hmwk 1 sol - EECS 314 HW1 SOLUTIONS Fall 2006 EECS 314 Fall...

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EECS 314 Fall 2006 HW 01 Problem 1 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) Alexander Ganago The big picture This is a KCL problem. Refer to the sample problems in Lecture 1 notes. Problem In the circuit shown on the diagram, determine the unknown currents labeled as X, Y, and Z. Show your work. Your answers: X = ____________ A Y = ____________ A Z = ____________ A EECS 314 HW1 SOLUTIONS Fall 2006 1 of 24
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Problem 1 (Alex Kuo) The strategy for solving any KCL problem is to locate the node with only 1 unknown value. In this case it’s the node I at the bottom. Using the rules of KCL, with current going into the node as the positive orientation, we get 1A + Y + 8A + 4A = 0 Y = - 13A. Next we want to find the current W flowing through the circuit element at the bottom right, so we use KCL on node II. This gives us - 4 A + 20A + W = 0 W = - 16A. Notice that it is imperative to assign a direction to the current flow. I assigned the current W of the circuit element to flow into node II as positive. Naturally this assignment is arbitrary. This allows us to do another KCL at node III 3A + X + -Y + -W = 0. Since we know Y and W we get 3A + X + 13A + 16A = 0 X = -32A. To calculate the value of Z, we need to do KCL on nodes IV and V to get the values of U and V. -U + -15 + -3 = 0 I II W III IV U V V VI
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U = -18A U + -X + -1 + V = 0 V = -13A. Lastly we perform KCL at node VI and get Z + -V + -8A = 0 Z = -5A.
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EECS 314 Fall 2006 HW 01 Problem 2 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) © 2006 Alexander Ganago The big picture This is a KVL problem. Refer to sample solutions in Lecture 1 notes. Problem In the circuit shown on the diagram, determine the unknown voltages labeled as X, Y, and Z. Show your work. Your answers: X = ____________ V Y = ____________ V Z = ____________ V EECS 314 HW1 SOLUTIONS Fall 2006 4 of 24
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Problem 2 (Barry Mullins) KVL states that the voltages over elements arranged in a loop must sum to zero. It is important to pay attention to the orientation of the + and – labels, as these signs indicate how the voltages should be summed. 10 V 25 V 2 V 4 V – 16 V X Y Z A B C D E F Try to choose loops with only one unknown voltage. To find X: Start at node A and use KVL on loop ABCA. -10V + 2V + X = 0V X = 8V To find Y: Start at node E and use KVL on loop EABCDE. -4V – 10V + 2V +(-16V) – Y = 0V Y = -28V To find Z: Start at node E and use KVL on loop EAFE. -4V + 25V - Z = 0V Z = 21V
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EECS 314 Fall 2006 HW 01 Problem 3 Student's name ___________________________ Discussion section # __________ (Last name, first name, IN INK) © 2006 Alexander Ganago The big picture This problem requires KCL, KVL, and the passive sign convention. Refer to sample solutions in Lecture 2 notes.
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This note was uploaded on 09/23/2009 for the course EECS EECS314 taught by Professor Ganago during the Fall '09 term at University of Michigan.

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hmwk 1 sol - EECS 314 HW1 SOLUTIONS Fall 2006 EECS 314 Fall...

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