introductory_circuit_analysis_11th_ed_solution

introductory_circuit_analysis_11th_ed_solution

introductory_circuit_analysis_11th_ed_solution_manual-8860.pdf

목차 Chapter1~25 본문 introductory_circuit_analysis_11th_ed_solution b. I2 = 1.27 A, I3 = 0.26 A c. I R = I2 = 1.27 A,I R = I3 = 0.26 A 2 3 I = I2 ? R I3 = 1.27 A ? 0.26 A = 1.01 A 4 I = 3 A ? R I2 = 3 A ? 1.27 A = 1.73 A 1 31. I1(2 + 1) ? 1I2 = 10 I2(1 + 4 + 5) ? 1I1 ? 5I3 = 0 I3(5 + 3) ? 5I2 = ?6 ─ I2 =I R = ?63.69 mA (exact match with problem 18) 3 32. From Sol. 24(b) I1(6 + 4) ? 4I2 = ?12 I2(4 + 5 + 2) ? 4I1 ? 2I3 = 12 + 16 I3(2 + 3) ? 2I2 = ?16 ─ I5Ω = I2 = 1.95 A I3 = ?2.42 A, ∴ Va = (I3)(3 Ω) = (?2.42 A)(3 Ω) = ?7.26 V 33. (I): (2.2 kΩ + 9.1 kΩ)I1 ? 9.1 kΩI2 = 18 (9.1 kΩ+ 7.5 kΩ+ 6.8 kΩ)I2 ? 9.1 kΩI1 ? 6.8 kΩI3 = ?18 (6.8 kΩ+ 3.3 kΩ)I3 ? 6.8 kΩI2 = ?3 ─ I1 = 1.21 mA, I2 = ?0.48 mA, I3 = ?0.62 mA (II): (4 Ω + 4 Ω + 3 Ω)I1 ? 3 Ω I2 ? 4 Ω I3 = 16 ? 12 (4 Ω + 3 Ω + 10 Ω)I2 ? 3I1 ? 4 Ω I3 = 12 ? 15 (4 Ω + 4 Ω + 7 Ω)I3 ? 4I1 ? 4I2 = ?16 ─ I1 = ?0.24 A, I2 = ?0.52 A, I3 = ?1.28 A 34. a. I1(6.8 kΩ + 4.7 kΩ + 2.2 kΩ) ? 4.7 kΩ I2 ? 2.2 kΩ I4 = 6 I2(2.7 kΩ + 8.2 kΩ + 4.7 kΩ) ? 4.7 kΩ I1 ? 8.2 kΩ I3 = ?6 I3(8.2 kΩ + 1.1 kΩ + 22 kΩ) ? 22 kΩ I4 ? 8.2 kΩ I2 = ?9 I4(2.2 kΩ + 22 kΩ + 1.2 kΩ) ? 2.2 kΩ I1 ? 22 kΩ I3 = 5 ─ I1 = 0.03 mA, I2 = ?0.88 mA, I3 = ?0.97 mA, I4 = ?0.64 mA b. From Sol. 26(b): I1(2 + 4) ? 4I2 = ?6 I2(4 + 1) ? 4I1 ? 1I3 = ?6 I3(1 + 8) ? 1I2 = 6 ─ I1 = 3.8 A, I2 = ?4.20 A, I3 = 0.20 A 74 CHAPTER 8 본문내용 t L. Boylestad Upper Saddle River, New Jersey Columbus, Ohio __________________________________________________________________________________ Copyright2007 by Pearson Education, Inc., Upper Saddle River, New Jersey 07458. Pearson Prentice Hall. All rights reserved. Printed in the United States of America. This publication is protected by Copyright and permission should be obtained from 하고 싶은 말 직접보시라