Node voltage with dependent voltage source

[bnosam]

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.

 

[NascentOxygen]

(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.)

 

[bnosam]

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

 

[NascentOxygen]

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

 

[bnosam]

We have to use the node voltages method for this.

 

[NascentOxygen]

We have to use the node voltages method for this.

Assume a node voltage, then sum the currents into that node.

 

[bnosam]

So if I understand then I still get without adding the dependent source.
VB/(100) + VB/(5) + (VB-VC)/25 = .45 A

 

[NascentOxygen]

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.

 

[bnosam]

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?

 

[NascentOxygen]

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.

 

[bnosam]

VB/(100) + (VB-6.25I)/(5) + (VB-VC)/25 = .45 A

Is that what you mean?

 

[NascentOxygen]

VB/(100) + (VB-6.25I)/(5) + (VB-VC)/25 = .45 A

Is that what you mean?

Yes. And you know I in terms of V_B and V_C. And you know the value of V_C.
So that seems like enough equations to be able to solve for all unknowns.

 

[bnosam]

VC = 45 V right?

 

[NascentOxygen]

VC = 45 V right?

+45V, yes.

 

[bnosam]

So when I solve I get : VB = 5I + 9I'm not really sure what I had for I/what I is

 

[NascentOxygen]

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 - V_B) / 25