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The results of a ‘Slip-Test’ for determining direct-axis (Xd) and quadrature-axis (Xq) reactances of a star-connected, salient-pole alternator are given below:Phase values: Vmax = 108V; Vmin = 96V, Imax = 12A, Imin = 10A. Hence the two reactances will bea)Xd = 10.8 ohms and Xq = 8 ohmsb)Xd = 9 ohms and Xq = 9.6 ohmsc)Xd = 9.6 ohms and Xq = 9 ohmsd)Xd = 8 ohms and Xq = 10.8 ohmsCorrect answer is option 'A'. Can you explain this answer? for Electrical Engineering (EE) 2025 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about The results of a ‘Slip-Test’ for determining direct-axis (Xd) and quadrature-axis (Xq) reactances of a star-connected, salient-pole alternator are given below:Phase values: Vmax = 108V; Vmin = 96V, Imax = 12A, Imin = 10A. Hence the two reactances will bea)Xd = 10.8 ohms and Xq = 8 ohmsb)Xd = 9 ohms and Xq = 9.6 ohmsc)Xd = 9.6 ohms and Xq = 9 ohmsd)Xd = 8 ohms and Xq = 10.8 ohmsCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Electrical Engineering (EE) 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The results of a ‘Slip-Test’ for determining direct-axis (Xd) and quadrature-axis (Xq) reactances of a star-connected, salient-pole alternator are given below:Phase values: Vmax = 108V; Vmin = 96V, Imax = 12A, Imin = 10A. Hence the two reactances will bea)Xd = 10.8 ohms and Xq = 8 ohmsb)Xd = 9 ohms and Xq = 9.6 ohmsc)Xd = 9.6 ohms and Xq = 9 ohmsd)Xd = 8 ohms and Xq = 10.8 ohmsCorrect answer is option 'A'. Can you explain this answer?.

Solutions for The results of a ‘Slip-Test’ for determining direct-axis (Xd) and quadrature-axis (Xq) reactances of a star-connected, salient-pole alternator are given below:Phase values: Vmax = 108V; Vmin = 96V, Imax = 12A, Imin = 10A. Hence the two reactances will bea)Xd = 10.8 ohms and Xq = 8 ohmsb)Xd = 9 ohms and Xq = 9.6 ohmsc)Xd = 9.6 ohms and Xq = 9 ohmsd)Xd = 8 ohms and Xq = 10.8 ohmsCorrect answer is option 'A'. Can you explain this answer? in English & in Hindi are available as part of our courses for Electrical Engineering (EE). Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free.

Here you can find the meaning of The results of a ‘Slip-Test’ for determining direct-axis (Xd) and quadrature-axis (Xq) reactances of a star-connected, salient-pole alternator are given below:Phase values: Vmax = 108V; Vmin = 96V, Imax = 12A, Imin = 10A. Hence the two reactances will bea)Xd = 10.8 ohms and Xq = 8 ohmsb)Xd = 9 ohms and Xq = 9.6 ohmsc)Xd = 9.6 ohms and Xq = 9 ohmsd)Xd = 8 ohms and Xq = 10.8 ohmsCorrect answer is option 'A'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of The results of a ‘Slip-Test’ for determining direct-axis (Xd) and quadrature-axis (Xq) reactances of a star-connected, salient-pole alternator are given below:Phase values: Vmax = 108V; Vmin = 96V, Imax = 12A, Imin = 10A. Hence the two reactances will bea)Xd = 10.8 ohms and Xq = 8 ohmsb)Xd = 9 ohms and Xq = 9.6 ohmsc)Xd = 9.6 ohms and Xq = 9 ohmsd)Xd = 8 ohms and Xq = 10.8 ohmsCorrect answer is option 'A'. Can you explain this answer?, a detailed solution for The results of a ‘Slip-Test’ for determining direct-axis (Xd) and quadrature-axis (Xq) reactances of a star-connected, salient-pole alternator are given below:Phase values: Vmax = 108V; Vmin = 96V, Imax = 12A, Imin = 10A. Hence the two reactances will bea)Xd = 10.8 ohms and Xq = 8 ohmsb)Xd = 9 ohms and Xq = 9.6 ohmsc)Xd = 9.6 ohms and Xq = 9 ohmsd)Xd = 8 ohms and Xq = 10.8 ohmsCorrect answer is option 'A'. Can you explain this answer? has been provided alongside types of The results of a ‘Slip-Test’ for determining direct-axis (Xd) and quadrature-axis (Xq) reactances of a star-connected, salient-pole alternator are given below:Phase values: Vmax = 108V; Vmin = 96V, Imax = 12A, Imin = 10A. Hence the two reactances will bea)Xd = 10.8 ohms and Xq = 8 ohmsb)Xd = 9 ohms and Xq = 9.6 ohmsc)Xd = 9.6 ohms and Xq = 9 ohmsd)Xd = 8 ohms and Xq = 10.8 ohmsCorrect answer is option 'A'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice The results of a ‘Slip-Test’ for determining direct-axis (Xd) and quadrature-axis (Xq) reactances of a star-connected, salient-pole alternator are given below:Phase values: Vmax = 108V; Vmin = 96V, Imax = 12A, Imin = 10A. Hence the two reactances will bea)Xd = 10.8 ohms and Xq = 8 ohmsb)Xd = 9 ohms and Xq = 9.6 ohmsc)Xd = 9.6 ohms and Xq = 9 ohmsd)Xd = 8 ohms and Xq = 10.8 ohmsCorrect answer is option 'A'. Can you explain this answer? tests, examples and also practice Electrical Engineering (EE) tests.