Electric Circuit, drawing question

AI Thread Summary
The discussion focuses on simplifying an electric circuit involving resistances. The initial step involves combining resistors 2 and 4 in series to yield a total resistance of 14 ohms. This combined resistance is then treated as being in parallel with resistor 3, allowing for further simplification. The resulting equivalent resistance is then added in series with the remaining resistors 1 and 5. The conversation emphasizes the systematic approach to circuit simplification, highlighting the importance of understanding series and parallel combinations.
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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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