Solving for multiple unkowns using KVL and KCL equations

  • Thread starter fredro
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  • #1
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Homework Statement


I am asked to solve for vX, PDEV BY iS1, P ABS BY vS3 (please see attached figure). We have only covered KCL, KVL, Ohm's law and Power in class. No Thevinen theorem, so it can't be used to help solve.


The Attempt at a Solution


I know that current will not pass through the 4 Ohm resistor or vS4. My initial hunch was to solve for iQ in terms of vS1 (upper left of figure) and then use that equation to help solve for a KVL with the 3 and 7 ohm resistors and vS3. The result was not pretty and very wrong.

Homework Statement





Homework Equations





The Attempt at a Solution

 

Attachments

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Answers and Replies

  • #2
gneill
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Yeah, it's going to get messy. There are a lot of interrelationships to account for with all the controlled sources. However, to make a start, choose a ground node, label the other important nodes, and see what you can do about writing equations for their voltages in terms of the given information. Here's a suggestion:
attachment.php?attachmentid=38590&stc=1&d=1315258755.gif


The voltage at node D (which is the same as -vQ) will be important. You can write a KCL equation for that node.
 

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  • #3
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Here's what I have so far. Unfortunately, I keep running into nonsensical, trivial answers where my systems keep having coefficients of zero. :confused: I've spent several hours on this problem.
 

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  • #4
gneill
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Watch out for the polarities of things. The diagram shows vQ to be the negative of the voltage at your KCL1 node (note the +/- sign locations).

Suppose you take the voltage at node D to be vD. Then [itex] v_Q = -v_D[/itex]. Can you write expressions for the node voltages at A and B (with respect to the ground node that I indicated) in terms of this vD ? (You may need to keep vx as a variable for now).
 
  • #5
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Here's what I have so far. Unfortunately, I keep running into nonsensical, trivial answers where my systems keep having coefficients of zero. :confused: I've spent several hours on this problem.
 

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