EE2111_A03_SOL2

EE2111_A03_SOL2 - EE2111 Homework #2 A03 Assigned 04...

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EE2111 Homework #2 A03 Assigned 04 September 03 Due 10 September 03 Problem 1 . Hambly P2.37 Solution: First, we can write 10 1 v i x = Then since x i v v 5 2 1 = , we have 2 10 5 1 1 2 1 v v v v = = or 2 1 2 v v = . Now we will do nodal analysis at nodes I and 2. For node 1 we have 0 10 5 12 1 = + + i v Applying KCL at node 2 we have 0 3 20 2 12 = + v i Adding these 2 equations we get 0 3 20 10 5 2 1 = + + v v or 160 2 2 1 = + v v
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Thus we solve the 2 equations 160 2 0 2 2 1 2 1 = + = v v v v to get 64 1 = v V and 32 2 = v V. Finally 4 . 6 10 1 = = v i x A. Problem 2. Hambly P2.39 Solution: Label the left node 1 and the right node 2 and insert a 1 A current source between a and b. There is a dependent current source in this problem. We have 5 5 2 1 v v v i x s = = Now apply KCL at nodes 1 and 2 0 5 5 10 1 10 20 2 1 2 1 2 2 1 1 = + + = + v v v v v v v v or 0 20 2 3 1 2 2 1 = + = v v v v Solve these 2 equations to find 4 1 = v V The equivalent resistance is equal in value to 1 v hence = 4 eq R .
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This note was uploaded on 10/13/2009 for the course ECE 2111 taught by Professor Labonte during the Spring '09 term at WPI.

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EE2111_A03_SOL2 - EE2111 Homework #2 A03 Assigned 04...

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