circuit analysis
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?
First off, mesh analysis would be simpler because you would only have one unknown (the current in the right mesh).
As you know, in nodal analysis you are summing currents at the nodes. Current into the nodes positive and current out of the nodes negative. The way you have defined the problem you should have 3 equations and three unknowns. Your first equations seems to do this correctly for node B. It kind of falls apart after that. For example, node A equation should just be .005 = (Va-Vb)/3000. You already know that Vc = 16 volts.
So simplifying the node A equation gives you what Andy aka said, then just substitute it back into your node B equation.
I don't know how to proceed with this question or where to start.
If there is (net) negative feedback, then you proceed by setting the voltage across the op-amp input terminals equal to zero:
$$v_+ = v_-$$
Note that with zero volts across the input terminals, the 2k resistor in parallel with the current source is irrelevant; there is zero volts across it so there is zero current through it. You may remove it from the circuit without changing the solution.
This should get you started.
@AlfredCentauri I still don't see the bottom loop, do you mean the loop v+ connected to VB then connected to the voltage source and then the resistor and finally VA. Is that considered a loop even with the op-amp? And when I do I still don't get your equation.
simulate this circuit – Schematic created using CircuitLab
This is the bottom-most loop and KVL clock-wise 'round the loop starting with the voltage across the 1k resistor is:
$$i_1 \cdot 1k\Omega -2V + V_B - V_A = 0 $$
rearranging yields
$$V_B = V_A - i_1 \cdot 1k\Omega + 2V$$
If the presence of the voltage source above is puzzling, recall that the output of the ideal op-amp is an ideal (controlled) voltage voltage source referenced to ground which I've shown explicitly here.
Best Answer
Jim's comment is correct. Your calculated answer is right but your simulation has CCCS1 going in the wrong direction. I'll give you a few ways of thinking about this intuitively so you don't have to rely on blind calculations.
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.