Homework 3 Questions

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University Of Connecticut *

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2000

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Electrical Engineering

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Jan 9, 2024

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1. Award: 10.00 points Consider the circuit below. All voltage sources have the same value, V, and all resistors have the same value, R. The voltage across R5 is what percentage of V? Enter your answer as a whole number without the % sign. (Example : If the voltage across R5 is 15% of V, then enter 15 in the box.) 80 ± 2%

2. Award: 15.00 points In the circuit given below, R = 6 Ω. Find v 0 using nodal analysis. The equation at node v 1 is: where A = v1 – 100 and B = v1 – vo . The equation at node v 0 is: where C = vo – v1 , D = 4 , and E = 6 . The value of v 0 in the circuit is 44.44 ± 2% V. Explanation: + + 12.5 = 0 1 2 − 12.5 + + = 0 2 v 0 + 50 v 0

At node 1, At node 0 , Solving Eq.(1) and Eq. (2), v 1 = 73.15 V and v 0 = 44.44 V. The equation at node 1 is . The equation at node 0 is . The value of v 0 in the circuit is 44.44 V. References Worksheet Difficulty: Medium Learning Objective: Understand Kirchhoff’s current law. + + 12.5 A = 0 − 100 V v 1 1 Ω − v 1 v 0 2 Ω 1 + 0.5 − 0.5 = 100 A − 12.5 A v 1 v 0 1.5 − 0.5 = 87.5 (1) v 1 v 0 − 12.5 A + + = 0 − v 0 v 1 2 Ω − 0 V v 0 4 Ω + 50 V v 0 6 Ω + + − 0.5 = 12.5 A − 1 2 Ω 1 4 Ω 1 6 Ω v 0 v 1 50 V 6 Ω −0.5 + 0.917 = 4.167 (2) v 1 v 0 + + 12.5 A = 0 − 100 V v 1 1 Ω − v 1 v 0 2 Ω − 12.5 A + + = 0 − v 0 v 1 2 Ω − 0 V v 0 4 Ω + 50 V v 0 6 Ω

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3. Award: 15.00 points In the circuit given below, R = 18 Ω. Determine the node voltages. The equation at node 1 is where A = 18 , B = 40 , C = (v1 + 300) – v2 , and D = 40 . The equation at node 2 is where E = 40 and F = v2 – (v1 + 300) . The value of node voltage v 1 in the circuit is 93.10 ± 2% V. The value of node voltage v 2 in the circuit is 543.10 ± 2% V. The value of node voltage v 3 in the circuit is 393.10 ± 2% V. Explanation: First we identify the unknown nodes and find that there really are only two unknown nodes v 1 and v 2 since v 3 = v 1 + 300 V (essentially a supernode). At node 1, + + + = 0 v 1 − v 1 v 2 40 + 300 v 1 − 15 + = 0 − v 2 v 1 40 + + + = 0 − 0 V v 1 18 Ω − v 1 v 2 40 Ω + 300 V − v 1 v 2 40 Ω + 300 V − 0 V v 1 40 Ω 0.131 − 0.05 = − 15 (1) v 1 v 2

At node 2, Solving Eq. (1) and Eq. (2), v 1 = 93.1 V and v 2 = 543.1 V. Finally, v 3 = v 1 + 300 V = 93.1 V + 300 V = 393.1 V The equation at node 1 is . The equation at node 2 is . The value node voltage v 1 in the circuit is 93.1 V. The value node voltage v 2 in the circuit is 543.1 V. The value node voltage v 3 in the circuit is 393.1 V. References Worksheet Difficulty: Medium Learning Objective: Understand Kirchhoff’s current law. − 15 A + = 0 − v 2 v 1 40 Ω − + 300 V v 2 v 1 40 Ω −0.05 + 0.05 = 22.5 (2) v 1 v 2 + + + = 0 − 0 V v 1 18 Ω − v 1 v 2 40 Ω + 300 V − v 1 v 2 40 Ω + 300 V − 0 V v 1 40 Ω − 15 A + = 0 − v 2 v 1 40 Ω − + 300 V v 2 v 1 40 Ω

4. Award: 15.00 points Using node voltage analysis, find the voltages V 1 and V 2 for the circuit of the given figure. The voltages V 1 and V 2 are 5.41 ± 2% V and 3.24 ± 2% V. Explanation: Applying KCL at each node, Rearranging, Solving, V 1 = 5.41 V and V 2 = 3.24 V The voltages . References Numeric Response Difficulty: Easy + + = 0 −20 V 1 30 V 1 20 − V 1 V 2 10 + + = 0 V 2 30 V 2 30 − V 2 V 1 10 5.5 − 3 = 20 V 1 V 2 − 3 + 5 = 0 V 1 V 2 = 5.41 V and = 3.24 V V 1 V 2

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5. Award: 15.00 points Use nodal analysis to find node voltages V 1 , V 2 , and V 3 in the given figure. Let R 1 = 16 Ω, R 2 = 12 Ω, R 3 = 13 Ω, R 4 = 10 Ω, I 1 = 2 A, I 2 = 1 A. The voltages are: V 1 = 19.8823 ± 2% V V 2 = 12.3529 ± 2% V V 3 = -5.2941 ± 2% V Explanation: Using the node analysis: The linear system of equations are solved using the matrix form and the values are found to be: V 1 = 19.88235 V V 2 = 12.35294 V V 3 = -5.29412 V The voltages are: V 1 = 19.88235 V V 2 = 12.35294 V V 3 = -5.29412 V + + = 0 −I 1 V 1 R 3 − V 1 V 2 R 1 + = − V 2 V 1 R 1 − V 2 V 3 R 2 I 2 + = − − V 3 V 2 R 2 V 3 R 4 I 1

References Numeric Response Difficulty: Medium