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Solution 4

# Solution 4 - ECE 205 Section A B and C Homework#4 Solutions...

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ECE 205 Section A, B, and C. Homework #4 Solutions 2.22 At the node, KCL requires that 0 0 v 2 10 4 v + + = 0 v 0 = –4.444V The current through the controlled source is i = 2V 0 = -8.888A and the voltage across it is v = (6 + 4) i 0 (where i 0 = v 0 /4) = 10 111 . 11 4 v 0 = Hence, p 2 v i = (-8.888)(-11.111) = 98.75 W 2.26, If i 16 = i o = 2A, then v = 16x2 = 32 V 8 4 A 8 v i = = , 4 2 8 A, i 16 4 2 v v i = = = = 2 4 8 16 16 8 4 2 30 A x i i i i i = + + + = + + + = 2 2 2 2 2 16 2 8 4 4 8 2 16 960 W P i R x x x x = = + + + = or 30 32 960 W x P i v x = = = 6 Ω 2v 0 + v 0 - 10A

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2.30. We start by combining the 6-ohm resistor with the 2-ohm one. We then end up with an 8-ohm
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