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Unformatted text preview: ECE 1100 Homework #6 Shattuck Section Solve each of the circuits given below. Circuit #1: Find v X . i S1 = 2[A] R 1 = 33[ Ω ] R 2 = 22[ Ω ] v X + Solution: The first step in such problems is to define variables. We will do this in the diagram that follows. i S1 = 2[A] R 1 = 33[ Ω ] R 2 = 22[ Ω ] v X + i R1 i R2 v R1 + We have defined two additional currents, and one additional voltage. The voltage v R1 that has been defined will be addressed first. If we write KVL around the loop shown as a red dashed line below, we get 1 1 0, or . X R X R v v v v = = ECE 1100 – Homework #6 – Solution From this, we might recognize that the voltage across all three elements is the same, and is v X . This is true, because wires connect both ends of each element. These wires have zero voltage across them. i S1 = 2[A] R 1 = 33[ Ω ] R 2 = 22[ Ω ] v X + i R1 i R2 v R1 + Now, let’s write KCL for the closed surface shown as a green line. We get 1 2 1 0. S R R i i i + = Next, we substitute in for each of these terms. We know the value of the current source. We can substitute for the other two currents using Ohm’s Law for the resistors, and get 2[A] 0. 22[ ] 33[ ] X X v v +  = ÷ Ω Ω Here we should note that for resistor R 1 , we have defined the reference polarities in the active sign convention, and so there needs to be a negative sign in that Ohm’s Law equation. We can solve this equation for v X , and we get 1 1 2[A], or 22[ ] 33[ ] X v + =  ÷ Ω Ω 2 6 .4[ V ] . X v =  2 ECE 1100 – Homework #6 – Solution Circuit #2: Find i X . 2.2[k Ω ] 820[ Ω ] 10[V] + 4.7[k Ω ] 1[k Ω ] i X Again, our first step is to define some variables. We will choose to use some insight here, and only define one current through the 1[k Ω ] and 820[ Ω ] resistors, and one current through the voltage source and the 4.7[k Ω ] resistor. Really, we only know this is true after we write KCL at the node connecting the elements, but having done this a couple of times, we will choose to skip this KCL step. You may choose to take either approach; you get the same answer either way. We have the following circuit....
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This note was uploaded on 10/29/2009 for the course ECE 1100 taught by Professor Staff during the Spring '08 term at University of Houston.
 Spring '08
 Staff

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