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Equivalent resistance

  • Thread starter Dr.Phy
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  • #1
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Hi guys,I'm new here and maybe I'm posting this homework in a wrong part of the forum.I tried to find the equivalent resistance.It may be a simple homework but im learning first for the electricy physic.I tried to find Re first R2+R3=4 Ω and than found Re=1/4 Ω + 1/2 Ω + 1/2 Ω can i slove it this way? (P.s sorry for my bad english.)
 

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  • #2
ehild
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Hi guys,I'm new here and maybe I'm posting this homework in a wrong part of the forum.I tried to find the equivalent resistance.It may be a simple homework but im learning first for the electricy physic.I tried to find Re first R2+R3=4 Ω and than found Re=1/4 Ω + 1/2 Ω + 1/2 Ω can i solve it this way? (P.s sorry for my bad english.)

R1 is not parallel with the resultant of R2, R3, R4. Check.

ehild
 
  • #3
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Can it be formed so?
 

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  • #4
ehild
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  • #5
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Yes. :smile:

ehild
Then I did it so:R23=2Ω +2Ω =4Ω and 1/Re=1/4Ω + 1/2Ω +1/2Ω = 5/4Ω Re=4Ω /5? Is this the right way of slowing it?
 
  • #6
ehild
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Then I did it so:R23=2Ω +2Ω =4Ω and 1/Re=1/4Ω + 1/2Ω +1/2Ω = 5/4Ω Re=4Ω /5? Is this the right way of solwing it?
No. What does parallel connection mean? Explain which resistors are connected in series and which are parallel.


ehild
 
  • #7
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R2,R3 are connected in series.And R4 R1 in paralel?
 
  • #8
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R2,R3 are connected in series.
Yes.

Dr.Phy said:
And R4 R1 in paralel?
No. The equivalent resistance of R2 and R3 is in parallel with R4. Do you see why?
 
  • #9
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Yes.



No. The equivalent resistance of R2 and R3 is in parallel with R4. Do you see why?
Yes but i dont know how is it connected the R1,and how to find the eq reasistance of this circuit can u help me :/ .
 
  • #10
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Yes but i dont know how is it connected the R1,and how to find the eq reasistance of this circuit can u help me :/ .
Draw figures after each reduction. R2 and R3 are in series as you said. Replace them with a single resistor and draw the figure.
 

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