--- title: "Receiving end complex power and power circle" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:receiving-power-circle" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/receiving-power-circle site: BCT Engineering --- # Receiving end complex power and power circle Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2076 Bhadra, 6+2 marks - Video: [Receiving End Power Circle Diagram of Transmission line| Numerical Solved |Power System| #electrical](https://www.youtube.com/watch?v=AYmQjfNs6Eg) — TechnoSmart - 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/receiving-power-circle - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question Starting from a suitable point, derive the expression for receiving end complex power in terms of voltages and line parameters. Construct the power circle based on the expression. ## Short answer From VS = A VR + B IR, taking VR=|VR|∠0°, VS=|VS|∠δ, yields receiving complex power SR = (|VS||VR|/|B|)∠(β−δ) − (|A||VR|²/|B|)∠(β−α). This is a circle in the P-Q plane: centre at −(|A||VR|²/|B|)∠(β−α), radius |VS||VR|/|B|. ## Answer #### Derivation of receiving-end complex power For any line, $V_S = AV_R + BI_R$. Take $$ V_R = |V_R|\angle0^\circ,\ V_S = |V_S|\angle\delta,\ A = |A|\angle\alpha,\ B = |B|\angle\beta $$ Then $$ I_R = \frac{V_S - AV_R}{B} = \frac{|V_S|}{|B|}\angle(\delta-\beta) - \frac{|A||V_R|}{|B|}\angle(\alpha-\beta) $$ Receiving-end complex power (per phase; with line voltages it is three-phase): $$ \begin{aligned} S_R &= V_R I_R^* \\ &= \frac{|V_S||V_R|}{|B|}\angle(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\angle(\beta-\alpha) \end{aligned} $$ In rectangular form: $$ \begin{aligned} P_R &= \frac{|V_S||V_R|}{|B|}\cos(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha) \\ Q_R &= \frac{|V_S||V_R|}{|B|}\sin(\beta-\delta) - \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha) \end{aligned} $$ #### Construction of the receiving-end power circle Write the result as $$ \left(P_R + \frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha)\right)^2 + \left(Q_R + \frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha)\right)^2 = \left(\frac{|V_S||V_R|}{|B|}\right)^2 $$ This is a circle in the $P$–$Q$ plane with: - **Centre** $N$: $\left(-\frac{|A||V_R|^2}{|B|}\cos(\beta-\alpha),\ -\frac{|A||V_R|^2}{|B|}\sin(\beta-\alpha)\right)$, fixed for a fixed $|V_R|$. - **Radius** $NM = \frac{|V_S||V_R|}{|B|}$. Steps: 1. Draw $P$ (horizontal) and $Q$ (vertical) axes with origin $O$. 2. From $O$ draw line $ON$ of length $|A||V_R|^2/|B|$ at angle $(\beta-\alpha)$ below the negative $P$-axis; $N$ is the centre. 3. With centre $N$ draw an arc of radius $|V_S||V_R|/|B|$. 4. From $O$ draw the load line at angle $\phi_R$ above the $P$-axis (lagging); it cuts the circle at the operating point $M$. $M$ gives $P_R$ and $Q_R$. 5. Angle between $NM$ and the horizontal through $N$ is $(\beta-\delta)$, so $\delta$ can be read off. ```text Q | . M (P_R, Q_R) | . / | . / load line (phi_R) --------O-------/------------- P /| . / | . radius = |Vs||Vr|/|B| / |. N (centre) angle (beta-alpha) below -P axis ``` The maximum $P_R$ is at the point where the radius $NM$ is horizontal ($\delta = \beta$). Circles for different $|V_S|$ are concentric about $N$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. From which relation is the receiving-end complex power derived? A. I_S = C V_R + D I_R with S_R = V_S I_S B. V_S = V_R only C. V_S = A V_R + B I_R, giving I_R = (V_S - A V_R)/B and S_R = V_R I_R* D. V_R = V_S + Z I_S with S_R = V_S I_S* Answer: C. Solve V_S = A V_R + B I_R for I_R and form S_R = V_R I_R*, which gives magnitudes and angles (beta - delta) and (beta - alpha). (Key point: Starting equation) 2. Where is the centre N of the receiving-end power circle? A. At the origin B. At (+|A||V_R|^2/|B| cos(beta-alpha), +|A||V_R|^2/|B| sin(beta-alpha)) C. At (-|A||V_R|^2/|B| cos(beta-alpha), -|A||V_R|^2/|B| sin(beta-alpha)), fixed for fixed |V_R| D. At (-|V_S||V_R|/|B|, 0) Answer: C. The constants terms in P_R and Q_R are subtracted, so the centre lies at the negative of them; it depends only on A, B and |V_R|. (Key point: Centre of circle) 3. What is the radius of the receiving-end power circle? A. |A||V_R|^2/|B| B. |V_S||V_R|/|B| C. |V_S|^2/|B| D. |B|/(|V_S||V_R|) Answer: B. The right side of the circle equation is (|V_S||V_R|/|B|)^2, so the radius is |V_S||V_R|/|B|. (Key point: Radius) 4. How is the operating point and maximum power obtained? A. The load line is vertical; maximum P occurs at delta = 0 B. Circles for different V_S have different centres; maximum P occurs at delta = 90 degrees C. The operating point is the circle's centre D. Load line at angle phi_R from O cuts the circle at M; maximum P_R occurs when NM is horizontal (delta = beta) Answer: D. The load line from O at phi_R meets the circle at M giving P_R, Q_R; P_R is greatest when radius NM is horizontal, i.e. delta = beta. Different V_S circles are concentric. 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