title: Circuit Analysis II with MATLAB Applications
author:  Steven T. Karris 
description: Highly recommended. Lower- and upper-division undergraduates; two-year technical program students. -- <i>I. L. Kosow, emeritus, City University of New York, CHOICE Magazine, March 2004</i>&lt;br /&gt;&lt;br /&gt;Highly recommended. Lower- and upper-division undergraduates; two-year technical program students. --<i>I. L. Kosow, emeritus, City University of New York, CHOICE Magazine, March 2004</i>
zlib_text: 8
  Natural and Step Responses of RLC
  Circuits
  Assessment Problems
  AP 8.1 [a]
  1
  (2RC)2 =
  1
  LC ,
  therefore
  C = 500 nF
  [b] α = 5000 =
  1
  2RC ,
  therefore
  C = 1 µF
  s1,2 = −5000 ±
  �
  25 × 106 − (103)(106)
  20
  = (−5000 ± j5000) rad/s
  [c]
  1
  √
  LC = 20,000,
  therefore
  C = 125 nF
  s1,2 =
  �
  −40 ±
  �
  (40)2 − 202
  �
  103,
  s1 = −5.36 krad/s,
  s2 = −74.64 krad/s
  AP 8.2
  iL
  =
  1
  50 × 10−3
  � t
  0[−14e−5000x + 26e−20,000x] dx + 30 × 10−3
  =
  20
  �−14e−5000x
  −5000
  ����
  t
  0 + 26e−20,000t
  −20,000
  ����
  t
  0
  �
  + 30 × 10−3
  =
  56 × 10−3(e−5000t − 1) − 26 × 10−3(e−20,000t − 1) + 30 × 10−3
  =
  [56e−5000t − 56 − 26e−20,000t + 26 + 30] mA
  =
  56e−5000t − 26e−20,000t mA,
  t ≥ 0
  AP 8.3 From the given values of R, L, and C, s1 = −10 krad/s and s2 = −40 krad/s.
  [a] v(0−) = v(0+) = 0,
  therefore
  iR(0+) = 0
  8–1
  
  8–2
  CHAPTER 8. Natural and Step Responses of RLC Circuits
  [b] iC(0+) = −(iL(0+) + iR(0+)) = −(−4 + 0) = 4 A
  [c] C dvc(0+)
  dt
  = ic(0+) = 4,
  therefore
  dvc(0+)
  dt
  = 4
  C = 4 × 108 V/s
  [d] v = [A1e−10,000t + A2e−40,000t] V,
  t ≥ 0+
  v(0+) = A1