4-63 - 1 /30) 0.5v o = (1/30) 0.5x2/3 = 0.03333 0.33333 =...

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Chapter 4, Solution 63. Because there are no independent sources, I N = I sc = 0 A R N can be found using the circuit below. Applying KCL at node 1, v 1 = 1, and v o = (20/30)v 1 = 2/3 i o = (v
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Unformatted text preview: 1 /30) 0.5v o = (1/30) 0.5x2/3 = 0.03333 0.33333 = 0.3 A. Hence, R N = 1/(0.3) = 3.333 ohms 20 0.5v o v 1 + v o 1V + 10 i o...
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This note was uploaded on 08/27/2011 for the course EEL 3111C taught by Professor Srivastava during the Spring '08 term at University of Florida.

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