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F07 314 HW01 solutions all

F07 314 HW01 solutions all - EECS 314 Fall 2007 HW 01...

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Part 1 Current Y is the easiest one to solve first. Inspecting node N 1 , there is a complete set of currents with one unknown quantity Y . Using Kirchhoff’s current law (KCL): - 8A + 2A + 14A = Y Y = 8A Next, solve for the current I 1 which will be used to solve for current X . Current I 1 is found by applying KCL at node N 2 : 14A + ( - 24A) + I 1 = 0 I 1 = 10A Use KCL at node N 3 to find current X : X + Y + I 1 = 3A X + 8A + 10A = 3A X = - 15A Several more steps are required to find the last current Z . First, use KCL on node N 4 to find current I 2 , then use KCL on node N 5 to find current I 3 , and finally, solve for Z : 3A + ( - 15A) + I 2 = 0 I 2 = 12A I 2 + X + 2A = I 3 12A + ( - 15A) + 2A = I 3 I 3 = - 1A I 3 + ( - 8A) + Z = 0 ( - 1A) + ( - 8A) + Z = 0 Z = 9A EECS 314 Fall 2007 HW 01 Solutions Problem 1 – 15 A 3 A – 24 A 2 A 14 A – 8 A N 1 N 2 N 3 N 4 N 5 I 1 I 2 I 3 Z X Y
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Part 2 Voltage X is the easiest one to solve first. Inspecting loop L 1 , there is a complete set of voltages with one unknown quantity X . Using Kirchhoff’s voltage law (KVL): - 12V + 120V - X = 0 X = 108V Next, solve for voltage Y by applying KVL at loop L 2 : - 6V + 15V - Y = 0 Y = 9V With some creativity, the last current Z can be found even without X or Y by using loop L 3 : - ( - 16V) - 12V + 120V + 6V + Z = 0 Z = - 130V EECS 314 Fall 2007 HW 01 Solutions Problem 1 120 V 15 V 12 V 6 V – 16 V X Z Y L 1 L 2 L 3
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Part 3 For element 2, find the voltage across it using KVL and surrounding elements. Then use that information with the provided current to find power absorbed. Notice the polarity of V 2 is chosen such that the product yields absorbed power, according to the passive sign convention. V 2 + ( - 36V) + 8V - 15V = 0 V 2 = 43V P 2 = ( - 13A)(43V) = - 559W For element 3, find its voltage using KVL and its current using KCL, and then multiply to find the power. There are many ways to approach this, one being: 15A + ( - 13A) = I 1 I 1 = 2A 2A + I 3 = 6A I 3 = 4A 8V + ( - 15V) + V 3 = 0 V 3 = 7V P 3 = (4A)(7V) = 28W For element 5, use the prior calculation for I 3 to aid in finding the current I 5 so that the power can be calculated: - 13A + 4A + I 5 = 0 I 5 = 9A P 5 = (9A)(15V) = 135W EECS 314 Fall 2007 HW 01 Solutions Problem 1 15 A – 13 A 6 A 15 V – 36 V 8 V V 2 I 1 I 5 I 3 V 3
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Part 3 For element 2, find the voltage across it using KVL and surrounding elements. Then use that information with the provided current to find power absorbed. Notice the polarity of V 2 is chosen such that the product yields absorbed power, according to the passive sign convention. V 2 + ( - 36V) + 8V - 15V = 0 V 2 = 43V P 2 = ( - 13A)(43V) = - 559W For element 3, find its voltage using KVL and its current using KCL, and then multiply to find the power. There are many ways to approach this, one being: 15A + ( - 13A) = I 1 I 1 = 2A 2A + I 3 = 6A I 3 = 4A 8V + ( - 15V) + V 3 = 0 V 3 = 7V P 3 = (4A)(7V) = 28W For element 5, use the prior calculation for I 3 to aid in finding the current I 5 so that the power can be calculated: - 13A + 4A + I 5 = 0 I 5 = 9A P 5 = (9A)(15V) = 135W EECS 314 Fall 2007 HW 01 Solutions Problem 1 15 A – 13 A 6 A 15 V – 36 V 8 V V 2 I 1 I 5 I 3 V 3
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Part 1 In circuit C
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