--- title: "Load sharing of two transformers in parallel" subject: "Electrical Machines" chapter: 2 key: "semester_4/electrical_machines:parallel-transformers-load-sharing" source: https://www.bctengineer.com/q/semester_4/electrical_machines/parallel-transformers-load-sharing site: BCT Engineering --- # Load sharing of two transformers in parallel Electrical Machines · Chapter 2: Transformer - Asked in: 2067 Mangsir (old course), 8 marks - Video: [Lecture-26 Numerical Based On Load Sharing #gate #eee #parallel #transformers #machines#videos](https://www.youtube.com/watch?v=wTzZjzcVXvg) — Watt's Up - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/electrical_machines/parallel-transformers-load-sharing - Whole chapter: https://www.bctengineer.com/md/semester_4/electrical_machines/chapter-2 ## Question Two 1-phase transformers with equal number of turns have impedance of $(0.5 + j3)\ \Omega$ and $(0.6 + j10)\ \Omega$ with respect to the secondary. If they operate in parallel, determine how they will share total load of 100 kW at pf 0.8 lagging. ## Short answer Total kVA = 100 kW / 0.8 = 125 kVA. Load shares inversely with impedances ZA=0.5+j3 Ω and ZB=0.6+j10 Ω. Transformer A supplies 95.98 kVA (78.17 kW, pf 0.814 lag); transformer B supplies 29.14 kVA (21.83 kW, pf 0.749 lag). ## Answer Two transformers with equal turns ratio have the same secondary voltage, and they share the load current in inverse proportion to their impedances. **Given:** $Z_A = 0.5 + j3\ \Omega$, $Z_B = 0.6 + j10\ \Omega$ (referred to secondary), total load $P = 100$ kW at 0.8 pf lagging. **Total load** $$ S = \frac{100}{0.8} = 125\ \text{kVA},\qquad \phi = \cos^{-1}0.8 = 36.87^\circ\ \text{lagging} $$ **Current division.** Let the total current be $I = I\angle-36.87^\circ$ with $V_2$ as reference. For parallel branches: $$ I_A = I\frac{Z_B}{Z_A + Z_B},\qquad I_B = I\frac{Z_A}{Z_A + Z_B} $$ $$ \begin{aligned} Z_A + Z_B &= 1.1 + j13\ \Omega \\ \frac{Z_B}{Z_A + Z_B} &= 0.7676 + j0.0188 = 0.7679\angle1.40^\circ \\ \frac{Z_A}{Z_A + Z_B} &= 0.2324 - j0.0188 = 0.2331\angle-4.63^\circ \end{aligned} $$ **Load shared** (kVA = fraction of 125 kVA, since $V_2$ is common): $$ \begin{aligned} S_A &= 0.7679\times125 = 95.98\ \text{kVA at angle } 36.87^\circ - 1.40^\circ = 35.47^\circ \\ P_A &= 95.98\cos35.47^\circ = 78.17\ \text{kW},\quad \text{pf}_A = 0.814\ \text{lag} \\ S_B &= 0.2331\times125 = 29.14\ \text{kVA at angle } 36.87^\circ + 4.63^\circ = 41.50^\circ \\ P_B &= 29.14\cos41.50^\circ = 21.83\ \text{kW},\quad \text{pf}_B = 0.749\ \text{lag} \end{aligned} $$ Check: $P_A + P_B = 78.17 + 21.83 = 100$ kW. | Transformer | kVA | kW | Power factor | |---|---|---|---| | A | 95.98 | 78.17 | 0.814 lag | | B | 29.14 | 21.83 | 0.749 lag | **Answer: A supplies about 78.2 kW (96.0 kVA) and B about 21.8 kW (29.1 kVA).** The transformer of lower impedance (A) carries the greater share. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Equal-turns transformers in parallel share current: A. In inverse proportion to their impedances B. In proportion to impedances C. Equally D. In proportion to resistance only Answer: A. IA = I ZB/(ZA + ZB). (Key point: Current division) 2. Total load in kVA is: A. 100 kVA B. 125 kVA at 36.87 degrees lagging C. 80 kVA D. 160 kVA Answer: B. S = 100/0.8; phi = arccos 0.8. (Key point: Total load) 3. Transformer A (lower impedance) supplies about: A. 21.8 kW B. 50 kW C. 78.2 kW D. 100 kW Answer: C. SA = 0.7679 x 125 = 95.98 kVA at 35.47 degrees, P = 78.17 kW. (Key point: Share of A) 4. Transformer B supplies about: A. 78.2 kW B. 50 kW at pf 0.8 C. 29.1 kW at unity pf D. 21.8 kW at pf 0.749 lag Answer: D. SB = 29.14 kVA at 41.5 degrees; P = 21.83 kW; check sum is 100 kW. 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