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: 4
  Techniques of Circuit Analysis
  Assessment Problems
  AP 4.1 [a] Redraw the circuit, labeling the reference node and the two node voltages:
  The two node voltage equations are
  −15 + v1
  60 + v1
  15 + v1 − v2
  5
  =
  0
  5 + v2
  2 + v2 − v1
  5
  =
  0
  Place these equations in standard form:
  v1
  � 1
  60 + 1
  15 + 1
  5
  �
  +
  v2
  �
  −1
  5
  �
  =
  15
  v1
  �
  −1
  5
  �
  +
  v2
  �1
  2 + 1
  5
  �
  =
  −5
  Solving, v1 = 60 V and v2 = 10 V;
  Therefore, i1 = (v1 − v2)/5 = 10 A
  [b] p15A = −(15 A)v1 = −(15 A)(60 V) = −900 W = 900 W(delivered)
  [c] p5A = (5 A)v2 = (5 A)(10 V) = 50 W= −50 W(delivered)
  4–1
  
  4–2
  CHAPTER 4. Techniques of Circuit Analysis
  AP 4.2 Redraw the circuit, choosing the node voltages and reference node as shown:
  The two node voltage equations are:
  −4.5 + v1
  1 + v1 − v2
  6 + 2
  =
  0
  v2
  12 + v2 − v1
  6 + 2 + v2 − 30
  4
  =
  0
  Place these equations in standard form:
  v1
  �
  1 + 1
  8
  �
  +
  v2
  �
  −1
  8
  �
  =
  4.5
  v1
  �
  −1
  8
  �
  +
  v2
  � 1
  12 + 1
  8 + 1
  4
  �
  =
  7.5
  Solving, v1 = 6 V
  v2 = 18 V
  To ﬁnd the voltage v, ﬁrst ﬁnd the current i through the series-conn