# Kirchhoff's law problem

1. Jan 7, 2010

### Masterx00

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

Find I1,I2,I3

2. Relevant equations

Attached the circuit diagram

2. The attempt at a solution

I1 - I2 + I3 = 0

-120I1 - 60I2 + 0I3 = -1.8

0I1 -60I2 -20I3 = -1.2

I1 = 1/150 A

I2 = 1/60 A

I3 = 1/100 A

Is that correct ? The answer is my textbook is different :(

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2. Jan 7, 2010

### tiny-tim

Welcome to PF!

Hi Masterx00! Welcome to PF!
Nooo …

3. Jan 7, 2010

### Masterx00

Thanks :)
I took the 1st loop clockwise and the second loop anticlockwise, so why the 1st equation is wrong then ?
Can you please explain further ? :)

4. Jan 7, 2010

### tiny-tim

Your 1st equation (I1 - I2 + I3 = 0) doesn't involve loops, it's for a vertex, so it only involves "in" or "out".

According to the diagram, I1 is in, and both I2 and I3 are out.

5. Jan 7, 2010

### Masterx00

But, If I made it I1 - I2 - I3 = 0, I1 = 0, I2 = 0.03, I3=-0.03, that sounds correct ?

6. Jan 7, 2010

### tiny-tim

No … how did you get that?

7. Jan 7, 2010

### tomeatworld

Kirchhoff's first law says that the current into a juction is equal to the juntion out. So use that on the top middle junction and have another think about that first question (remembering the directions of currents!)

Could I also ask if the current directions were given or if you chose them?

8. Jan 7, 2010

### Masterx00

By substituting it with the other 2 equations (using the calculator eqn solver):
I1 - I2 - I3 = 0

-120I1 - 60I2 + 0I3 = -1.8

0I1 -60I2 -20I3 = -1.2

9. Jan 7, 2010

### tiny-tim

No, I2 and I3 are in opposite directions.

10. Jan 7, 2010

### Masterx00

aha
Now it gives correct answer (0.02,-0.01,0.03), I though direction of I3 was same like I1, because of the direction of the drawn battery poles :shy:
Thanks for help tiny-tim