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Hi I'm having a really bad time in analyzing this circuit

I can't seem to establish the node equations, I've been reading on articles on how to solve it but the majority use current sources and the methods differ quite a bit, any help is really appreciated.

I know that I have to use KCL on each node, for example for node B I said:

\$ I_{1}+ I_{2} - I_{3} = 0 \$

\$ I_{1} = \dfrac{A-B}{220} \$
\$ I_{2} = \dfrac{C-B}{100} \$
\$ I_{3} = \dfrac{B}{330} \$

Given that and solving for B

A = 12 v
C = 4.5 v

\$ \dfrac{A-B}{220} + \dfrac{C-B}{100} - \dfrac{B}{330} = 0 \$

\$ \dfrac{A}{220} - \dfrac{B}{220} + \dfrac{C}{100} - \dfrac{B}{100} - \dfrac{B}{330} = 0 \$

\$ \dfrac{A}{220} + \dfrac{C}{100} - B ( \dfrac{1}{220} + \dfrac{1}{100} + \dfrac{1}{330}) = 0 \$

\$ \dfrac{A}{220} + \dfrac{C}{100} = \dfrac{B}{\dfrac{1}{220} + \dfrac{1}{100} + \dfrac{1}{330}} \$

B = 650( A/220 + C/100 )

Substituting A=12, C=4.5

\$ B = \dfrac{ \dfrac{A}{220} + \dfrac{C}{100 } }{ \left( \dfrac{1}{200} + \dfrac{1}{330} + \dfrac{1}{100} \right) } \$

B = 5.66 v

because of independent voltage sources.

I know my problem is on establishing the equations as I don't fully grasp how the analysis is made.

share|improve this question
Come on, you know that 1/220 + 1/100 + 1/330 is not equal to 1/650! Should be \$\dfrac{100 \times 330 + 220 \times 330 + 220 \times 100}{220 \times 100 \times 300} \$ – stevenvh Apr 27 '12 at 8:56
OMG!!, such a silly mistake... and I kept doing it again and again.. – Triztian Apr 27 '12 at 9:00
Replace the 650 by the result of my division and you're there! – stevenvh Apr 27 '12 at 9:02
Done, kept making mistakes with the calculator..., \$ B = 5.66 v \$ – Triztian Apr 27 '12 at 9:10
up vote 2 down vote accepted

Your equations for I1, I2 and I3 are OK, and when you replace A and C with their resp. voltages, you have only 1 variable left: B. Filling in the equations in

\$I1 + I2 - I3 = 0\$

gives you a single linear equation in one variable, which you no doubt can solve. The calculation for D is exactly the same: there's a resistor from A, one from C and one to ground.

share|improve this answer
Solving for B I get B= ~64.7 volts, I highly doubt that its the correct voltage.. – Triztian Apr 27 '12 at 8:33
No, that's wrong. Intuition says that it must be between 4.5V and 12V. I have the correct solution here on paper. Could you post your equation? – stevenvh Apr 27 '12 at 8:40
Off course, I'll post it. – Triztian Apr 27 '12 at 8:43

In general you not only have to apply KCL but also KVL.

KCL gives you an equation for each node (A, B, C, D, E). The sum of currents in each node is 0
KVL gives you an equation for each loop. The sum of voltages in each loop is zero.

Some of the loops are (see added picture):

  1. R1 V1
  2. V1 R2 R7
  3. R7 R3 V2
  4. V2 R8 R6
  5. R2 R3 R4
  6. R4 R5 R8

enter image description here

All equations together result in an simultaneous linear equation system for all your currents and all your voltages.

You can solve the linear equation system e.g. by Gaussian Elimination.

share|improve this answer
What exactly do you mean by "loop"? – Triztian Apr 27 '12 at 8:24
You can perfectly solve this with only KCL. See my answer. – stevenvh Apr 27 '12 at 8:30
That is what I did initially, but as my above comment suggests I think that the value is way off. – Triztian Apr 27 '12 at 8:35
Thanks for taking the time to edit the image, clears the "loop" question.. – Triztian Apr 27 '12 at 8:39
@stevenvh: I agree totally for this particular case, but the other approach works in general, no matter how the circuit is connected. Solving it that way is so mechanical that it can be implemented easily in an algorithm. And I think being aware that solving ANY circuit containing only resistors, current and voltage sources (plus L and C if you use complex numbers) is equivalent to solving a set of linear equations is very valuable. – Curd Apr 27 '12 at 9:10

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