W09 314 HW04 solution

# W09 314 HW04 solution - EECS 314 Winter 2009 Homework set 4...

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EECS 314 Winter 2009 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 1 solution First: Choose non-reference nodes as nodes A, B, and C. The other node is the reference node. Part 1: Use KVL to find voltage at node A: -48V + (V A – V B ) + V B = 0 => -48V + V A = 0 => V A = 48V Part 2: Use KCL to get node voltage equations for nodes B and C For node B: (V B – V C )/30 – (0- V B )/20 - (V A – V B )/50 = 0 => (V B – V C )/30 + V B /20 - (48 - V B )/50 = 0 => V B *(1/20 + 1/50 - 1/30) -48/50 – V C /30 =0 =>V B *(15/300 + 6/300 + 10/300) - .96 – V C /30 = 0 => V B *(31/300) -.96 –V C /30 = 0 => 31*V B -288 – 10*V C = 0 => 31*V B – 10*V C = 288 For node C: 3 - (V A – V C )/70 - (V B – V C )/30 = 0 => 3 - (48 - V C )/70 - (V B -V C )/30 = 0 => 3 + V C *(1/70 + 1/30) – 48/70 – V B /30 = 0 => 3 + V C *(10/210) -48/70 –V B /30 = 0 => 630 + 10*V C -144 – 7V B = 0

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=> 10*V C – 7*V B = -486 This gives two equations and two unknowns 31*V B -10*V C = 288 -7*V B + 10*V C = -486 Add equations together: 24*V B = -198 => V B = -198/24 => V B = -8.25 V Substitute V B into one of the equations and solve Using Equation 2: => -7*V B + 10*V C = -486 => -7*(-8.25) + 10*V C = -486 => V C = (-486 – 57.75)/10 = -54.38 V Using Equation 1: => 31*V B - 10*V C = 288 => 31*(-8.25) - 10*V C = 288 => V C = -(288 + 31*(8.25))/10 = -(288 + 255.75)/10 = -54.38 V Part 3: P 48V = V*I 48V = 48 V*-((48V – -54.38 V)/70 + (48 V – -8.25 V)/50 ) = 48 V*-(1.4626 A + 1.125 A) = 48 V*-2.5876 A = -124.2 W P 40 = I 2 *R = (3) 2 *40 = 360 W Part 4: => I 20 = (0 - V B )/20 = (0 - -8.25)/20 = .4125 A I 40 = 3 A I 70 = (V A – V C )/70 = (48V – -54.38 V)/70 = 1.4626 A I 50 = (V A - V B )/50 = (48 – -8.25)/50 = 1.125 A I 30 = I 50 + I 20 = 1.125 A + .4125 A = 1.5375 A I 48V = I 70 + I 50 = 2.5876 A I 70 + I 30 = 1.4626 A + 1.5375 A = 3.0001 A =3 A (Everything Matches Up)
EECS 314 Winter 2009 Homework set 4 Student’s name ___________________________ Discussion section # _______ (Last, First, write legibly, use ink) (use ink) Instructor is not responsible for grading and entering scores for HW papers lacking clear information in the required fields above Problem 2 The Mesh Current Method for solving circuits The Big Picture If connections of the circuit elements cannot be reduced to a combination of series and/or parallel, we have to use KCL, KVL, and Ohm's law to solve for the currents and voltages. (Note that we always use KCL, KVL, and Ohm's law: for example, to derive the familiar formulas for the voltage and current division in circuits with series/parallel connections.) Here we use the mesh current equations. The strategy is, step by step: (1) Identify and carefully label the meshes in the circuit. Note that, for mesh current solution, there is no need to identify the reference (ground) node in the circuit. (2) If a mesh contains a current source, determine the mesh current from KCL. (3) For each mesh that does not contain a current source, write a KVL equation for the voltages around the mesh. (4) For each resistor that belongs to two neighboring meshes, find the current through KCL and find the voltage through Ohm’s law. Note that, at this moment, you use mesh currents as reference marks, without any worry of which way the currents actually flow.

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## This note was uploaded on 10/28/2009 for the course EECS 314 taught by Professor Ganago during the Winter '07 term at University of Michigan.

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W09 314 HW04 solution - EECS 314 Winter 2009 Homework set 4...

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