Generator and load complex power conventions

Generator and load complex power

Complex power is always written as $S = VI^* = P + jQ$. What it means depends on the assumed current direction.

• Generator (source) convention: the current leaves the + terminal. $S_G = VI^*$ is the power delivered. For an over-excited generator supplying a lagging load, $P_G > 0$ and $Q_G > 0$.
• Load convention: the current enters the + terminal. $S_L = VI^*$ is the power absorbed. For an inductive load $Z = R + jX$: $S_L = |I|^2Z = P_L + jQ_L$ with $Q_L > 0$.

At any bus, the net injected power is $S_i = S_{Gi} - S_{Li} = V_iI_i^*$. This is the form used in load-flow studies.

If the conjugate is not taken

Let $V = |V|\angle\theta_v$, $I = |I|\angle\theta_i$ and $\phi = \theta_v - \theta_i$.

Case 1: $S = V^*I$

$$V^*I = |V||I|\angle(\theta_i - \theta_v) = |V||I|\cos\phi - j|V||I|\sin\phi = P - jQ$$

P is correct, but the sign of Q is reversed. An inductive (lagging) load would show negative Q, as if it were supplying vars like a capacitor. Generators and loads would then disagree with the standard convention used in meters, capability charts and load-flow programs. The power triangle would also become the mirror image of the impedance triangle ($S = |I|^2(R - jX)$).

Case 2: $S = VI$ (no conjugate at all)

$$VI = |V||I|\angle(\theta_v + \theta_i)$$

The angle now depends on the sum of the phase angles. If the reference is changed by $\alpha$, then $VI$ becomes $|V||I|\angle(\theta_v + \theta_i + 2\alpha)$. So P and Q would change just because a different phasor was chosen as reference, which is physically meaningless. Taking the conjugate makes S depend only on the difference $\theta_v - \theta_i$, which is independent of the reference.

Example: $V = 100\angle30^\circ$ V and $I = 10\angle0^\circ$ A (lagging by 30°).

• $VI^* = 1000\angle30^\circ = 866 + j500$ VA. Correct: an inductive load absorbs 500 var.
• $V^*I = 866 - j500$ VA. Wrong sign of Q.
• $VI = 1000\angle30^\circ$ here, but with V as reference ($V = 100\angle0^\circ$, $I = 10\angle-30^\circ$) it becomes $1000\angle-30^\circ$. The result changes with the reference.

So the conjugate of current must be taken to get a unique, reference-independent S with the standard sign of Q.