Circuit with Dependent Voltage source

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SUMMARY

The discussion focuses on calculating the current ix in a circuit with a dependent voltage source. The correct value of ix is established as 1.176 A, derived using Thevenin's Theorem and Kirchhoff's Voltage Law (KVL). Participants explore various approaches, including the application of KCL and KVL, and the relationship between currents in different branches of the circuit. The confusion arises from differing interpretations of the circuit's behavior and the application of KVL.

PREREQUISITES
  • Thevenin's Theorem
  • Kirchhoff's Voltage Law (KVL)
  • Kirchhoff's Current Law (KCL)
  • Basic circuit analysis techniques
NEXT STEPS
  • Study Thevenin's Theorem applications in dependent source circuits
  • Review Kirchhoff's Voltage Law with complex circuits
  • Practice circuit analysis problems involving dependent sources
  • Explore current division in parallel circuits
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing circuits with dependent sources will benefit from this discussion.

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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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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.
 
Are you sure that´s the right answer? Because if it is then KVL isn't true...

10(1.176)+5(2.824)-2(1.176) isn't 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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