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Circuit Analysis - Source Transformation

  • Engineering
  • Thread starter timeforplanb
  • Start date
  • #1

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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Answers and Replies

  • #2
gneill
Mentor
20,792
2,770
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:
  • #3
thank you very much for the verification sir!
 

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