Simplifying Complex Circuit Diagrams

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SUMMARY

This discussion focuses on simplifying complex circuit diagrams, specifically transforming a black diagram into series and parallel circuits. The user seeks guidance on combining resistors effectively, utilizing the equations for equivalent resistance in series (R_{eq} = R_1 + R_2 + ... + R_n) and parallel (\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + ...). Key insights include the importance of Kirchhoff's Law in analyzing current flow and achieving an accurate equivalent resistance. The final equivalent resistance for the circuit is determined to be approximately 3.575 ohms.

PREREQUISITES
  • Understanding of series and parallel resistor configurations
  • Familiarity with Kirchhoff's Law
  • Knowledge of calculating equivalent resistance
  • Basic circuit diagram interpretation skills
NEXT STEPS
  • Study advanced circuit analysis techniques using Kirchhoff's Laws
  • Learn about Thevenin's and Norton's theorems for circuit simplification
  • Explore simulation tools like LTSpice for circuit modeling
  • Investigate practical applications of equivalent resistance in real-world circuits
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Electrical engineering students, hobbyists working on circuit design, and professionals involved in circuit analysis and troubleshooting will benefit from this discussion.

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i want to transform the black diagram into a series and parellel circuit (as u a see that the black diagram isn't exactly series or parellel). i made the two top left resistors into one (blue circle), and i want to do the same thing for the other two in green circle. however, if i make the top right resistors into one just like the top left ones, i won't get the right answer need for the problem. therefor, can someone please show me to right way to simplify this diagram?
 

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I can't see the diagram but

For series: R_{eq} = R_1 + R_2 + ... + R_n

For parallel: \frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ... + \frac{1}{R_n}
 
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why can't you see it? i can see it no problem...hmmm...

i know those equations...anways..you' ll have to see my diagram to see what i am asking.
 
It says its pending approval.
 
oh...(it's my first time posting pics)
 
Why don't you go ahead and post the problem.
 
What are the resistances of each one?
What are you trying to find?
 
well, that isn't really my problem, but if you want to know, R=2.2 (for every one).
i am trying to find the net resistance.
 
  • #10
Equivalent resistance in the blue circle is 4.4ohms. You can't reduce the two in the green circle alone because they are in parallel with the one just above the green circle. Ithink the equivalent resistance would be 0.733ohms for all 3.


edit: If you do the blue resistor with the one between the two, you should be able to simplify that section. The resistance would be 1/(1/4.4+1/2.2) = 1.466ohms.

Your circuit is now a 1.466 ohm resistor in parallel with (two 2.2 resistors in series), in series with a 2.2 ohm resistor.
 
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  • #11
if the answer is 1/.733...then it's right...
 
  • #12
why you can't combine the ones in the green circle?
 
  • #13
Because you have to keep in consideration Kirschoff's Law. The net current into a node is the net current out of a node. The net current coming out of the node on the bottom right of the green circle doesn't have all of the incoming current accounted for.
 
  • #14
k, thanks a lot whozum.
 
  • #15
No problem. I got a equivalent resistance for the whole circuit of 2.57ohms. What were you trying to find the resistance of initially?
 
  • #16
actually, the answer is 3.6 ohms...
 
  • #17
Now I am getting 3.52..
 
  • #18
lol...the exact answer in the book is 3.575...

and i still can't get the answer...don't know what i am missing...
 
  • #19
Im drawing it all out. Hold on.
 
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  • #20
http://www.public.asu.edu/~hyousif/circuitreduction.JPG
 
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  • #21
whew, holycow, THANKS A LOT MAN. i finally get it, turns out i am missing some thing i "think" i would of thought about...

again, thanks very much for your time and your elaborate drawings.
 
  • #22
No problem. Once I got it wrong I wouldn't rest without figuring it out :D
 

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