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TomLikesPhysics

  • 3 years ago

How do I find the total resistance of this network? Somehow I can not figure this simple network out.

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  1. TomLikesPhysics
    • 3 years ago
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  2. TomLikesPhysics
    • 3 years ago
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    I know how to deal with resistors that a parallel or in a row but this configuration somehow confuses me.

  3. TomLikesPhysics
    • 3 years ago
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    Hmmmm I ended up with 50/3 Ohms for the total resistance.

  4. henpen
    • 3 years ago
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    Given that R3=R4, I think no electrons will flow down resistor 5.

  5. TomLikesPhysics
    • 3 years ago
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    But R5 is not as big as R3 so woudn´t some electrons prefer R5 over R3?

  6. henpen
    • 3 years ago
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    But they 'know' (because of repulsion of electrons in front of them, possibly) that R4 exists.

  7. eSpeX
    • 3 years ago
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    To find the equivalent resistance for this circuit you need to do a delta|wye conversion and then do your calculations.

  8. TomLikesPhysics
    • 3 years ago
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    I have never heard of a delta wye conversion.

  9. Mashy
    • 3 years ago
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    tommy.. this network indeed is very complicated and you need to do a delta conversion .. however if R1/R3= R2/R4 then the its called a balanced circuit.. and it can be proved that.. no current flows through R5 and you can chuck it!

  10. TomLikesPhysics
    • 3 years ago
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    Ok, but this network ist not a balaced circuit because that resistor-ratio does not fit this network.

  11. Vincent-Lyon.Fr
    • 3 years ago
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    Use Δ→Y transform (fast), or use 2 loops and write down all Kirchhoff's circuit laws that apply (slow). http://en.wikipedia.org/wiki/Delta_Y

  12. Vincent-Lyon.Fr
    • 3 years ago
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    You can also remove R5 and use Thevenin's equivalent to find the fictitious generator between those terminals, then put R5 back in again. But this is longer.

  13. ghazi
    • 3 years ago
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    using thevnin will give the load voltage and i don't think we can apply thevnin here, if you remove R5 then yes we can think of it

  14. TomLikesPhysics
    • 3 years ago
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    Okay, I used Kirchhoff and got now this:

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  15. Mashy
    • 3 years ago
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    ok this is out of my league now :D

  16. ghazi
    • 3 years ago
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    i guess its 300/11 ohm =27.27 ohm

  17. TomLikesPhysics
    • 3 years ago
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    Are my equations alright?

  18. Vincent-Lyon.Fr
    • 3 years ago
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    Answer is 685/24 = 28.54 Ω.

  19. Vincent-Lyon.Fr
    • 3 years ago
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    There is a mistake : J5 cannot have a minus sign in eq.1 and eq.2

  20. Vincent-Lyon.Fr
    • 3 years ago
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    Besides, I do not understand what UR means. Are you multiplying resistance with voltage? This will not lead you anywhere.

  21. TomLikesPhysics
    • 3 years ago
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    -.- of course. My mistake, the Us have to be Is. For the Equations one and two I was looking at the nodes in the middle.

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