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Hey guys I cannot wrap my head around this damn circuit and how I'm supposed to solve it using nodal analysis. I chose the bottom node as my reference node and I got to these equations:

Node A is the left one, node B is the middle one, node C is the right one and the reference node is the bottom one.

Node A : (Va/100) -4 + (Va-Vb)/50 + (Va-Vc)/20 = 0

Node B : -(Va - Vb)/50 + 4 - 10 -2 + (Vb-Vc)/40 = 0

Node C: 2 + (Vc/25) -(Vb-Vc)/40 -(Va-Vc)/20 = 0

I am supposed to find Vx.

Are my equations right? How do I find Vx?

enter image description here

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4 Answers 4

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Since the bottom node is your reference node, Vx = Vb. Now you've got three equations and three unknowns, and that's do-able.

Check your equations, as well. You mix up current flowing into the node as positive sometimes (e.g. Node B), negative other times (e.g. Node A). You need to be consistent. Almost everyone uses the convention that current flowing out of the node is positive.

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  • \$\begingroup\$ Yeah sometimes I notice I'm not being consistent with the conventions. Anyway I did Cramer's to solve the equations and I switched two elements of the equation before and that's why I got wrong results. Now I got 397.4 V for Vb = Vx. \$\endgroup\$ Apr 10, 2015 at 21:00
  • \$\begingroup\$ @crgrace (1/2): Yes, he needs to be consistent but there are many ways to be consistent. I wouldn't say that "Almost everyone uses the convention that current flowing out of the node is positive". Especially in nodal analysis it is much more useful to use following convention: \$\endgroup\$
    – Curd
    Dec 29, 2015 at 10:28
  • \$\begingroup\$ @crgrace (2/2): Put all voltage dependent currents on the left side of the equation with currents out of the node defined as positive and put all independent currents on the right side of the equation with currents into the node defined as positive (i.e. dependent currents equal independent currents). That results automatically in a nice equation system that can be written as a 3x3 dimensional conductance matrix times a 3 dimensional nodal voltage vector equaling a 3 dimensional independent current source vector. That's just Ohm's law in matrix form. \$\endgroup\$
    – Curd
    Dec 29, 2015 at 10:28
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Your equations look right to me. Have you tried solving them?

According to your schematic, \$V_X\$ is equal to \$V_B\$. So once you find \$V_B\$, you'll have the answer.

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Edit (I made it all consistent - current leaving node +ve):

Node A : -4A + (Va/100) + (Va-Vb)/50 + (Va-Vc)/20 = 0

Node B : -10A +4A +(-2A) + (Vb - Va)/50 + (Vb-Vc)/40 = 0

Node C: +2A + (Vc/25) + (Vc-Vb)/40 + (Vc-Va)/20 = 0

Equations:

Node A : 0.08 Va - 0.02 Vb - 0.05 Vc = 4 (1)

Node B : -0.02 Va + 0.045 Vb - 0.025 Vc = 8 (2)

Node C : -0.05 Va - 0.025 Vb - 0.125 Vc = -2 (3)

So your equations are correct. Only change I would make is how you do your referencing. Node - Other Node. It's the same as you have, just more consistent. You flip-flop on signs.

I get Va = 237.3V, Vb = 368V, and Vc = 152.5V. Check your math. I used a TI83+. Cramer's rule is messy.

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schematic

simulate this circuit – Schematic created using CircuitLab

equations are fine just slightly confusing i calculated solution and verified it with simulation they do match.

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