Solving Circuits with Kirchoff's Laws

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Homework Help Overview

The discussion revolves around solving a circuit problem using Kirchhoff's laws. The original poster seeks a shorter method than the one they have learned, which involves setting up equations based on current and voltage in the circuit.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster shares their attempts at formulating equations based on Kirchhoff's laws but expresses frustration with the length of the solution. Some participants question whether the original poster has fully understood the laws and suggest showing their work. Others mention the possibility of using determinants for solving simultaneous equations.

Discussion Status

The discussion is ongoing, with participants exploring different approaches to solving the circuit problem. There is a suggestion to set equations equal to zero to create a solvable system, but no consensus on the shortest method has been reached.

Contextual Notes

The original poster has provided an attachment with their work, but it is noted that their solution is incomplete. There is also a mention of potential constraints related to homework rules and the desire for a more efficient method of solving the problem.

meczs
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http://www.pa.msu.edu/courses/1997spring/PHY232/lectures/kirchoff/examples.html"

how can you solve my circuit (that i attached) using that procedure... in that site.. i saw it last week... that's the shortest way i seen... can you help me.. thankss
 

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Did you learned Kirchoff's laws ? Show your work.
 
yah i learned it... but it is the longest solution... i hate that... i want the shortest method... that i found in that site.. can you solve it using that method?

i put the answer @ my attachment

here is the solution I've learned

@node B
i1 + 12 - i3 = 0
i3 = i1 + i2 that is for eq. 1

@loop abefa
E1 + E2 -15i1 +26i2
40v + 50v -15i1 +26i2
90 + 26i2 / 15i1 eq. 2

@loop bcdeb
-E2 -E3 -26i2 -37i3
-50 -60 -26i2 -37i3
-110 -37i3 / 26i2 eq. 3

that is the my sample solution I've learnd... but it is not finish.. like i said; that is the longest solution... hoping that you understand it...

(sorry bad grammar)
 
You need to set the second two things equal to zero to make them equations. That gives you 3 equations and 3 unknowns -- just solve away. There's not really a quicker way to solve it that I know of, unless you mean using determinants to solve the simultaneous equations. In this circuit, the equations are simple enough that solving with determinants won't save you any time.
 

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