# Solving a circuit using superposition (strange case)

My professor gave us a problem that we are free to do by ourselves that is a typical exam problem. I was wondering if half of my solution can be verified (the other half I am fine with). The circuit, along with the half that is of concern is attached below.

If there is anything of concern please let me know. The problem statement in case it is not clear is solve for Vo using superposition; in this case I only am wondering if I found Vo1 correctly.

EDIT: More neat work.

$$Vo = Vo1 + Vo2\\ Va = -Vx \\$$ Nodal Analysis at Node A

$$Therefore: \\ {(-Vx-Vc)\over R1} + {(-Vx-Vc)\over R2} - {Vx\over 2} = 0\\ -{Vx\over R1} - {Vx\over R2} - {Vx\over 2} = 0\\ Assuming: {1\over R1} +{1\over R2} + {1\over 2} ≠ 0\\ Then\\ Vx({1\over R1} + {1\over R2} + {1\over 2}) = 0\\ Vx =0\\$$ If Vx =0, the dependent current source acts as an open circuit as no current flows through it; also, there is no current flowing through the loop with Vx, that is, Ix =0, therefore the dependent voltage source has no voltage across it and is thus short circuited. The circuit therefore now can be simplified to:

$$Vo1 = (-Is)(+R3) = -IsR3$$

• Please see electronics.stackexchange.com/editing-help#latex and edit your work into the question so we can read it. Jul 10, 2015 at 19:06
• Your notation isn't the greatest.... Vc... you should use C if you got a cap. Jul 10, 2015 at 22:34
• Also, you should use other method of solving linear system like nodal analysis or Mesh analysis. Because direct analysis is possibly the worst solving method when you got commanded source. Jul 10, 2015 at 22:38
• Va != 0 that for sure. R2 make a current divider with R3 therefore there is a tension level at Va. Jul 10, 2015 at 22:41
• I have reverted your edit as you deleted large portions of the question, rendering it unintelligible. Please do not remove necessary information (like schematics) from a question. Oct 6, 2022 at 3:48