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

schematic

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.

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  • \$\begingroup\$ Asking about simulators and nodal analysis in one question is a step too far. \$\endgroup\$
    – Andy aka
    Mar 25 at 9:35
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    \$\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
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    \$\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

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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.

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

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