I am stuck at # 29 (chapter 19) and I really don't know where I went wrong. I made 2 loops, 1 with R1 R2 and another with R2 and R3 and then I had one junction equation.
Well, from what you said it seems like you are on the right track. Using two loop rules, and one junction rule, I get the following equations:
loop 1: E1-I1R1-I2R2=0 loop 2: E2-I2R2-I3R3=0 junction: I1+I3=I2 |  |
This does give you 3 equations with 3 unknowns. Solving these systems by substitution can be a bit tricky, but with perserverance can be done. I show an example on the help home solving both by substitution and a matrix method (which can be a quick way to solve if you have access to a graphing calculator, or are good with matrix reduction). To see the example and a review of Kirchoff's laws go to: Kirchoff's laws and their application.
Let me know if these are not the equations you got, and we will backtrack some more and find out where the problem is. (Remember, you can post follow-up questions as comments to this post.)
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