--- title: "Generator and load complex power conventions" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:generator-load-complex-power-convention" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/generator-load-complex-power-convention site: BCT Engineering --- # Generator and load complex power conventions Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2068 Bhadra, 2+3 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/generator-load-complex-power-convention - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question How are generator and load complex powers expressed? What will happen if the convention of taking the conjugate of current is not followed? Explain with necessary derivation. ## Short answer Complex power is S=VI*; for a generator current leaves the terminal so S is delivered, for a load current enters so S is absorbed. If the conjugate is not used, V*I gives P but reverses the sign of Q, while VI depends on the chosen reference angle and is physically meaningless. ## Answer #### 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. In the generator convention, what does S = VI* represent? A. Power absorbed with Q < 0 B. Power delivered by the source; an over-excited generator supplying a lagging load has P > 0 and Q > 0 C. Only the apparent power D. Zero net power Answer: B. Current leaves the + terminal. (Key point: Generator convention) 2. In the load convention, what does S = VI* represent? A. Power absorbed; inductive load has Q > 0 B. Power delivered C. Power lost in the source D. Always Q < 0 Answer: A. Current enters the + terminal. (Key point: Load convention) 3. What happens if S = V*I is used? A. P changes sign B. Both are correct C. S becomes real D. P is right but Q has the wrong sign, so an inductive load shows negative Q Answer: D. V*I = P - jQ. (Key point: Without conjugate: V*I) 4. Why is S = VI unacceptable? A. It has no magnitude B. It always equals zero C. Its angle is the sum theta_v + theta_i, so P and Q change with the reference phasor D. Its angle is the power factor angle Answer: C. A reference shift of alpha adds 2 alpha. 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