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ICE_Chapter_02

# ICE_Chapter_02 - CHAPTER 2 Solutions for Exercises E2.1(a...

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

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V 600 . 9 1 = = i R v eq eq A 480 . 0 / 2 2 = = R v i eq A 320 . 0 / 3 3 = = R v i eq A 240 . 0 / 4 4 = = R v i eq (b) R 1 , and R 2 are in series . Furthermore, R 3 , and R 4 are in series . Finally, the two series combinations are in parallel. = + = 20 2 1 1 R R R eq = + = 20 4 3 2 R R R eq 10 / 1 / 1 1 2 1 = + = eq eq eq R R R V 20 2 = × = eq eq R v A 1 / 1 1 = = eq eq R v i A 1 / 2 2 = = eq eq R v i (c) R 3 , and R 4 are in series . The combination of R 3 and R 4 is in parallel with R 2 . Finally the combination of R 2, R 3 , and R 4 is in series with R 1 . = + = 40 4 3 1 R R R eq 20 / 1 / 1 1 2 1 2 = + = R R R eq eq A 1 2 1 1 = + = eq s R R v i V 20 2 1 2 = = eq R i v A 5 . 0 / 2 2 2 = = R v i A 5 . 0 / 1 2 3 = = eq R v i E2.3 (a) V 10 4 3 2 1 1 1 = + + + = R R R R R v v s . V 20 4 3 2 1 2 2 = + + + = R R R R R v v s . Similarly, we find V 30 3 = v and V 60 4 = v . 2
(b) First combine R 2 and R 3 in parallel: . 917 . 2 ) 1 / 1 ( 1 3 2 = + = R R R eq Then we have V 05 . 6 4 1 1 1 = + + = R R R R v v eq s . Similarly, we find V 88 . 5 4 1 2 = + + = R R R R v v eq eq s and V 07 . 8 4 = v . E2.4 (a) First combine R 1 and R 2 in series: R eq = R 1 + R 2 = 30 . Then we have A. 2 30 15 30 and A 1 30 15 15 3 3 3 3 1 = + = + = = + = + = eq eq s eq s R R R i i R R R i i (b) The current division principle applies to two resistances in parallel. Therefore, to determine i 1 , first combine R 2 and R 3 in parallel: R eq = 1/(1/ R 2 + 1/ R 3 ) = 5 . Then we have . A 1 5 10 5 1 1 = + = + = eq eq s R R R i i Similarly, i 2 = 1 A and i 3 = 1 A. E2.5 Write KVL for the loop consisting of v 1 , v y , and v 2. The result is -v 1 - v y + v 2 = 0 from which we obtain v y = v 2 - v 1 . Similarly we obtain v z = v 3 - v 1 . E2.6 Node 1: a i R v v R v v = - + - 2 2 1 1 3 1 Node 2: 0 4 3 2 3 2 2 1 2 = - + + - R v v R v R v v Node 3: 0 1 1 3 4 2 3 5 3 = + - + - + b i R v v R v v R v E2.7 Using determinants we can solve for the unknown voltages as follows: V 32 . 10 04 . 0 35 . 0 2 . 0 3 5 . 0 2 . 0 2 . 0 7 . 0 5 . 0 1 2 . 0 6 1 = - + = - - - = v V 129 . 6 04 . 0 35 . 0 2 . 1 7 . 0 5 . 0 2 . 0 2 . 0 7 . 0 1 2 . 0 6 7 . 0 2 = - + = - - - = v 3

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Many other methods exist for solving linear equations. E2.8 First write KCL equations at nodes 1 and 2: Node 1: 0 10 5 2 10 2 1 1 1 = - + + - v v v v Node 2: 0 10 5 10 10 1 2 2 2 = - + + - v v v v Then simplify the equations to obtain: 50 8 2 1 = - v v and 10 4 2 1 = + - v v Solving, we find v 1 = 6.77 V and v 2 = 4.19 V.
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ICE_Chapter_02 - CHAPTER 2 Solutions for Exercises E2.1(a...

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