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Example3.2.4

# Example3.2.4 - At node C I 2 = I 3 I 4 ⇒ V B-V C R 2 = V...

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Circuit Analysis, 2008, Dr. J. Liu Example 3.2.4: problem—p. 1 Example 3.2.4 Consider the circuit below for which we know the following: V S 1 = 2 V and V S 2 = 4 V; R 1 = 1 Ω, R 2 = 2 Ω, R 3 = 3 Ω, R 4 = 4 Ω, and R 5 = 5 Ω. Find the power consumed by R 1 via node analysis. R 3 R 1 R 2 R 4 R 5 V S 1 V S 2 Figure 1:

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Circuit Analysis, 2008, Dr. J. Liu Example 3.2.4: solution—p. 1 Solution: R 3 R 1 R 2 R 4 R 5 I 3 A B C G + - + - + - + - I 4 V 3 V 4 I 5 V 5 D + - V 1 I 1 V S 1 V S 2 I 2 V 2 Figure 2: ND1 N N = 5: A, B, C, D, and G (ground); M V = 2: V S 1 and V S 2 . ND2 N N - M V - 1 = 2 independent variables: V B and V C ; M V = 2 dependent variables: V A = V S 1 , V D = V B - V S 2 . ND3 At the supernode containing nodes B, C, and D: I 1 = I 4 + I 5 V 1 R 1 = V 4 R 4 + V 5 R 5 (Component vtgs) V A - V B R 1 = V C R 4 + V D R 5 (Node vtgs) V S 1 - V B R 1 = V C R 4 + V B - V S 2 R 5 (Unknowns) ± 1 R 1 + 1 R 5 V B + V C R 4 = V S 1 R 1 + V S 2 R 5 . (1)
Circuit Analysis, 2008, Dr. J. Liu Example 3.2.4: solution—p. 2
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Unformatted text preview: At node C: I 2 = I 3 + I 4 ⇒ V B-V C R 2 = V C-V D R 3 + V C R 4 (Node vtgs) ⇒ V B-V C R 2 = V C-V B + V S 2 R 3 + V C R 4 (Unknowns) ⇒ ± 1 R 2 + 1 R 3 ¶ V B-± 1 R 2 + 1 R 3 + 1 R 4 ¶ V C = V S 2 R 3 . (2) ND4 Plugging the values into Eqns (1) and (2), we have ± 1 + 1 5 ¶ V B-V C 4 = 2 + 4 5 , ± 1 2 + 1 3 ¶ V B-± 1 2 + 1 3 + 1 4 ¶ V C = 4 3 . ⇒ V B = 2 . 473 V , V C = 0 . 672 V . ND5 P 1 = V 1 I 1 = V 2 1 R 1 = ( V A-V B ) 2 R 1 = 0 . 224 W . 2...
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Example3.2.4 - At node C I 2 = I 3 I 4 ⇒ V B-V C R 2 = V...

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