I seem to be unable to properly calculate the proper voltage for what I've called Node B. Would someone mind helping me discover where I've turned a wrong corner here. Schematic below (top right corner and top of 3Vx source -- ie. I drew an ideal wire -- it's the same node : Node B): 2i .--------. |dependent |current | Node B ------------ source +----------------------------,| | | | ,-´ | | '--------' -´ | | ,´ | | 5 ohm 2 ohm ,-´ | | ___ ___ ,-´ | Node A-----|___|--------|___|--------,|´ | | ,-´ | | | -Vx + ,-´ | | | ,-´ | | | ,-´ | | | ,-´ | | .---|----. |´ i | | | Indep. | /+\ .-. .-. 10 A | current| ( ) 3Vx | | | 5 ohm | | 1 ohm | source | \-/ \|/| | | | | | | '-' '-' '---|----' | | | | | | | | | | | | | | | | | | | | | | | | | | | |--------------|-----------------|----------------- - | | GND (created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de) Here's the math I did: 5ohm || 1 ohm = 5/6 ohm 5ohm in series 2 ohm = 7 ohm i=Vb / 5 ohm Vb=3Vx KCL @ A: 10+((Vb-Va)/7) - 2i =0 10+(Vb/7)-(Va/7)-2Vb/5=0 Vb(1/7 - 2/5) - 1/7 * Va = -10 -0.25714Vb -0.14285Va = -10 (I simulated the problem and this part is correct) KCL @ B: 2i - ((Vb-Va)/7 - (6/5)*Vb = 0 (2/5)Vb - (Vb/7) + (Va/7) - (6/5)*Vb =0 (2/5 - 1/7 - 6/5)Vb = -1/7 Va Va = 6.6Vb (this is incorrect based on my simulation) I expect the answer to be voltage a node b to be 30v and at node a to be 16 volts. What have I done incorrectly? Thanks.

If you mean additional current contributions (due to the 3Vx voltage source) in the legs I've already accounted for in the above equation -- then... I'm not sure I know what to do with it. How should I account for it in the KCL equation since it's a voltage and not a current.

Don't write KCL at a node whose voltage is already fixed by a voltage source. You can't, because you don't know the current through the voltage source, and you don't need to, because you already know that Vb = 3*Vx. Your KCL at A is fine, so you have one equation: -0.25714Vb -0.14285Va = -10 Now Vb = 3*Vx, and Vx is the drop across the 5 ohm resistor, which is equal to 5 ohms times the current through that branch: Vx = 5*((Vb - Va)/7) Vb/3 = 5/7*(Vb - Va) 0.3333*Vb = 0.7143*Vb - 0.7143*Va -0.3810Vb + 0.7143Va = 0 Now you have two equations in two variables. Solve to get Vb = 30 V Va = 16 V which is what you were hoping for. Jonathan

Ah, very good. I kept thinking I had to use KCL for both nodes. That makes much more sense now. Thanks.