs)
Capacitor0− (supplies)
Induction motor++

Example: V=100∠0∘V = 100\angle0^\circ V and the current leaving the + terminal is I=5∠−30∘I = 5\angle-30^\circ A. Then S=500∠30∘=433+j250S = 500\angle30^\circ = 433 + j250 VA, so the source delivers 433 W and 250 var.

✓ Answer checked by Claude Opus 5.5 (high effort)

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