ECE201_Lecture07

ECE201_Lecture07 - 8 1 8 . 1 . 1 8 . 5 . 3 . 6 . 4 . 2 . ....

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1 EE201 Linear Circuit Analysis I Lecture 7. Topics: Nodal Analysis 1. Nodal analysis with floating sources — supernodes Example 1 . (1) What is a floating source? Neither terminal is connected to ground node E CCVS, 8i y The problem with a floating source: can’t write Ohm’s law around a floating source The solution: use Gaussian surface and supernode around it (2) Known: V A = 50 v Unknown: V B . V C need 2 eqns. At the Supernode: Inside the Supernode: In matrix form: 3 1 5 1 V B V C = 125 0 V B V C = 1 8 1 1 5 3 125 0 = 1 8 125 625 = 125 / 8 625 / 8 + - A B C E 50 V + V x 0.5 mho + 0.5 mho 8i y 1 mho 2V x i y
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2 Example 2 . Find V a , V b , V c . At Node a: At the Supernode: Inside the Supernode: Matrix form: =
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Unformatted text preview: 8 1 8 . 1 . 1 8 . 5 . 3 . 6 . 4 . 2 . . 1 c b a V V V , From Matlab: = V 8 . 17 V . 22 V 5 . 12 c b a V V V 3A b c E 10 V + +-a 2I x 5A 0.4 mho 0.1 mho 0.2 mho 0.4 mho I x 0.1 mho 3 2. Practice Problems (from old exams) (1) For the circuit shown above: (a) Find V A (b) Find V C (c) Write a node equation at node B. Group like-terms and simplify. (d) Write a node equation at node D. Group like-terms and simplify. (e) Put in matrix form and solve for V B and V D . A B C D 12 V 2 A 20 V 4 A E (reference node) G1=0.25 mho G2=0.1 mho G3=0.75 mho G4=1 mho G5=1 mho G6=0.5 mho 4 (2) Find R eq in the circuit shown below. 5 (3) Find the equivalent resistance below as seen from A-B....
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ECE201_Lecture07 - 8 1 8 . 1 . 1 8 . 5 . 3 . 6 . 4 . 2 . ....

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