--- title: "Zero regulation for 40 MW leading load" subject: "Power System Analysis I" chapter: 6 key: "semester_4/power_system_analysis_i:zero-regulation-40mw-leading" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/zero-regulation-40mw-leading site: BCT Engineering --- # Zero regulation for 40 MW leading load Power System Analysis I · Chapter 6: Performance Analysis - Asked in: 2071 Bhadra, 6 marks - Video: [Condition for Maximum/Zero Voltage Regulation | Analysis of Objective-2 | GATE/ESE/ISRO/BARC](https://www.youtube.com/watch?v=VsmL4LIlY2E) — ELECTRICAL ENGINEERING CONCEPTS - 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/zero-regulation-40mw-leading - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-6 ## Question A 50 Hz, 3-phase, 132 kV overhead line is supplying a load of 40 MW at 0.9 p.f. leading. The line has series impedance of 12 + j36 Ω/phase. How can the voltage regulation of the line be made zero? ## Short answer The present regulation is −0.78%. To make it zero, absorb approximately 3.93 MVAr using a shunt reactor at the receiving end, bringing the net power factor to about 0.933 leading. ## Answer Short line: $Z = 12 + j36\ \Omega$, $V_R$ = 132 kV, load 40 MW at 0.9 pf **leading**. #### Present regulation $$ \begin{aligned} I &= \frac{40\times10^6}{\sqrt3\times132\,000\times0.9} = 194.39\ \text{A},\quad \phi = 25.84^\circ\ \text{lead} \\ V_S &= 76\,210.2 + 194.39\angle25.84^\circ\,(12 + j36) \\ &= 75\,613.9\angle5.55^\circ\ \text{V} \Rightarrow 130.97\ \text{kV (line)} \\ \%VR &= \frac{130.97 - 132}{132}\times100 = -0.78\% \end{aligned} $$ The regulation is **negative**: the load is too leading, so the receiving-end voltage is higher than the sending-end voltage. #### Condition for zero regulation Approximate condition: $R\cos\phi - X\sin\phi = 0$, i.e. $\tan\phi = R/X = 1/3$, giving pf $= 0.949$ leading. Exact condition $|V_S| = |V_R| = 132$ kV with $P$ = 40 MW (line kV, MW, MVAr): $$ (V_R^2 + RP + XQ)^2 + (XP - RQ)^2 = V_R^4 $$ $$ (17\,424 + 480 + 36Q)^2 + (1440 - 12Q)^2 = 17\,424^2 $$ Taking the root nearer zero: $$ Q = -15.44\ \text{MVAr}\quad(\text{pf} = 0.933\ \text{leading}) $$ #### How to make it zero The load now supplies $Q_L = -40\tan25.84^\circ = -19.37$ MVAr (leading). The leading VAr must be reduced to 15.44 MVAr, so connect a **shunt reactor** at the receiving end of $$ Q_{reactor} = 19.37 - 15.44 = 3.93\ \text{MVAr} $$ $$ L_{ph} = \frac{V_L^2}{\omega Q} = \frac{132\,000^2}{314.16\times3.93\times10^6} = 14.1\ \text{H (star)} $$ (Alternatively, reduce the excitation of a synchronous load/condenser so that it draws less leading current.) **Answer: Regulation is now −0.78 %. It is made zero by absorbing ≈ 3.9 MVAr with a shunt reactor at the receiving end, bringing the net pf to ≈ 0.933 leading.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the regulation -0.78 % negative? A. Load is lagging B. Line has high resistance C. V_S is zero D. Load is too leading so V_R exceeds V_S Answer: D. A strongly leading load raises voltage at receiving end above sending voltage. (Key point: Present regulation) 2. What is added to make regulation zero? A. Shunt capacitor B. Shunt reactor to absorb some leading VAr C. Series resistor D. Larger conductor Answer: B. The load is too leading, so VAr must be absorbed. (Key point: Remedy) 3. How much leading Q is allowed for zero regulation, and rating of reactor? A. Q = -40; reactor 20.6 MVAr B. Q = -15.44 MVAr (pf 0.933 leading); reactor about 3.93 MVAr C. Q = -19.37; reactor 0 D. Q = 0; reactor 19.37 MVAr Answer: B. Solving |V_S| = |V_R| gives Q = -15.44; load supplies 19.37, difference 3.93 MVAr. (Key point: Exact condition) 4. What does the approximate condition give? A. pf = 1 B. pf = 0.5 C. tan(phi) = R/X = 1/3, pf = 0.949 leading D. pf = 0.9 lagging Answer: C. R cos(phi) = X sin(phi) gives tan(phi) = 12/36. (Key point: Approximate method) ## Close variants set in other exams - [33 kV, 12 MW leading load: efficiency, zero regulation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A33kv-12mw-leading-efficiency) — Power System Analysis I, chapter 6, asked 2× ## Related in Power System Analysis I - [Efficiency at 0.995 leading pf and zero regulation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3Aleading-pf-efficiency-zero-regulation) — Power System Analysis I, chapter 6 - [Capacitor for zero regulation, 70 km 220 kV line](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A70km-220kv-zero-regulation-capacitor) — Power System Analysis I, chapter 6 - [70 km, 40 MW line: regulation and zero-regulation pf](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A70km-40mw-regulation) — Power System Analysis I, chapter 6 - [12 km 33 kV line: losses, zero-regulation capacitor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A12km-33kv-losses-zero-regulation) — Power System Analysis I, chapter 6 - [33 kV, 15 MW line: efficiency and regulation](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A33kv-15mw-efficiency-regulation) — Power System Analysis I, chapter 6 - [12 km line at 32.5 kV: regulation and capacitor](https://www.bctengineer.com/question.md?key=semester_4%2Fpower_system_analysis_i%3A12km-32-5kv-regulation-capacitor) — Power System Analysis I, chapter 6 ## Related in other subjects - [Conductor selection, 100 MW 70 km 132 kV](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Aconductor-selection-100mw-70km-132kv) — Transmission and Distribution System Design, chapter 4 - [Currents and efficiency of 132/11 kV transformer at 5 MW](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Astar-delta-132-11kv-5mw-currents) — Electrical Machines I, chapter 2, asked 3× - [Currents and efficiency of 132/11 kV transformer at 4 MW](https://www.bctengineer.com/question.md?key=semester_4%2Felectrical_machines_i%3Astar-delta-132-11kv-4mw-currents) — Electrical Machines I, chapter 2, asked 3×