F07 314 HW01 solutions all

F07 314 HW01 solutions all - Part 1 Current Y is the...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 Kirchhoffs 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 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 Kirchhoffs 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 = 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 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 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...
View Full Document

This note was uploaded on 10/25/2008 for the course EECS 314 taught by Professor Ganago during the Fall '07 term at University of Michigan.

Page1 / 15

F07 314 HW01 solutions all - Part 1 Current Y is the...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online