Lab4

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Defense Acquisition University *

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EET310

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Electrical Engineering

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Dec 6, 2023

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docx

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5

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Lab 4 - Node voltage analysis Please download the Multisim Lab file. Download Multisim Lab file. Solve for node voltages and then connect your measuring instruments to verify your calculations. Objective: Solve the circuit in the lab using Node Voltage Analysis following the below steps: Identify the number of nodes There are two nodes B and C. Assign each node a voltage notation. V B and V C I 2 I 1 I L I 4 I 3 I 3
Write the characteristic equations for each node by applying KCL and assuming all currents are leaving the node. At node “C”: I 2 + I L = I 4 At node “B”: I 1 – I L – I 2 I 1 = (12V – V B ) / 0.330 I 2 = (V B – 0) / 0.300 I 3 = (12V – V C ) / 0.360 I 4 = (V C – 0) / 0.390 I L = (V B – V C ) / 1 Solve the simultaneous equations to calculate the node voltages. And then calculate the current in each resistor. I 3 + I L = I 4 (12V – V C ) / 0.360 + (V B – V C ) / 1 = (V C – 0) / 0.390 = 0 (12V / 0.360) – (V C / 0.360) + V B – V C – (V C /0.390) = 0 (12V / 0.360) + V B – V C (1 / 0.360 + 1 + 1 / 0.390) = 0 33.33 + V B – V C (2.78 + 1 + 2.56) = 0 33.33 + V B – V C (6.34) = 0 33.33 + V B – 6.34V C = 0 – 33.33 = V B – 6.34V C I 1 = I 2 – I L (12V – V B ) / 0.330 = (V B – 0) / 0.300 – (V B – V C ) / 1 (12V / 0.330) – (V B / 0.330) – (V B /0.300) – V B + V C = 0 36.36 – V B (1 / 0.330 – 1 / 0.300) – V B + V C = 0 36.36 – V B (3.03 + 3.33 + 1) + V C = 0 36.36 – V B (7.36) + V C = 0 36.36 = 7.36V B – V C – 33.33 = V B – 6.34V C 36.36 = 7.36V B – V C [ 1 6.34 7 . 36 1 ] [ v B v C ] = [ 33.33 36.36 ] Matrix A Matrix B 1 -6.34 -33.33 7.36 -1 36.36 INV of A sol of X -0.0219 0.13885 5.7783 - 0.16118 0.0219 6.1685
VB = 5.78V VC = 6.17V I 1 = (12V – V B ) / 0.330 I 1 = (12V – 5.78V) / 0.330 I 1 = 6.22 / 0.330 I 1 = 18.9 mA I 2 = (V B – 0) / 0.300 I 2 = (5.78V – 0) / 0.300 I 2 = 5.78V / 0.300 I 2 = 19.3 mA I 3 = (12V – V C ) / 0.360 I 3 = (12V – 6.17V) / 0.360 I 3 = 5.83 / 0.360 I 3 = 16.2 mA I 4 = (V C – 0) / 0.390 I 4 = (6.17V – 0) / 0.390 I 4 = 6.17V / 0.390 I 4 = 15.8 mA I L = (V B – V C ) / 1000 I L = (5.78V – 6.17V) / 1 I L = – 0.39 / 1000 I L = – 0.39 µA
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