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

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

henpen
 3 years ago
Best ResponseYou've already chosen the best response.0Given that R3=R4, I think no electrons will flow down resistor 5.

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

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

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

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0I have never heard of a delta wye conversion.

Mashy
 3 years ago
Best ResponseYou've already chosen the best response.0tommy.. 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!

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0Ok, but this network ist not a balaced circuit because that resistorratio does not fit this network.

VincentLyon.Fr
 3 years ago
Best ResponseYou've already chosen the best response.2Use 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

VincentLyon.Fr
 3 years ago
Best ResponseYou've already chosen the best response.2You 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.

ghazi
 3 years ago
Best ResponseYou've already chosen the best response.1using 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

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0Okay, I used Kirchhoff and got now this:

Mashy
 3 years ago
Best ResponseYou've already chosen the best response.0ok this is out of my league now :D

ghazi
 3 years ago
Best ResponseYou've already chosen the best response.1i guess its 300/11 ohm =27.27 ohm

TomLikesPhysics
 3 years ago
Best ResponseYou've already chosen the best response.0Are my equations alright?

VincentLyon.Fr
 3 years ago
Best ResponseYou've already chosen the best response.2Answer is 685/24 = 28.54 Ω.

VincentLyon.Fr
 3 years ago
Best ResponseYou've already chosen the best response.2There is a mistake : J5 cannot have a minus sign in eq.1 and eq.2

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

TomLikesPhysics
 3 years ago
Best 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.
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