W08 314 HW01 solutions

W08 314 HW01 solutions - EECS 314 Winter 2008 Problem 1 HW...

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EECS 314 Winter 2008 HW 01 Solutions Problem 1 KCL, KVL, and the Passive Sign Convention Consider the circuit shown on this diagram. Note that nodes A, C and B are all connected with wires thus they are in fact one node! Some currents and some voltages in the circuit are given in the table below and on the diagrams on the following page; your challenge is to calculate the unknown voltages and currents, as well as the power absorbed by each element. Element 1 2 3 4 5 6 7 Current through 4 A 2 A 8 A 3 A 5 A 6 A 6 A Voltage across 5 V – 1 V – 2 V 8 V – 8 V 6 V 5 V Power absorbed 20 W – 2 W –16 W 24 W 40 W –36 W –30 W Note to the grader: In the table above, the currents through elements 3, 6, and 7 can be shown with any sign, because the students choose the reference marks themselves. However, the signs of all voltages and of all powers should be exactly as shown above. On the diagram to the right, the elements supplying electric power to the circuit are circled. The answers can be obtained in many ways (for example, you can start writing KCL equation for another node); the solutions below show one possibility. © 2008 Alexander Ganago
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EECS 314 Winter 2008 HW 01 Solutions Part 1 (10 points) Currents through 4 elements out of 7 are given (see the diagram below, on the right). Use KCL to calculate the unknown currents. Record your results in the table above (do not omit the units of measure). Show the directions of the currents on the diagram below (on the left). Many solutions are possible. Below is an example of step-by-step (substitution) method. At the top node, two outgoing currents 4 A and 2 A should match the incoming current, which must be 6 A. At the bottom left node, two incoming currents 4 A and 3 A must match the outgoing current of 7 A. At the bottom right node, 7 A is incoming and 6 A is outgoing (toward the top node) thus another outgoing current (toward the center of the circuit) must equal 1 A. At the node, from which 3 A and 5 A are outgoing, the incoming current must be 8 A. At node B, where the incoming current of 5 A is given, we already found the other incoming current of 1 A thus the outgoing current must equal 6 A. Note that, besides the currents through circuit elements, we had to find the current through the wires (such as the bottommost in the circuit). © 2008 Alexander Ganago Page 2 of 21
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EECS 314 Winter 2008 HW 01 Solutions hown Part 2 (10 points) Voltages across 3 elements out of 7 are given on the diagram. Use KVL to calculate the unknown voltages labeled V1, V2, V3, and V5, with the reference marks already s on the diagram. Record your results in the table above (do not omit the units of measure). Again, many solutions are possible. For example, we can start at the bottom mesh: – 8 V – V5 = 0 thus V5 = – 8 V (shown on the solution diagram with positive voltage value, which required the change of reference marks).
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This homework help was uploaded on 04/10/2008 for the course EECS 314 taught by Professor Ganago during the Winter '07 term at University of Michigan.

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W08 314 HW01 solutions - EECS 314 Winter 2008 Problem 1 HW...

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