Applying Kirchhoff's Rules to Solve Circuit Problems

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Homework Help Overview

The discussion revolves around applying Kirchhoff's rules to analyze a circuit problem. Participants are attempting to set up equations based on the circuit diagram provided, focusing on the relationships between currents and voltages in the circuit.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster expresses confusion about identifying loops and setting up equations for the circuit. Some participants suggest using Kirchhoff's first rule to relate currents at junctions and propose setting up equations based on different loops in the circuit.

Discussion Status

Participants are actively discussing the setup of equations and relationships between currents. There is an indication of guidance being offered regarding the use of substitution in solving the equations, but no consensus on a complete solution has been reached.

Contextual Notes

The original poster indicates a lack of understanding of the problem setup, which may affect their ability to proceed. The discussion includes references to multiple loops in the circuit and the need for clarity on how to apply Kirchhoff's rules effectively.

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Homework Statement



http://img245.imageshack.us/my.php?image=physicsef4.jpg

Homework Equations



The change in Voltage = 0.

The Attempt at a Solution



Ok I have no clue where the loops go from. I want to say something like the current for I1 starts at the 10V battery in the middle and the current would look something like this:

8ohm + 2ohm +15v +I1 +3ohm - I2 = 0...

I just really don't understand these problems and could use some help thanks.
 
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First, making use of Kirchhoff's first rule :
The algebric sum of current at a junction is zero.
We have I3 = I1+I2.
Then , by considering two loops of the circuit,(there are a total of three circuits, choose two you like)

Setting up equation
20-I3(10)-8I2+10-I3(4)=0
20-10I3-I1(2)+15-3I1-4I3 =0

30-22I2-14I1=0
35-19I1-14I2=0
and solve it.
 
mysqlpress said:
First, making use of Kirchhoff's first rule :
The algebric sum of current at a junction is zero.
We have I3 = I1+I2.
Then , by considering two loops of the circuit,(there are a total of three circuits, choose two you like)

Setting up equation
20-I3(10)-8I2+10-I3(4)=0
20-10I3-I1(2)+15-3I1-4I3 =0

30-22I2-14I1=0
35-19I1-14I2=0
and solve it.

Thanks, so do I just do substitution from there?
 
jcpwn2004 said:
Thanks, so do I just do substitution from there?
Yes.
and I3=I2+I1.
all currents are solved
 

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