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Simple KCL help

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
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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.

Reply to
R.Spinks
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Which term accounts for the current due to the dependent voltage source?

Jonathan

Reply to
Jonathan Westhues

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.

Reply to
R.Spinks

source)

I'm

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

Reply to
Jonathan Westhues

source?

KCL

equal

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

Reply to
R.Spinks

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