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I'm supposed to use nodal analysis.

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

So I got these equations:

Node A: \$ -5 + \dfrac{V_a-V_b}{20} + \dfrac{V_a-V_c}{50} = 0 \$

Node B: \$ -\dfrac{V_a-V_b}{20} + \dfrac{V_b-V_c}{30} = 0 \$

Node C: \$ -\dfrac{V_b-V_c}{30} - 0.01V_1 - \dfrac{V_a-V_c}{50} = 0 \$

\$V_b = 0.4V_1\$

I tried solving the system with these equations but I didn't get anywhere.

What am I doing wrong?

The answer is: \$V_1 = 148.15 V\$

schematic

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  • \$\begingroup\$ Is the equation derived from the VCVS Vb = 0.4V1 ? \$\endgroup\$ Commented Apr 11, 2015 at 15:49
  • \$\begingroup\$ Yes, sorry I missed that you already had that one written down. \$\endgroup\$
    – The Photon
    Commented Apr 11, 2015 at 22:53

1 Answer 1

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Node A: \$ -5 + \dfrac{V_a-V_b}{20} + \dfrac{V_a-V_c}{50} = 0 \$

Node B: \$ -\dfrac{V_a-V_b}{20} + \dfrac{V_b-V_c}{30} = 0 \$

Node C: \$ -\dfrac{V_b-V_c}{30} - 0.01V_1 - \dfrac{V_a-V_c}{50} = 0 \$

\$V_b = 0.4V_1\$

You have four equations but only three unknowns, so you know you have a problem.

Also, your node B equation is incorrect because it doesn't account for current through the VCVS.

I would throw away the node B equation and use the other three.

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  • \$\begingroup\$ If I used this wouldn't I have 4 unknowns? Node A: -5 + (Va-Vb)/20 + (Va-Vc)/50 = 0; Node C: -(Vb-Vc)/30 - 0.01V1 - (Va-Vc)/50 = 0; Vb = 0.4V1 \$\endgroup\$ Commented Apr 11, 2015 at 15:56
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    \$\begingroup\$ @StefanBurnett, fourth equation: V1 = Va. \$\endgroup\$
    – The Photon
    Commented Apr 11, 2015 at 18:41

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