电路-电路原理-尼尔森Riedel(第九版)(2011)课后习题答案第6章.pdf

6 Inductance, Capacitance, and Mutual Inductance Assessment Problems AP 6.1 [a] ig = 8e−300t − 8e−1200tA di v = L g = −9.6e−300t + 38.4e−1200tV, t > 0+ dt v (0+) = −9.6 + 38.4 = 28.8 V [b] v = 0 when 38.4e−1200t = 9.6e−300t or t = (ln 4)/900 = 1.54 ms [c] p = vi = 384e−1500t − 76.8e−600t − 307.2e−2400t W dp 1800t 900t [d] = 0 when e − 12.5e + 16 = 0 dt Let x = e900t and solve the quadratic x2 − 12.5x + 16 = 0 ln 1.45 x = 1.44766, t = = 411.05 µs 900 ln 11.05 x = 11.0523, t = = 2.67 ms 900 p is maximum at t = 411.05 µs [e] pmax = 384e−1.5(0.41105) − 76.8e−0.6(0.41105) − 307.2e−2.4(0.41105) = 32.72 W [f] W is max when i is max, i is max when di/dt is zero. When di/dt = 0, v = 0, therefore t = 1.54 ms. [g] imax = 8[e−0.3(1.54) − e−1.2(1.54)] = 3.78 A wmax = (1/2)(4 × 10−3 )(3.78)2 = 28.6 mJ © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyri