Electric Circuit, drawing question

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Homework Help Overview

The discussion revolves around simplifying an electric circuit involving multiple resistors. Participants are exploring the process of combining resistances in series and parallel configurations, specifically focusing on resistors labeled 2, 3, 4, and 5.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss different methods for combining resistors, including converting a 'pie' configuration to a star and the implications of combining resistors in series versus parallel. Some express confusion about the steps involved and seek clarification on the correct approach.

Discussion Status

There are various interpretations of how to simplify the circuit, with some participants suggesting specific combinations of resistors while others express uncertainty. Guidance has been offered regarding the series and parallel combinations, but no consensus has been reached on the best method to proceed.

Contextual Notes

Some participants mention communication with their teacher regarding the combination of specific resistors, indicating that there may be differing opinions on the approach to take. The complexity of the circuit and the need for careful consideration of resistor configurations are highlighted.

NYROCKFAN
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I have to simplify this circuit step by step, and I'm stuck on how the first step after the drawing shown should look (how it looks after the resitances labeled 2 & 4 are combined).
 

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It is the 3 resistances at right which are 'combined ' you must convert the 'pie' to a star ( standard method ) then two arms of the star are combined with the existing arms , this leaves a star of which the right most is redundant the other two gives
the net resistance, as viewed from the source.
 
rayjohn01 said:
It is the 3 resistances at right which are 'combined ' you must convert the 'pie' to a star ( standard method ) then two arms of the star are combined with the existing arms , this leaves a star of which the right most is redundant the other two gives
the net resistance, as viewed from the source.


I didn't really understand any of that...

I emailed my teacher and he said the resistances 2 & 4 are combined, then I have to figure out the rest. They either combine and resistance2,4 is shown on the far right side, and then is considered parallel to resitance 3, making the circuit a series that can be simplified and solved for. Or, they combine and are next to resistance 1, which combines with it, and forms a series which can be solved.
 
That's right. The easy way is to combine resistors 2 and 4 in series (7 + 7 = 14 ohms) Then you have the new combined resistor in parallel with resistor 3 so the resistance of that is (7 * 14) / (7 + 14). Then you can just add the resulting resistor in series with the remaining two (numbers 1 and 5).
 
ceptimus said:
That's right. The easy way is to combine resistors 2 and 4 in series (7 + 7 = 14 ohms) Then you have the new combined resistor in parallel with resistor 3 so the resistance of that is (7 * 14) / (7 + 14). Then you can just add the resulting resistor in series with the remaining two (numbers 1 and 5).

Thank you! I got it!
 
I apologise , I over complicated it .
The two right resistances are combined in 'series ' rt = rx + ry. This result is in parallel with the next one left 1/rt = 1/rx + 1/ry
( in this rt is the result of the other two rx , ry whatever they are ).
the remaining 3 resistances are in series rt = rx + ry + rz
it takes a little practice but there is no magic.
Ray.
 

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