--- title: "Active power between machines, 200∠0° and 200∠15° V" subject: "Power System Analysis I" chapter: 3 key: "semester_4/power_system_analysis_i:two-machines-200v-15deg" source: https://www.bctengineer.com/q/semester_4/power_system_analysis_i/two-machines-200v-15deg site: BCT Engineering --- # Active power between machines, 200∠0° and 200∠15° V Power System Analysis I · Chapter 3: Computational Technique - Asked in: 2068 Magh, 6 marks - 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/two-machines-200v-15deg - Whole chapter: https://www.bctengineer.com/md/semester_4/power_system_analysis_i/chapter-3 ## Question Two 3-phase synchronous machines A and B are connected by a link having series impedance of 0 + j10 Ohms/phase. The terminal voltages at machines A and B are 200∠0° V and 200∠15° V respectively. Compute the magnitude and direction of active power flow through the link. ## Short answer Active power per phase P = (200×200/10) sin(0°−15°) = −1035.28 W, so 1.035 kW flows from machine B to A (three-phase 3.106 kW). B is generator, A motor because V_B leads V_A by 15°. ## Answer For a purely reactive link the real power transfer is $$ P = \frac{|V_A||V_B|}{X}\sin(\delta_A - \delta_B) $$ and it flows from the machine with the **leading** angle to the one with the lagging angle. #### Calculation (per phase) $$ \begin{aligned} P_{A\to B} &= \frac{200\times200}{10}\sin(0^\circ - 15^\circ)\\ &= 4000\times(-0.2588) = -1035.28\ \text{W} \end{aligned} $$ The negative sign means the power actually flows from B to A. Check with phasors: $$ \begin{aligned} I_{A\to B} &= \frac{200\angle0^\circ - 200\angle15^\circ}{j10} = 5.221\angle-172.5^\circ\ \text{A}\\ S_A &= V_AI^* = 200\times5.221\angle172.5^\circ = -1035.28 + j136.30\ \text{VA}\\ S_B &= V_BI^* = -1035.28 - j136.30\ \text{VA} \end{aligned} $$ So A receives 1035.28 W and B delivers 1035.28 W. The link has no resistance, so there is no real-power loss. Each machine supplies 136.30 var to the link ($I^2X = 27.26\times10 = 272.59$ var). | Quantity | Per phase | Three-phase | |---|---|---| | Real power B → A | 1035.28 W | 3105.83 W | | Reactive loss in link | 272.59 var | 817.78 var | **Answer:** About **1.035 kW per phase (3.106 kW three-phase)** flows from machine **B to machine A**. B acts as the generator and A as the motor, because $V_B$ leads $V_A$ by 15°. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a purely reactive link between two bus voltages, which formula gives the real power transfer? A. P = |VA||VB| cos(dA - dB) / X B. P = |VA||VB| sin(dA - dB) / X C. P = (|VA| - |VB|)^2 / X D. P = |VA||VB| X sin(dA - dB) Answer: B. With only reactance in the link, real power depends on the sine of the angle difference divided by X. (Key point: Power transfer formula) 2. Real power across a reactive link flows from the machine with which angle? A. The lagging angle to the leading angle B. The leading angle to the lagging angle C. The larger voltage magnitude to the smaller, regardless of angle D. From the machine with the smaller current Answer: B. Active power goes from the leading-angle end to the lagging-angle end. (Key point: Direction of power flow) 3. With VA = 200 angle 0 and VB = 200 angle 15 deg, X = 10 ohm, the formula for A to B gives -1035.28 W per phase. What does this mean? A. 1035 W flows from A to B B. 1035 W flows from B to A C. The link loses 1035 W as heat D. No real power flows; only reactive power does Answer: B. A negative P(A to B) means the power actually flows from B to A, since VB leads VA by 15 deg. (Key point: Calculation and sign) 4. Why is there no real-power loss in the link, and what is its three-phase real power B to A? A. The link has no resistance; about 3.106 kW B. The link has no reactance; about 1.035 kW C. The link has no resistance; about 1.035 kW D. Losses cancel between phases; about 3.106 kW Answer: A. The link is purely reactive so I^2R = 0, and three times 1035.28 W is 3105.83 W. 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