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Voltage Division - AC cct.

  • Thread starter Ammar w
  • Start date
  • #1
28
0

Homework Statement


Hi,
We have this cut. :

Screen%20Shot%202013-02-26%20at%204.24.05%20PM.png


and in the image below, I10 is calculated by using current division :

Screen%20Shot%202013-02-26%20at%204.24.30%20PM.png


The Attempt at a Solution



I tried to find it using voltage division but the answer is wrong :
Vx(of the upper node) = 100 * [tex]\frac{j5}{4+j5}[/tex] = 60.98 + j48.78 V

=> I10 = [tex]\frac{Vx}{10 - j5}[/tex] = 2.93 + 6.34 = 6.98 [itex]\ 65.2[/itex]

which is wrong, WHY?

thanks
 

Answers and Replies

  • #2
gneill
Mentor
20,795
2,773

Homework Statement


Hi,
We have this cut. :

Screen%20Shot%202013-02-26%20at%204.24.05%20PM.png


and in the image below, I10 is calculated by using current division :

Screen%20Shot%202013-02-26%20at%204.24.30%20PM.png


The Attempt at a Solution



I tried to find it using voltage division but the answer is wrong :
Vx(of the upper node) = 100 * [tex]\frac{j5}{4+j5}[/tex] = 60.98 + j48.78 V

=> I10 = [tex]\frac{Vx}{10 - j5}[/tex] = 2.93 + 6.34 = 6.98 [itex]\ 65.2[/itex]

which is wrong, WHY?
Your voltage division isn't taking into account the impedance of the 10Ω and -5j capacitor branch that's also connected at the Vx node. You might try applying nodal analysis to find Vx...
 
Last edited:
  • #3
28
0
oh
thank you gneill
 

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