I have a circuit below and I need to determine the potential at \$v_1\$ and \$v_2\$. I have 3 questions:


simulate this circuit – Schematic created using CircuitLab

  1. How do I use nodal analysis at \$v_2\$, or how do I solve this without nodal analysis?
  2. I have tried to do this:

At node \$v_1\$: \$-1+\frac{v_1-v_2}{10}+\frac{v_1}{5}+\frac{v_1-v_2}{15}=0\$

At node \$v_2\$: \$v_2 = 2v_x = 2\frac{v_2-v_1}{10} = \frac{v_2-v_1}{5}\$

$$=> \frac{11}{30}v_1-\frac{1}{6}v_2=1; \frac{1}{5}v_1+\frac{4}{5}v_2=0$$

$$=> v_1=\frac{120}{49}V; v_2=-\frac{30}{49}V$$

This is wrong, but when I ran a simulator and I accidentally set the function in VCVS is \$0.2(v_a-v_b)\$, then I have the same result. Why does this happen?

  1. (This question is unrelate to 2 questions above, but just for curious) When I tried to substitute the \$I_2\$ source with a voltage source in a simulator, the simulation stop immediately, but this does not happen with \$I_1\$ source. Why does this happen?

Any help would be appreciated.

  • \$\begingroup\$ Asking about simulators and nodal analysis in one question is a step too far. \$\endgroup\$
    – Andy aka
    Mar 25 at 9:35
  • 2
    \$\begingroup\$ Is a and b and v1 all connected together, as it looks they are to me? \$\endgroup\$
    – jonk
    Mar 25 at 9:38
  • \$\begingroup\$ @jonk a and b are on 2 side of the 10 ohm resistor. \$\endgroup\$
    – Becker
    Mar 25 at 9:40
  • 2
    \$\begingroup\$ I see a lot of dotted connections in your schematic. Way too many, to my eye. \$\endgroup\$
    – jonk
    Mar 25 at 9:41
  • \$\begingroup\$ @jonk sorry, this is my first time using circuit lab, let me edit the schematic. \$\endgroup\$
    – Becker
    Mar 25 at 9:42

1 Answer 1


You said that \$v_x = \frac{v_2 - v_1}{10}\$

The left-hand side is a voltage, the right-hand side is a current.

  • \$\begingroup\$ Oh, I didn't realize that. How about question #3? \$\endgroup\$
    – Becker
    Mar 25 at 10:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.