Chapter 02

# Chapter 02 - CHAPTER 2 Exercises E2.1(a R2 R3 and R4 are in...

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1 CHAPTER 2 Exercises E2.1 (a) R 2 , R 3 , and 4 are in parallel. Furthermore 1 is in series with the combination of the other resistors. Thus we have: = + + + = 3 / 1 / 1 / 1 1 4 3 2 1 eq (b) 3 and 4 are in parallel. Furthermore, 2 is in series with the combination of 3 and 4 . Finally 1 is in parallel with the combination of the other resistors. Thus we have: = + + + = 5 )] / 1 / 1 /( 1 /[ 1 / 1 1 4 3 2 1 (c) 1 and 2 are in parallel . Furthermore, 3 and 4 are in parallel . Finally, the two parallel combinations are in series. = + + + = 52.1 / 1 / 1 1 / 1 / 1 1 4 3 2 1 (d) 1 and 2 are in series . Furthermore, 3 is in parallel with the series combination of 1 and 2 . = + + = k 1.5 ) /( 1 / 1 1 2 1 3 E2.2 (a) First we combine 2 , 3 and 4 in parallel. Then 1 is in series with the parallel combination. = + + = 231 . 9 / 1 / 1 / 1 1 4 3 2 A 04 . 1 231 . 9 10 20 V 20 1 1 = + = + = i V 600 . 9 1 = = v A 480 . 0 / 2 2 = = A 320 . 0 / 3 3 = = A 240 . 0 / 4 4 = =

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2 (b) R 1 and 2 are in series . Furthermore, 3 , and 4 are in series . Finally, the two series combinations are in parallel. = + = 20 2 1 1 eq = + = 20 4 3 2 10 / 1 / 1 1 2 1 = + = V 20 2 = × = v A 1 / 1 1 = = i A 1 / 2 2 = = (c) 3 and 4 are in series . The combination of 3 and 4 is in parallel with 2 . Finally the combination of 2, 3 and 4 is in series with 1 . = + = 40 4 3 1 20 / 1 / 1 1 2 1 2 = + = A 1 2 1 1 = + = s V 20 2 1 2 = = A 5 . 0 / 2 2 2 = = A 5 . 0 / 1 2 3 = = E2.3 (a) V 10 4 3 2 1 1 1 = + + + = . V 20 4 3 2 1 2 2 = + + + = . Similarly, we find V 30 3 = and V 60 4 = .
3 (b) First combine R 2 and 3 in parallel: . 917 . 2 ) 1 / 1 ( 1 3 2 = + = eq Then we have V 05 . 6 4 1 1 1 = + + = v s . Similarly, we find V 88 . 5 4 1 2 = + + = and V 07 . 8 4 = . E2.4 (a) First combine 1 and 2 in series: = 1 + 2 = 30 . Then we have A. 2 30 15 30 and A 1 30 15 15 3 3 3 3 1 = + = + = = + = + = i (b) The current division principle applies to two resistances in parallel. Therefore, to determine 1 , first combine 2 and 3 in parallel: = 1/(1/ 2 + 1/ 3 ) = 5 . Then we have . A 1 5 10 5 1 1 = + = + = Similarly, 2 = 1 A and 3 = 1 A. E2.5 Write KVL for the loop consisting of 1 , y , and 2. The result is - 1 - + 2 = 0 from which we obtain = 2 - 1 . Similarly we obtain z = 3 - 1 . E2.6 Node 1: a = + 2 2 1 1 3 1 Node 2: 0 4 3 2 3 2 2 1 2 = + + Node 3: 0 1 1 3 4 2 3 5 3 = + + + b E2.7 Following the step-by-step method in the book, we obtain = + + + + 0 1 1 1 0 1 1 1 1 1 0 1 1 1 3 2 1 5 4 4 4 4 3 2 2 2 2 1 E2.8 Instructions for various calculators vary. The MATLAB solution is given in the book following this exercise.

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4 E2.9 (a) Writing the node equations we obtain: Node 1: 0 10 5 20 2 1 1 3 1 = + + v Node 2: 0 5 10 10 3 2 1 2 = + + Node 3: 0 5 10 20 2 3 3 1 3 = + + (b) Simplifying the equations we obtain:
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## This note was uploaded on 11/02/2010 for the course ECE 3710 taught by Professor Haris during the Spring '07 term at Georgia Tech.

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Chapter 02 - CHAPTER 2 Exercises E2.1(a R2 R3 and R4 are in...

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