--- title: "Reactance diagram 1000 kVA base and generator currents" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:reactance-diagram-1000kva-load-current" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/reactance-diagram-1000kva-load-current site: BCT Engineering --- # Reactance diagram 1000 kVA base and generator currents Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2073 Bhadra, 10 marks - Video: [Impedance and Reactance diagram Problem-3 | Power System Analysis | PSA | per unit Reactance diagram](https://www.youtube.com/watch?v=TQ8W4-l9S2I) — Simple n Easy Learning - 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/reactance-diagram-1000kva-load-current - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Develop the reactance diagram of the following network and express all the parameters in p.u. values based on power 1000 kVA and base voltage of 11 kV at low voltage side: G1: 1000 kVA, 11 kV, X = 2.5%; G2: 500 kVA, 11 kV, X = 0.1%; Tr-1: 1000 kVA, 11 kV/66 kV, X = 2%; Tr-2: 500 kVA, 12.5 kV/75 kV, X = 2%; Tr-3: 1500 kVA, 66 kV/400 kV, X = 2%; TL: 10 Ω/phase. If the load draws a current of 1200 A, calculate the current supplied by G1 and G2. ## Short answer On 1000 kVA base: X_G1=0.025, X_G2=0.002, X_Tr1=0.02, X_Tr2=0.0517, X_TL=0.0023, X_Tr3=0.0133 pu. Assuming equal emfs, load current divides inversely with branch impedances: G1 supplies 652.6 A, G2 supplies 547.4 A. ## Answer No figure is printed, so the usual arrangement is assumed: G1–Tr1 and G2–Tr2 feed a common 66 kV bus; from it the line TL and Tr3 (66/400 kV) supply the load. The two generator emfs are taken as equal, so the load current divides between the two generator branches in inverse ratio of their impedances. The 1200 A load current is taken as referred to the 11 kV level, since its voltage level is not stated. ```text G1--Tr1(11/66)--+ |66 kV bus G2--Tr2(12.5/75)+--TL(10 ohm)--Tr3(66/400)--Load ``` #### Base values ($S_b$ = 1000 kVA) - G1 side: 11 kV (given). 66 kV bus: $11\times\frac{66}{11} = 66$ kV. - G2 side through Tr2: $66\times\frac{12.5}{75} = 11$ kV. - Load side: $66\times\frac{400}{66} = 400$ kV. - $Z_b$ at 66 kV $= \frac{66^2}{1} = 4356\ \Omega$. $I_b$ at 11 kV $= \frac{1000}{\sqrt3\times11} = 52.49$ A. #### Per-unit reactances $$ \begin{aligned} X_{G1} &= 0.025\ \text{pu}\\ X_{G2} &= 0.001\times\frac{1000}{500}\times\left(\frac{11}{11}\right)^2 = 0.002\ \text{pu}\\ X_{Tr1} &= 0.02\ \text{pu}\\ X_{Tr2} &= 0.02\times\frac{1000}{500}\times\left(\frac{12.5}{11}\right)^2 = 0.0517\ \text{pu}\\ X_{TL} &= \frac{10}{4356} = 0.0023\ \text{pu}\\ X_{Tr3} &= 0.02\times\frac{1000}{1500} = 0.0133\ \text{pu} \end{aligned} $$ | Element | X (pu, 1000 kVA) | |---|---| | G1 | 0.0250 | | G2 | 0.0020 | | Tr-1 | 0.0200 | | Tr-2 | 0.0517 | | TL | 0.0023 | | Tr-3 | 0.0133 | #### Reactance diagram ```text j0.025 j0.02 +-/\/\/--/\/\/--+ | G1 Tr1 | j0.0023 j0.0133 (E1) +---/\/\/----/\/\/---> Load | G2 Tr2 | TL Tr3 +-/\/\/--/\/\/--+ j0.002 j0.0517 (E1, E2 return to the common reference bus) ``` #### Current shared by G1 and G2 Branch impedances: $Z_1 = 0.025 + 0.02 = 0.045$ pu and $Z_2 = 0.002 + 0.0517 = 0.0537$ pu. $$ \begin{aligned} I_{G1} &= I_L\frac{Z_2}{Z_1+Z_2} = 1200\times\frac{0.0537}{0.0987} = 1200\times0.5439 = 652.6\ \text{A}\\ I_{G2} &= I_L\frac{Z_1}{Z_1+Z_2} = 1200\times0.4561 = 547.4\ \text{A} \end{aligned} $$ (In per unit the load current is $1200/52.49 = 22.86$ pu. G1 carries 12.43 pu and G2 carries 10.43 pu.) **Answer:** G1 supplies about **652.6 A** and G2 about **547.4 A** (54.4 % and 45.6 % of the load current). Note: if the G2 reactance of 0.1 % is a misprint for 10 %, then $X_{G2} = 0.2$ pu, $Z_2 = 0.2517$ pu, and the shares become 1018.0 A (G1) and 182.0 A (G2). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What is the base voltage at the G2 side through Tr2 (12.5/75 kV) from 66 kV bus? A. 12.5 kV B. 11 kV C. 75 kV D. 66 kV Answer: B. 66 x 12.5/75 = 11 kV. (Key point: Base on G2 side) 2. Why is X_Tr2 = 0.02 x 1000/500 x (12.5/11)^2 = 0.0517? A. Tr2 rated voltage 12.5 kV differs from base 11 kV, and rating is 500 kVA B. Tr2 is a delta unit C. 66 kV bus D. G2 is 0.1% Answer: A. Both rating and voltage corrections apply. (Key point: Tr2 correction) 3. How does the load current divide between G1 and G2 branches with equal emfs? A. Equally B. In direct ratio of impedances C. In inverse ratio of impedances D. Only G1 Answer: C. I_G1 = I_L Z2/(Z1+Z2). (Key point: Current sharing) 4. Z1 = 0.045 and Z2 = 0.0537 pu. What is I_G1 for 1200 A? A. 547.4 A B. 652.6 A C. 600 A D. 1018 A Answer: B. 1200 x 0.0537/0.0987 = 652.6 A. 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