# Simple Circuit Help

1. Sep 17, 2009

### glover_m

1. The problem statement, all variables and given/known data

R1 = 6Ohms
R2 = 4Ohms

2. Relevant equations

Find ia and ib.

3. The attempt at a solution

Ia = V/R = 27V/6Ohms = 9/2A
Ib = I'm guessing that 3Ia = 27/2 and since that branch and ib are in parallel, Ib = 27/4A

2. Sep 18, 2009

### Redbelly98

Staff Emeritus
Not quite. The voltage across R1 is not 27V.

We have to introduce another variable here, the voltage at the point where R1, R2, and the current source meet. We could call this voltage V1.

You can use Kirchoff's voltage and current laws to set up 3 equations for the 3 unknowns Ia, Ib, and V1.

3. Sep 18, 2009

### mikelepore

You can get ia with one equation. A loop equation for the left loop has only one unknown.

(EDIT: Deleted equation.)

Last edited: Sep 18, 2009
4. Sep 18, 2009

### Redbelly98

Staff Emeritus
This would be giving the OP too much help -- if it were correct. Remember, we give hints here to encourage students to think and work things out.

Are you implying that the voltage drop across the current source is 3ia?

5. Sep 18, 2009

### mikelepore

It's not a current source. It's a dependent voltage source, for which the thing that it depend on is a current. The diamond symbol means dependent. The +- inside of the diamond, instead of an arrow, means voltage source.

Based on your advice that I gave out too much information, I am editing my post to leave the sentence and delete the equation.

6. Sep 18, 2009

### Redbelly98

Staff Emeritus
Thanks, I didn't know about that convention.

So, if they were to actually use units properly, that voltage source should really be labeled

3Ω · ia?​

Because, strictly speaking, 3·ia is a current, not a voltage.

7. Sep 18, 2009

### mikelepore

I agree with you about units, but this notation is traditional. I'm looking now at Hayt and Kemmerly, _Engineering Circuit Analysis_, McGraw-Hill, 1971.

Page 61, they have a VOLTAGE source labeled: ix/8

Conversely, page 63, here's a CURRENT source labeled: 2vx

8. Sep 19, 2009

### Redbelly98

Staff Emeritus
Thanks for clearing that up (for me at least).

I agree with your post #4. KVL applied to the left loop will give ia