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# 3-21 - Chapter 3 Solution 21 4 k v1 2 k v3 3v0 3v0 v2 v0 3...

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Chapter 3, Solution 21 Let v 3 be the voltage between the 2k Ω resistor and the voltage-controlled voltage source. At node 1, 12 = 3v 1 - v 2 - 2v 3 (1) At node 2, 3v 1 - 5v 2 - 2v 3 = 0 (2) Note that v 0 = v 2 . We now apply KVL in Fig. (b)
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