# Fundamentals of Electric Circuits, 2nd Edition - Solutions by Charles K. Alexander, Matthew N. O. Sadiku

By Charles K. Alexander, Matthew N. O. Sadiku

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Extra resources for Fundamentals of Electric Circuits, 2nd Edition - Solutions Manual

Example text

73 A 4 (2) Chapter 3, Solution 18 –+ v2 v1 2Ω 5A v3 2Ω 8Ω 4Ω 10 V + + v1 v3 – (a) At node 2, in Fig. (a), 5 = At the supernode, – (b) v 2 − v1 v 2 − v3 + 2 2 10 = - v1 + 2v2 - v3 v 2 − v1 v 2 − v 3 v1 v 3 + = + 2 2 4 8 From Fig. 267 V Chapter 3, Solution 20 Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence V1 V2 V3 + + =0  → V1 + 4V2 + V3 = 0 (1) 4 1 4 . V1 4Ω . V2 1Ω 2Ω V3 4Ω Between nodes 1 and 3, − V1 + 12 + V3 = 0  → V3 = V1 − 12 Similarly, between nodes 1 and 2, V1 = V2 + 2i But i = V3 / 4 .

100 Ω a 100 Ω a 100 Ω 100 Ω 100 Ω 100 Ω b 100 Ω 200 Ω 100 Ω 100 Ω 200 Ω b R1 = 100 Ω 100 Ω 100 Ω 100 Ω 100 Ω 100 Ω R3 R2 100x 200 + 200x 200 + 200 x100 80000 = = 800Ω 100 100 R2 = R3 = 80000/(200) = 400 100x 400 But 100 400 = = 80Ω 500 We connect the ∆ to Y. 100 Ω a b 100 Ω a 100 Ω 100 Ω 80 Ω 100 Ω 100 Ω 80 Ω 800 Ω b 100 Ω 100 Ω Rb 100 Ω 100 Ω Rc 80 x800 64,000 400 = = Ω 80 + 80 + 800 960 3 80x80 20 = Ω Rb = 960 3 Ra = Rc = We convert T to ∆ . 36 Converting the Ts to ∆ s, we have the equivalent circuit below.

267 V Chapter 3, Solution 20 Nodes 1 and 2 form a supernode; so do nodes 1 and 3. Hence V1 V2 V3 + + =0  → V1 + 4V2 + V3 = 0 (1) 4 1 4 . V1 4Ω . V2 1Ω 2Ω V3 4Ω Between nodes 1 and 3, − V1 + 12 + V3 = 0  → V3 = V1 − 12 Similarly, between nodes 1 and 2, V1 = V2 + 2i But i = V3 / 4 . Combining this with (2) and (3) gives . 5V, V3 = −15V Chapter 3, Solution 21 4 kΩ v1 2 kΩ v3 3v0 + 3v0 v2 + v0 3 mA – 1 kΩ + + + v3 v2 – – (b) (a) Let v3 be the voltage between the 2kΩ resistor and the voltage-controlled voltage source.