# Issue with node voltage method

For the following circuit, I'm trying to find the voltage at node B, so I write some KCL equations:

at node A:

$$v_A=20$$

and,

at node B: $$\frac{v_B-20}{50}+\frac{v_B}{5}-9(\frac{20-v_B}{50})=0$$

And I get that vB = 10 V. I run this in Circuit Lab and get the vB is -80 V though. Where did I go wrong?

• You might want to include a diagram of the circuit – tangrs Mar 25 '16 at 11:58
• What circuit? Use the circuit editor Ctrl-M to add it. – Tyler Mar 25 '16 at 11:58
• Sorry about that! I always forget to add the attachment. – user104243 Mar 25 '16 at 12:59
• Ia is (20 - Vb)/50 and the current is downwards through R2 and not additive to Ia. – Andy aka Mar 25 '16 at 13:04
• If you're getting -80V then CCCS1 is taken in the wrong direction. Look for an error in the sign. Error in = error out – JIm Dearden Mar 25 '16 at 15:28

• If $V_B$ is negative (or just less than $V_A$), $I_A$ must be positive -- that's just Ohm's Law. Because of the direction of the current source, R2's current must flow from node B to ground, so node B must be at a higher voltage than ground -- Ohm's Law again. Thus, $0 \mathrm V < V_B < 20 \mathrm V$.
• If $I_A$ is negative, node B's voltage must be higher than node A's. But now R2's current flows from ground to node B, which means node B's voltage must be negative! This means $20 \mathrm V < V_B < 0 \mathrm V$, which is impossible. So $I_A$ must be positive, which means that $V_B$ must be positive.
• Notice that the current going through R2 is $I_A + 9I_A = 10I_A$. So R2's current is ten times higher than R1's, but R2's resistance is one tenth of R1's. Since $V = IR$, this suggests that they have the same voltage drop. Think of it as a sort of active voltage divider. Splitting the 20V between the two resistors gives $V_B = 10 \mathrm V$.
• If the current source were reversed, then you'd have $I_A$ flowing into node B through R1 and $9I_A$ flowing out of node B through CCCS1. That means $8I_A$ has to come from the R2 branch, which (again) means node B's voltage is negative.
• You can kind of see just by looking at the equation that $V_B$ has to be positive -- separate out the $V_B$ terms from the constant terms: $$\frac {V_B}{50} - \frac {20}{50} + \frac {V_B}{5} - 9\frac {20}{50} + 9\frac{V_B}{50} = 0$$ See how all the $V_B$ terms are positive and all the constant terms are negative? When you move the constant terms to the other side of the equation, you'll get: $$Positive\ number \cdot V_B = Positive\ number$$ Now look at what happens if you switch the direction of CCCS1: $$\frac {V_B}{50} - \frac {20}{50} + \frac {V_B}{5} + 9\frac {20}{50} - 9\frac{V_B}{50} = 0$$ It's harder to see, but you end up with both the $V_B$ and constant terms being positive. That implies they have opposite signs.