# Simple KCL help

Discussion in 'Electronic Basics' started by R.Spinks, Jun 6, 2005.

1. ### R.SpinksGuest

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.

2. ### Jonathan WesthuesGuest

Which term accounts for the current due to the dependent voltage source?

Jonathan

3. ### R.SpinksGuest

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.

4. ### Jonathan WesthuesGuest

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

5. ### R.SpinksGuest

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