# Basic KCL Problem

Discussion in 'Homework Help' started by SilverKing, May 4, 2014.

1. ### SilverKing Thread Starter Member

Feb 2, 2014
72
0
1. The problem statement, all variables and given/known data

2. Relevant equations

KCL

3. The attempt at a solution

At node a: 6= io + Is
At node b: Is= io/4 + I8

Couldn't go much further.

Apr 5, 2008
15,798
2,384
Hello,

What would be the current in the 8 Ω resistor, when the current in a 2 Ω resistor is io ?
Both resistors are parallel and will have the same voltage accross them.

Bertus

3. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
In terms of the voltage on your top node (relative to the bottom node), what is Io and I8?

4. ### SilverKing Thread Starter Member

Feb 2, 2014
72
0
Vo=I8*8, Vo=Io*2 --> Vo=Vo, I8*8=Io*2 --> I8=Io/4?

5. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Keep going....

You now have:

6A= io + Is
Is= io/4 + I8
I8=Io/4

That's three equations and three unknowns (io, Is, and I8).

6. ### SilverKing Thread Starter Member

Feb 2, 2014
72
0
I've got it all, thanks.

Another simple question:

For the following example:

I understand how to get equation (2.6.1), the terminals of the resistors were defined depending on the 12V voltage source, but for the equation (2.6.2), shouldn't be as same as the "6i" in (2.6.1)?

7. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
Read up on the "passive sign convention".

If vo is positive, that means that the current is flowing through the device from the positive terminal to the negative terminal. But this is in the opposite direction to the current i.

8. ### SilverKing Thread Starter Member

Feb 2, 2014
72
0
But the current flows in the clockwise direction (depending on the 12V voltage source), and therefore the terminals of 6 ohms resistors would opposite, or I'll just have to stick to the assumed terminals from the problem?

___________________________________________________________________________

For this problem:

Why wouldn't the currents flows like these:

So, at the node: 10+2Vo=io --> 10+2Vo=Va-Vo-0/6+4
Where Va is the voltage at the node a.

9. ### WBahn Moderator

Mar 31, 2012
18,085
4,917
You can define the currents to be in any direction you want. Flip a coin.

You mention node 'a' but never indicate which node is node 'a'.

You equation is nonsensical, which would be obvious to you if you would start tracking units.

You have

10 + 2Vo = io

10 is a number, 2Vo is a voltage, and io is a current. You can't add a number and a voltage and, even if you could, the result would not be a current. Instead, what you should have is

10 A + (2 A/V)*Vo = io

The coefficient of the VCCS is not '2', but '2 A/V' (and even then I am having to guess what the other meant since it could be 2 mA/V or 2 A/mV or any of a dozen other possibilities -- engineering is not about guessing, but that is something that most textbook authors haven't figured out because very few of them have ever actually worked as practicing engineers).

What you have on the right side is even more nonsensical -- so much so that I can't tell what you were thinking.

Va - Vo - 0/6 + 4

You have a voltage minus a voltage minus 0/6 plus 4. Even recognizing that the 6 and the 4 are resistances and thus have units of ohms, it makes no sense.

Assuming that Va is the voltage on the node at the top of the current sources (relative to the bottom node), the wouldn't

io = Va/(4Ω+6Ω)

Also, note the relationship between Vo and io. It is NOT Vo = io*R because you have defined io to be going through the resistor from negative to positive. Instead, you have

Vo = -io*4Ω

It is a very good habit to avoid defining a voltage, Vx, and a corresponding current, Ix, that do not adhere to the passive sign convention. It's fine to define your current the way you did, but don't call it io when you already have a Vo that would be inconsistent with it. It's asking for trouble.