- #1
Masterx00
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Applying Kirchhoff's Laws to Solve Circuit Problems
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Homework Help Overview
The discussion revolves around applying Kirchhoff's Laws to analyze a circuit, specifically focusing on determining the currents I1, I2, and I3 based on a provided circuit diagram. Participants are exploring the relationships between these currents and the equations derived from Kirchhoff's Laws.
Discussion Character
- Exploratory, Conceptual clarification, Assumption checking
Approaches and Questions Raised
- Participants discuss the setup of equations based on Kirchhoff's Laws, questioning the correctness of initial equations and the assumptions regarding current directions. There is an exploration of the distinction between vertex and loop equations.
Discussion Status
The discussion is active, with participants providing feedback on each other's equations and reasoning. Some guidance has been offered regarding the application of Kirchhoff's first law and the importance of current direction, leading to a reevaluation of initial assumptions.
Contextual Notes
There is an indication that the original poster's textbook provides different answers, prompting questions about the correctness of their approach. Participants are also considering whether the current directions were predetermined or chosen arbitrarily.
- #2
tiny-tim
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Welcome to PF!
Hi Masterx00! Welcome to PF!
Nooo …
Hi Masterx00! Welcome to PF!
Masterx00 said:
Nooo …
- #3
Masterx00
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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 ? :)
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
tiny-tim
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Masterx00 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.
- #5
Masterx00
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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.
But, If I made it I1 - I2 - I3 = 0, I1 = 0, I2 = 0.03, I3=-0.03, that sounds correct ?
- #6
tiny-tim
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Masterx00 said:
No … how did you get that?
- #7
tomeatworld
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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?
Could I also ask if the current directions were given or if you chose them?
- #8
Masterx00
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tiny-tim said:No … how did you get that?
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
tiny-tim
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Masterx00 said:
No, I2 and I3 are in opposite directions.
- #10
Masterx00
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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
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
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