Finding 3 currents in 3 different branches

[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 - i_a - i_b + 0.1v_x - i_c - 2A = 0

Here, substituting for 0.1v_x as 1.8i_a, I get:

5.6A + 0.8i_a - i_b - i_c - 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?

 

[gneill]

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

 

[judas_priest]

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

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

I'll still be left with two unknowns in that case. i_A and i_b

The equation will look like this:
5.6 + 0.8 i_A = i_b + 2 i_A + 2

 

[gneill]

you mean
i_A*18 = i_c*9 ?

And then substitute i_c in KCL as i_A*18 /9 i.e 2i_A?

I'll still be left with two unknowns in that case. i_A and i_b

The equation will look like this:
5.6 + 0.8 i_A = i_b + 2 i_A + 2

Why not replace the V_x in the controlled source, too? i_b depends on V_x.

 

[judas_priest]

Why not replace the V_x in the controlled source, too? i_b depends on V_x.

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

 

[gneill]

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

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

 

[judas_priest]

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

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

 

[gneill]

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

i_b is determined by the controlled source. i_b = -0.1V_x. You've said that the controlled source current is 1.8 i_A...

 

[judas_priest]

i_b is determined by the controlled source. i_b = -0.1V_x. You've said that the controlled source current is 1.8 i_A...

How is i_b = -0.1V_x? 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.
Please help me out with this.

 

[gneill]

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.

 

[judas_priest]

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.

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

 

[gneill]

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

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 i_b thanks to the labeling on the diagram. That branch consists of a current source. That current source produces current 0.1V_x. Noting the direction associated with the current variable i_b as compared with the direction of the source, we conclude that i_b = -0.1V_x.

(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)