# How can I figure out the voltage in this circuit? [duplicate]

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: $V_1 = 148.15 V$

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

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.

• 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 Commented Apr 11, 2015 at 15:56
• @StefanBurnett, fourth equation: V1 = Va. Commented Apr 11, 2015 at 18:41