Exam1_solution

Exam1_solution - We first combine the two resistors in...

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ENGR 204/F Spring 2009 Wednesday, February 11, 2009 EXAM 1 Name: _____________________________________________ SSN#: -_ _ _ _ Last First Problem 1 Calculate v and i x in the circuit below. Solution: For loop 1, –12 + v +2 = 0, v = 10 V For loop 2, –2 + 8 + 3i x =0, i x = –2 A + 12 V 3 i x + 2 V _ + v 8 V 12 Ω + _ i x + _
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Problem 2 Find V x in the circuit below. Figure 2.85 For Prob. 2.21. Applying KVL, -15 + (1+5+2)I + 2 V x = 0 But V x = 5I, -15 +8I + 10I =0, I = 5/6 V x = 5I = 25/6 = 4.167 V 1 Ω 2 Ω + 5 Ω + _ 15 V V x 2 V x _ +
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Problem 3 Find v 1 , v 2 , and v 3 in the circuit below, if R 1 =14 , R 2 ,=15 , R 3 =10 and V s =40V. Solution
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Unformatted text preview: We first combine the two resistors in parallel = 10 15 6 We now apply voltage division, v 1 = = + ) 40 ( 6 14 14 28 V v 2 = v 3 = = + ) 40 ( 6 14 6 12 V Hence, v 1 = 28 V , v 2 = 12 V, v 3 = 12 V Problem 4 Find V in the circuit below and the power dissipated by the controlled source Chapter 2, Solution 22 At the node, KCL requires that v 2 10 4 v + + = 0 v = 4.444V The current through the controlled source is i = 2V = -8.888A and the voltage across it is v = (6 + 4) i (where i = v /4) = 10 111 . 11 4 v-= Hence, p 2 v i = (-8.888)(-11.111) = 98.75 W...
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This note was uploaded on 02/20/2010 for the course EE Electrical taught by Professor Unknown during the Spring '10 term at City.

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Exam1_solution - We first combine the two resistors in...

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