--- title: "Consequences of computing S = V*I" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:complex-power-v-conjugate-i" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/complex-power-v-conjugate-i site: BCT Engineering --- # Consequences of computing S = V*I Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2080 Chaitra, 4 marks; 2079 Chaitra, 4 marks; 2075 Bhadra, 4 marks; 2070 Magh, 6 marks (asked 4 times) - 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/complex-power-v-conjugate-i - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question What will be the consequences if AC complex power is calculated as S = V*I? Explain with necessary phasor diagrams. ## Short answer Using S = V*I leaves real power unchanged but reverses the sign of reactive power, making inductive loads appear to supply vars. This contradicts the standard convention where inductive loads absorb positive Q, causing confusion in load flow and compensation studies. ## Answer The standard definition of complex power is $S = VI^* = P + jQ$. With this convention, a lagging (inductive) load absorbs positive $Q$. Using $S = V^*I$ gives the same $P$ but reverses the sign of $Q$. #### Derivation Let $V = |V|\angle\theta_v$ and $I = |I|\angle\theta_i$, with $\phi = \theta_v - \theta_i$ (positive for a lagging current). $$ \begin{aligned} VI^* &= |V||I|\angle(\theta_v-\theta_i) = |V||I|\cos\phi + j|V||I|\sin\phi\\ V^*I &= |V||I|\angle(\theta_i-\theta_v) = |V||I|\cos\phi - j|V||I|\sin\phi \end{aligned} $$ So $V^*I = (VI^*)^* = P - jQ$. #### Phasor diagrams (inductive load, current lags) ```text S = V I* S = V* I Q>0 S V ^ / / | / phi / phi (V ref) | / /------> P | / \ |/-------> P \ S v Q<0 Inductive load: Inductive load: Q positive (absorbs) Q appears negative ``` #### Consequences of using $S = V^*I$ - The real power $P$ is unchanged, since $\cos\phi$ is even. - The sign of $Q$ is reversed: an inductor (lagging current) would appear to supply reactive power, and a capacitor to absorb it. - This contradicts the standard convention (IEEE/IEC) that inductive loads absorb VAr. Generator, motor and load data would be read wrongly. - The power factor angle and the power triangle are flipped, so lagging and leading would be confused. - In load flow and compensation studies, capacitor banks would be sized in the wrong direction. **Example:** Take $V = 100\angle0^\circ$ V and $I = 10\angle-30^\circ$ A (inductive load). - $VI^* = 1000\angle30^\circ = 866 + j500$: the load absorbs 500 var, which is correct. - $V^*I = 1000\angle-30^\circ = 866 - j500$: this wrongly shows the inductive load as a source of 500 var. $S = VI^*$ is used so that inductive Q is positive. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. The standard definition of complex power is A. S = V*I B. S = VI* C. S = VI D. S = V/I Answer: B. S = VI* = P + jQ, which makes inductive Q positive. (Key point: Standard definition) 2. If S = V*I is used, the real power P A. Changes sign B. Becomes zero C. Is unchanged D. Doubles Answer: C. cos(phi) is even, so P is the same. (Key point: Effect on P) 3. Using S = V*I A. Reverses the sign of Q, so an inductive load appears to supply VAr B. Doubles Q C. Leaves Q unchanged D. Makes Q zero Answer: A. V*I = (VI*)* = P - jQ, so Q is reversed. (Key point: Effect on Q) 4. A practical consequence in compensation studies is that A. Capacitor banks would be sized in the wrong direction B. P would be misread C. Voltage would be doubled D. Nothing changes Answer: A. Lagging and leading would be confused, so compensation sizing would be wrong. (Key point: Practical consequence) ## Related in Power System Analysis I - [Derivation and phasor diagram of complex power](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Acomplex-power-derivation-phasor) — Power System Analysis I, chapter 3 - [Generator and load complex power conventions](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Agenerator-load-complex-power-convention) — Power System Analysis I, chapter 3 - [Complex power and its significance](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Acomplex-power-significance) — Power System Analysis I, chapter 3, asked 2× - [Complex power and sign conventions](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Acomplex-power-sign-conventions) — Power System Analysis I, chapter 3 - [Complex power and why VI* rather than V*I](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Acomplex-power-vi-conjugate-why) — Power System Analysis I, chapter 3 - [Complex power, impedance and power triangles](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Acomplex-power-impedance-power-triangle) — Power System Analysis I, chapter 3 ## Related in other subjects - [DT complex exponential periodic if rational](https://www.bctengineer.com/question.md?key=semester_4%2Fsignals_and_systems%3Adt-exponential-rational-frequency) — Signals and Systems, chapter 1 - [Limiting error in calculated power](https://www.bctengineer.com/question.md?key=semester_4%2Finstrumentation_and_measurement%3Apower-error-limiting-error) — Instrumentation and Measurement, chapter 1 - [Transformer on load, phasor diagrams](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machine%3Atransformer-phasor-resistive-capacitive) — Electrical Machine, chapter 2 - [Transformer on load and no load, phasor diagrams](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines%3Atransformer-load-phasor-resistive-capacitive) — Electrical Machines, chapter 2