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Node voltage with dependent voltage source

  • Thread starter bnosam
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  • #1
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Homework Statement


http://i62.tinypic.com/245fwxu.jpg[/B]


Homework Equations

& Attempt to solve
[/B]
http://i62.tinypic.com/142v6mv.jpg

Bottom: A
Top left: B
Top Right: C


(1/100 + 1/5 + 1/25)VB - (1/25)VC + 6.25I = 450 mA

(20.8)VB - 0.04 VC + 6.25I = .450 A

VC = 45 + 6.25I

I know I'm doing something wrong here. VC doesn't seem right to me.
 

Answers and Replies

  • #2
NascentOxygen
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(1/100 + 1/5 + 1/25)VB - (1/25)VC + 6.25I = 450 mA
➊ This line is full of mistakes. Start again, explaining each term. If you are trying to do a lot in one line to save writing, then that's why you are making mistakes.

➋ Is the 6.25I a dependent voltage source or current source? (It's difficult to deduce from your working.)
 
  • #3
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➊ This line is full of mistakes. Start again, explaining each term. If you are trying to do a lot in one line to save writing, then that's why you are making mistakes.

➋ Is the 6.25I a dependent voltage source or current source? (It's difficult to deduce from your working.)
The way we always did it in my circuits class was put the thing straight into a matrix from the circuit by inspection. But that doesn't work with dependent sources as easily so in this class we abandoned that.

6.25I is a dependent voltage source.


So I thought from node A at the bottom to the first Node B at the top:

The resistors connected to it would be:

(1/100 + 1/5 + 1/25) *VB

And the common resistor is -(1/25) VC
 
  • #4
NascentOxygen
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The resistors connected to it would be:

(1/100 + 1/5 + 1/25) *VB

And the common resistor is -(1/25) VC
I don't understand this.

Try applying KCL clearly, expressing current leaving the node = current entering the node
 
  • #5
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We have to use the node voltages method for this.
 
  • #6
NascentOxygen
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We have to use the node voltages method for this.
Assume a node voltage, then sum the currents into that node.
 
  • #7
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So if I understand then I still get without adding the dependent source.
VB/(100) + VB/(5) + (VB-VC)/25 = .45 A
 
  • #8
NascentOxygen
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So if I understand then I still get without adding the dependent source.
VB/(100) + VB/(5) + (VB-VC)/25 = .45 A
It will be something like that. But you haven't correctly expressed the current through the 5 ohm resistor.
 
  • #9
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It will be something like that. But you haven't correctly expressed the current through the 5 ohm resistor.
would the 5 also contain a term for the dependent source?
 
  • #10
NascentOxygen
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would the 5 also contain a term for the dependent source?
Ohm's law always applies: current through a resistor = voltage difference across the resistor / R
You need to involve the dependent source, yes, it controls the voltage at one end of the resistor.
 
  • #11
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VB/(100) + (VB-6.25I)/(5) + (VB-VC)/25 = .45 A

Is that what you mean?
 
  • #12
NascentOxygen
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VB/(100) + (VB-6.25I)/(5) + (VB-VC)/25 = .45 A

Is that what you mean?
Yes. And you know I in terms of VB and VC. And you know the value of VC.
So that seems like enough equations to be able to solve for all unknowns.
 
  • #13
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VC = 45 V right?
 
  • #14
NascentOxygen
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  • #15
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So when I solve I get : VB = 5I + 9


I'm not really sure what I had for I/what I is
 
  • #16
NascentOxygen
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I is indicated on the diagram as the current through the 25Ω, with a direction going from R to L.
So, by Ohms Law, I = (the voltage on the right - voltage on the left) / 25
i.e., I = (45 - VB) / 25
 

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