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### Network # Simple KCL help

R

#### R.Spinks

Jan 1, 1970
0
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.

J

#### Jonathan Westhues

Jan 1, 1970
0
R.Spinks said:
KCL @ B:
2i - ((Vb-Va)/7 - (6/5)*Vb = 0

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

Jonathan

R

#### R.Spinks

Jan 1, 1970
0
Jonathan Westhues said:
Which term accounts for the current due to the dependent voltage source?

Jonathan

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.

J

#### Jonathan Westhues

Jan 1, 1970
0
R.Spinks said:
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

R

#### R.Spinks

Jan 1, 1970
0
Jonathan Westhues said:
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.

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