Why I can't solve this simple circuit using node analysis?

Thread Starter

babaliaris

Joined Nov 19, 2019
44
Let's forget about that 1Ω resistor and suppose that we have a circuit that is the same as in the picture up to the 4Ω resistor. If I try to
solve that circuit I have two unknown node voltages and only one equation! Why I can't find a second equation? But if I include that 1Ω
resistor, KCL will give me one more equation that includes \( V_{A} \) and \( V_{B} \) and I will be able to solve for \( V_{B} \) .

I know that I can find \( V_{th} \) by taking into consideration that 1Ω resistor, but the question is why I can't solve the circuit without that
1Ω resistor.

So assume that the 6 and 4Ω are in series and the 1Ω is disconnected from the circuit.

Thank you.


IMG_20191125_201508.jpg
 

WBahn

Joined Mar 31, 2012
24,969
For some reason you are treating Nodes A and B as a supernode. If that's the case, then you need an equation relating them to each other.

(Va-Vb)/(6 Ω) = i2 = Vb / 4 Ω

If you worth though this, you'll see that Vb, in terms of Va, is just the voltage-divider equation.
 

WBahn

Joined Mar 31, 2012
24,969
Why does he need to use superposition?

If he wants to treat both A and B as nodes and use nodal analysis, he can do that. All he has to do is write the node equations for each node.

Since B is not an essential node (with the load resistor removed), He can solve for Va using a single node equation and then get Vb using voltage division.
 

dl324

Joined Mar 30, 2015
9,321
If he wants to treat both A and B as nodes and use nodal analysis, he can do that. All he has to do is write the node equations for each node.
My mistake. It's been so long since I've done that type of analysis that my first choice would have been superposition.
 

Thread Starter

babaliaris

Joined Nov 19, 2019
44
For some reason you are treating Nodes A and B as a supernode. If that's the case, then you need an equation relating them to each other.

(Va-Vb)/(6 Ω) = i2 = Vb / 4 Ω

If you worth though this, you'll see that Vb, in terms of Va, is just the voltage-divider equation.
I just saw it. \[ \frac{V_{A}-V_{B}}{6} = \frac{V_{B}}{4} \]

Well I knew that \( i_{2} \) could be expressed by those 2 resistors but I had in mind that I still don't know \( i_{2} \) and didn't think to just work with their equals.

Thank you!!! Probably it's also due to too many hours of solving exercises. Right now I tried to solve it 3 times including the load and I'm finding wrong values for \( V_{th} \) ... (compared with the answer). The answer is \( V_{th} \)=9
 

Thread Starter

babaliaris

Joined Nov 19, 2019
44
Can also someone tell me what am I doing wrong here?
(I'm trying to calculate the \( V_{B}=V_{th} \)) including the load in my calculations. The correct answer is \( V_{th}=9V \)
but I found V_{th}=2.25V

IMG_20191125_211908.jpg
 

Thread Starter

babaliaris

Joined Nov 19, 2019
44
Can also someone tell me what am I doing wrong here?
(I'm trying to calculate the \( V_{B}=V_{th} \)) including the load in my calculations. The correct answer is \( V_{th}=9V \)
but I found V_{th}=2.25V
Well, it seems that I misunderstood the Thevenin theorem. I MUST not include the load when calculating the Thevenin voltage, correct?
 
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