# Node voltage with dependent voltage source

• bnosam
In summary, the conversation discusses solving a circuit problem using the node voltages method. The problem involves a dependent voltage source and the equations used to solve for the unknown variables are provided. The final solution is VB = 5I + 9 and VC = 45V. The value of I is also determined using Ohm's Law.

## Homework Statement

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

## Homework Equations

& Attempt to solve
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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.

bnosam said:
(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.)

NascentOxygen said:
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

bnosam said:
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

We have to use the node voltages method for this.

bnosam said:
We have to use the node voltages method for this.
Assume a node voltage, then sum the currents into that node.

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

bnosam said:
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.

NascentOxygen said:
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?

bnosam said:
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.

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

Is that what you mean?

bnosam said:
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.

VC = 45 V right?

bnosam said:
VC = 45 V right?
+45V, yes.

So when I solve I get : VB = 5I + 9

I'm not really sure what I had for I/what I is

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