Circuit with Dependent Voltage source

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  • Thread starter BBJ
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BBJ
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Homework Statement



Find ix
i tried answering it i wasn't able to get it. The answer is given for basis by the way. (ix = 1.176 A) The image is attached to this post. Sorry for the inconvenience, it's my first time to post here.

Homework Equations



KCL(?)
KVL(?)

The Attempt at a Solution



I am not really sure on how to tackle this problem. my professor said we could use Thevenin's Theory to get Ix. What i did was to make the independent current source open, then I was left with the circuit with the 2 resistors and the dependent voltage source. I used KVL giving me loop 1 and an equation 5I1 + 10I1 +2Ix = 0.. that's where i get stuck.. i don't know what to do next..

i hope someone can help me out! thank you very much!
 

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Answers and Replies

  • #2
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In the circuit a current of 4A is given. Since there are two branches, and one branch has ix and the other has 2ix, the total current is 3ix. If 4A is divided by 3, then ix=4/3.
 
  • #3
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Are you sure that´s the right answer? Because if it is then KVL isnt true...

10(1.176)+5(2.824)-2(1.176) isnt equal to 0.

To me Ix = I1 - I2
I1 being th current around the first loop and I2 the current of the second loop.
That way Ix = -6.66666666666666666667 Amps and KVL stands.
 

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