Basic circuit analysis

[Assafl]

I think I understand you.
I asked the teacher and he said that I should use KCL at the node that joins R1 and Rx but even doing that I can't come close to that equation, so I'm blind here.

Let the current in the loop be I. Now express I in terms of Vc, R1 and Vx. Lastly eliminate I by noting that VR1 = I x R1. You should get what you need to find.

Since it is DC analysis that is the approach you beat me to.

However, one would probably be safe showing how the equation work in the case of, say, a capacitor or an inductor. So how this also works when the R approaches infinity (charged capacitor) and when it approaches zero (inductor happy with its magnetic field) (or open / short circuit).

Yet another case is what if X is a voltage/current source (with/without a resistor). Even then it should work (albeit Thevenin would have made it easier...).

Perhaps a more general approach would be to let v(x)=f(I).



   

[Nexo]

I did it! Thank you very much you all.

Now, let me show you the last one, I was able to solve all of them and this one is giving me troubles:

I did it by solving the equations I have from KCL (solving for each unknown variable; first method) and the second method is to use matrixes (Kramer rule) but I get one answer with the first method and completely different one with the second method, so my questions are these:
1. Which answer is correct?
a) Va = 1.11v, Vb = -9.77v, Vc = -20.65v (first method)
b) Va = 6v, Vb = 0v, Vc = -6v (second method)

2. Which method do you prefer and why?

Thanks!

[sibeen]

Knowing the value of V1 and V2 would possible be of some assistance

Also, why did you put the reference node (earth) where you did?

[Nexo]

My bad, v1 = vb = 12v. The reference point was already given in the problem.

[sibeen]

Nexo, neither of your results are correct.

For this problem I'd go and use KVL (mesh analysis). Work out what the current flows are and then use some simple maths.

[Nexo]

We can only use KCL

[sibeen]

Just simplify the circuit to a 24V supply, a 7.52k resistor (10K||30K) in series with a 560k. Work out the voltage across the 7.52k and you can then just use the voltage divider equations to work out the individual nodes. Then just add 12 volts to each result to give you the result as a reference to the battery mid point.

[Ian.M]

That will work to check your results, (if you don't goof and come up with something ridiculous like 7.52 instead of 7.5, and I get Vc=11.68V to 2 d.p. so any solution that differs significantly or has Va or Vb outside the range Vc to +12V is bad^*) but to do it purely with KCL (+ Ohm's Law), you have to solve it by nodal analysis, without first simplifying the circuit. 

As you have tried to solve the same set of equations by two methods and got conflicting results, neither of which match the check method you'll need to show your workings.

*I know I haven't had a major brain fart and goofed up a simple potential divider calculation <paranoid> because  LTspice .op analysis gives:
          V(c):    11.6828    voltage ,
</paranoid>

[Nexo]

Give me an hour, I'll try to solve it again and post my results.

EDIT:

This are my answers:
Va = 11.894
Vb = 11.789
Vc = 11.683

At least Vc is pretty close to your answer.

[Nexo]

I think I understand you.
I asked the teacher and he said that I should use KCL at the node that joins R1 and Rx but even doing that I can't come close to that equation, so I'm blind here.

Let the current in the loop be I. Now express I in terms of Vc, R1 and Vx. Lastly eliminate I by noting that VR1 = I x R1. You should get what you need to find.

I did this exercise but I didn't do it using KCL, any idea on how to do it using it?

[sibeen]

This are my answers:
Va = 11.894
Vb = 11.789
Vc = 11.683

They are correct

[Nexo]

Thank you all!