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TomLikesPhysics
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How do I find the total resistance of this network? Somehow I can not figure this simple network out.
 one year ago
 one year ago
TomLikesPhysics Group Title
How do I find the total resistance of this network? Somehow I can not figure this simple network out.
 one year ago
 one year ago

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TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
I know how to deal with resistors that a parallel or in a row but this configuration somehow confuses me.
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Hmmmm I ended up with 50/3 Ohms for the total resistance.
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
Given that R3=R4, I think no electrons will flow down resistor 5.
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
But R5 is not as big as R3 so woudn´t some electrons prefer R5 over R3?
 one year ago

henpen Group TitleBest ResponseYou've already chosen the best response.0
But they 'know' (because of repulsion of electrons in front of them, possibly) that R4 exists.
 one year ago

eSpeX Group TitleBest ResponseYou've already chosen the best response.0
To find the equivalent resistance for this circuit you need to do a deltawye conversion and then do your calculations.
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
I have never heard of a delta wye conversion.
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
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!
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Ok, but this network ist not a balaced circuit because that resistorratio does not fit this network.
 one year ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.2
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
 one year ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.2
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.
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Okay, I used Kirchhoff and got now this:
 one year ago

Mashy Group TitleBest ResponseYou've already chosen the best response.0
ok this is out of my league now :D
 one year ago

ghazi Group TitleBest ResponseYou've already chosen the best response.1
i guess its 300/11 ohm =27.27 ohm
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
Are my equations alright?
 one year ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.2
Answer is 685/24 = 28.54 Ω.
 one year ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.2
There is a mistake : J5 cannot have a minus sign in eq.1 and eq.2
 one year ago

VincentLyon.Fr Group TitleBest ResponseYou've already chosen the best response.2
Besides, I do not understand what UR means. Are you multiplying resistance with voltage? This will not lead you anywhere.
 one year ago

TomLikesPhysics Group TitleBest ResponseYou've already chosen the best response.0
. 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.
 one year ago
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