# Finding 3 currents in 3 different branches

1. Jun 25, 2013

### judas_priest

Has to be done using KVL and KCL ONLY

The attempt at a solution

I first wrote down the KCL for the node 1 i.e the node above the 18 ohm resistor.

Here's what I get
5.6A - ia - ib + 0.1vx - ic - 2A = 0

Here, substituting for 0.1vx as 1.8ia, I get:

5.6A + 0.8ia - ib - ic - 2A = 0

Substituting V/R for each current, I get;

5.6A + 0.8V/18 - V/R (How do I find this R?) - V/9 -2A = 0

So that's 2 unknowns 1 equation.

How do I find R to substitute in the equation and get the value of V, which will then allow me to find the currents in each branch?

#### Attached Files:

• ###### Untitled.png
File size:
93.3 KB
Views:
54
2. Jun 25, 2013

### Staff: Mentor

If you choose to use the expression Vx = iA*18Ω, why not use this expression for Vx to find the current iC (via Ohm's law) and replace it in your KCL equation? How many unknowns will you be left with in your KCL equation?

3. Jun 25, 2013

### judas_priest

you mean
iA*18 = ic*9 ?
And then substitute ic in KCL as iA*18 /9 i.e 2iA?

I'll still be left with two unknowns in that case. iA and ib

The equation will look like this:
5.6 + 0.8 iA = ib + 2 iA + 2

4. Jun 25, 2013

### Staff: Mentor

Why not replace the Vx in the controlled source, too? ib depends on Vx.

5. Jun 25, 2013

### judas_priest

I didn't quite get you.
Could you elaborate?

6. Jun 26, 2013

### Staff: Mentor

ib is due to a controlled source. The controlled source current has a constant multiplied by Vx. Replace ib by the controlled source expression, and replace its Vx with your chosen expression.

7. Jun 26, 2013

### judas_priest

For the dependent source, I'll get i as 1.8 iA. I don't see how to relate ib to anything. How do I replace ib by the controlled source expression?

8. Jun 26, 2013

### Staff: Mentor

ib is determined by the controlled source. ib = -0.1Vx. You've said that the controlled source current is 1.8 iA...

9. Jun 26, 2013

### judas_priest

How is ib = -0.1Vx? Do the magnitudes really match? Or are we just equating with opposite signs because they're in opposite directions?
I'm sorry if I sound like I'm missing out on some very basic concept.

10. Jun 26, 2013

### Staff: Mentor

The current produced by the source cannot be different from the current in the branch... it produces the branch current. They must be the same. The only thing to be wary of is the chosen direction for the branch current variable. Adjust the sign to compensate for the fact that the source is driving current "against" the direction of the labeled current.

11. Jun 26, 2013

### judas_priest

Okay, so two currents in a branch in opposite directions are always equal in magnitude?

12. Jun 26, 2013

### Staff: Mentor

Sure, since they both describe the same current. It's just their chosen orientation that is different (the direction of the arrow describing the assumed current direction).

There is only one current in the branch. It's been assigned the variable name ib thanks to the labeling on the diagram. That branch consists of a current source. That current source produces current 0.1Vx. Noting the direction associated with the current variable ib as compared with the direction of the source, we conclude that ib = -0.1Vx.

(I should point out that there is an analysis method, called mesh analysis, that assumes that the current in a branch shared by two loops can be decomposed into two separate currents, each contributed by a loop. In the end, though, there's only one physical current flowing in a given branch which happens to be equal to the sum of the currents that were invented for purposes of analysis)