Circuit Analysis - Source Transformation

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SUMMARY

The discussion focuses on using source transformation to find the voltage Vx in a circuit analysis problem. The participant applied Kirchhoff's Voltage Law (KVL) and derived the current I as 1.222 A. The final calculation for Vx was confirmed to be 6.11 V, correcting the initial confusion regarding its polarity and value. The source transformation process was validated, emphasizing the importance of correctly interpreting the direction of current sources when converting to voltage sources.

PREREQUISITES
  • Understanding of Kirchhoff's Voltage Law (KVL)
  • Familiarity with source transformation techniques in circuit analysis
  • Basic knowledge of Ohm's Law (V=IR)
  • Ability to analyze circuits with resistors and current sources
NEXT STEPS
  • Study advanced source transformation techniques in circuit analysis
  • Learn about mesh analysis and its application in circuit problems
  • Explore the implications of current source polarity in voltage source conversion
  • Review examples of KVL applications in complex circuits
USEFUL FOR

Electrical engineering students, circuit analysts, and anyone involved in solving circuit problems using source transformation and KVL.

timeforplanb
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Homework Statement


Find Vx using source transformation
attachment.php?attachmentid=41205&stc=1&d=1322126072.jpg


Homework Equations


V=IR
KVL

The Attempt at a Solution


After doing source transformation on the three pairs of current sources and resistors...
attachment.php?attachmentid=41207&stc=1&d=1322126953.jpg


KVL:
-6+2I-Vx+4I-3+3I+5I-2=0
14I-Vx-11=0
but Vx=5I (is it? I'm not so sure about this)
14I-5I-11=0
1I=11
I=1.222 A

Vx=-5I=-5(1.222)
Vx=-6.11 V

Did I do the source transformation right? Also, I'm confused with the value of Vx. Is it -5I or just 5I? Because in our class, our convention is that when a current source is transformed into a voltage source, the part where the arrow head in the current source becomes the positive terminal for the voltage source. But since we're only dealing with a resistor here, I just followed our convention for the KVL (the voltage across 5 ohms is positive since the mesh current enters the positive terminal)
 

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Last edited:
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Everything looks fine except for your last two lines! Vx is in fact 5I, since its specified polarity indicates that it's a drop in the same direction as your current. So Vx = 6.11 V.
 
Last edited:
thank you very much for the verification sir!
 

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