Applying Kirchhoff's Laws to Solve Circuit Problems

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The discussion focuses on applying Kirchhoff's Laws to solve circuit problems, specifically finding the currents I1, I2, and I3. The initial equations presented by the user were incorrect due to a misunderstanding of current directions at a junction. Clarification was provided that Kirchhoff's first law requires the sum of currents entering a junction to equal the sum of currents leaving it. After correcting the equations and considering the proper directions of currents, the user arrived at the correct values for I1, I2, and I3. The importance of accurately determining current directions in circuit analysis is emphasized throughout the conversation.
Masterx00
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Homework Statement



Find I1,I2,I3

Homework 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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Welcome to PF!

Hi Masterx00! Welcome to PF! :smile:
Masterx00 said:
I1 - I2 + I3 = 0

Nooo … :redface:
 
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 ? :)
 
Masterx00 said:
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 ? :)

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. :smile:
 
tiny-tim said:
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. :smile:

But, If I made it I1 - I2 - I3 = 0, I1 = 0, I2 = 0.03, I3=-0.03, that sounds correct ?
 
Masterx00 said:
I1 = 0, I2 = 0.03, I3=-0.03, that sounds correct ?

No … how did you get that? :confused:
 
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?
 
tiny-tim said:
No … how did you get that? :confused:

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
 
Masterx00 said:
0I1 -60I2 -20I3 = -1.2

No, I2 and I3 are in opposite directions.
 
  • #10
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 :smile:
 

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