How to Use Kirchoff's Rules to Solve Complex Circuits: Step-by-Step Guide

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To solve the circuit using Kirchhoff's Rules, start by identifying the loops and assigning voltages to points in the circuit. Assign a potential of 0 to a reference point, typically the negative side of a voltage source, and calculate the potentials at other points. Use Kirchhoff's voltage law to write equations for each loop, ensuring to account for the direction of currents and voltage drops correctly. Kirchhoff's current law can then be applied at junctions to relate the currents. This systematic approach helps in accurately determining the currents in each branch of the circuit.
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Homework Statement



I am trying to calculate the currents on this 2 power source circuit using Kirchoff's Rules.

http://img101.imageshack.us/img101/4151/kirchoffc1.jpg

Ignore the numbers on the lines connecting the components and their length

Homework Equations



I#1=I#2+I#3

ΣV=0

The Attempt at a Solution



I've got loop 1 on the leaf and loop 2 on the right.

Loop1: 6+(-8*I#2)+(6*I#1)=0

Loop2: 12+(-1*I#3)+(-8*I#2)=0

This does not seem right though. Could someone take a look at the circuit and help me with the solution. For some reason I am having much difficulty using Kirchoff's Rules.
 
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Show with arrows the estimated current direction in each branch of the network. Go round each loop and add the potential differences along the path. ehild
 
I do not use the loop rule much directly. you usually get the direction
of some of the currents wrong (as you did here).

It's usually better assign a potential of 0 to some point of the circuit,
usually on the negative side of the voltage source, and then find any
other potentials you can, and assign unknowns for the one you can't

Then work out all of the currents, and use Kirchhof's current rule in all
the points with an unknown potential.

In this case I assign a potential of 0 to point 4. Point 1 must then be at
6 V, and point 2 at 18 V.
Assign an unknown potential of V to point 3.
The current in R2 is I1 = (18-V)/2
The current in R1 is I2 = ...
THe current in R3 is I3 = ...
you can take all currents in the direction away from point 3.
Now you can find V with I1+I2+I3 = 0.
 
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