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Equivalent resistance

  • Thread starter Saitama
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  • #1
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Homework Statement


http://i50.tinypic.com/w0ggl.png

Homework Equations





The Attempt at a Solution


I am stuck at part (a) (i haven't yet tried the part (b)).
I really have no clue on how should i go on finding the equivalent resistance, i don't see a way to reduce it to simpler parallel and series combinations. Any help is appreciated.
 

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  • #2
SammyS
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Homework Statement


http://i50.tinypic.com/w0ggl.png

Homework Equations



The Attempt at a Solution


I am stuck at part (a) (i haven't yet tried the part (b)).
I really have no clue on how should i go on finding the equivalent resistance, i don't see a way to reduce it to simpler parallel and series combinations. Any help is appreciated.
Try redrawing the circuits.

Label each node and resistor.



Use any symmetry you can find to your advantage.
 
Last edited:
  • #3
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Try redrawing the circuits.

Label each node and resistor.



Use any symmetry you can find to your advantage.
I see that i can apply delta star transformation, but i don't think it would be a good choice.
I don't see any symmetry here, can i have some hints?
 
  • #4
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Are you aware of Kirchhoff's circuit laws?
 
  • #5
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Are you aware of Kirchhoff's circuit laws?
Yes. :)
 
  • #6
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Yes. :)
So... did you apply them (making use of any symmetries that are apparent)?
 
  • #7
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So... did you apply them (making use of any symmetries that are apparent)?
I thought of doing so but i cannot find the appropriate path.
And, still i don't see any symmetries. :(
 
  • #8
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That's not much of an effort... yet. ;)
 
  • #9
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That's not much of an effort... yet. ;)
Effort? I don't have an idea on how should i start. :)
I suppose i should post a pic of what i am thinking, the current in the resistors R1 and R2 should be the same as coming out of R3 and R4 due to symmetry, is that right?
 

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  • #10
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Through each resistor some unknown current Ik flows.
At each node between resistors you have some unknown Vk.

Some of those Ik might be the same due to symmetry.
Some of those Vk might be the same due to symmetry.

Set up a system of equations?
 
  • #11
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Through each resistor some unknown current Ik flows.
At each node between resistors you have some unknown Vk.

Some of those Ik might be the same due to symmetry.
Some of those Vk might be the same due to symmetry.

Set up a system of equations?
I still don't get it. I can make the equation but i am not sure if i took the right use of symmetry. Can you review the attached pic? I have mentioned the current flowing through each resistor and took use of the symmetry of circuit, is it right?
 

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  • #12
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Your use of symmetry looks overall correct. :)

However, in your leftmost I2-I3 you seem to have ignored the current through the bottom most resistor?
 
  • #13
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Your use of symmetry looks overall correct. :)

However, in your leftmost I2-I3 you seem to have ignored the current through the bottom most resistor?
Oops, sorry, i forgot about, i was focusing on the middle part. :P

I think i will have to again label the currents, i haven't sent any current in the topmost resistor too.
 
  • #14
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Yeah well, if you designate the bottom current, the topmost one will be the same due to symmetry! :)
 
  • #15
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Yeah well, if you designate the bottom current, the topmost one will be the same due to symmetry! :)
Yes, but i will have to reassign the currents to the diagonally placed resistor.
 
  • #16
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Yes, but i will have to reassign the currents to the diagonally placed resistor.
Yep.
 
  • #17
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Yep.
Can you help me finding the right loop please? I have attached the figure with nodes marked.
I attached the battery. I could find one loop useful, AYHIJEBA, from here i get an equation,
V-xR-I2R=0, where x=I-I1-I2.
V=IR+I1R
Now which loop should i select? I am stuck here. :(

EDIT:Sorry, i forgot to mention I in the pic, its the current sent by the battery.
 

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  • #18
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That looks correct! :)

Why would you select a single loop?
Pick any loop you like.

For instance pick the one through the top.
The one through the bottom will be the same, so you can leave that one out.

Now keep picking loops until you have as much variables as you have equations (avoiding those that will duplicate existing equations).
 
  • #19
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That looks correct! :)

Why would you select a single loop?
Pick any loop you like.

For instance pick the one through the top.
The one through the bottom will be the same, so you can leave that one out.

Now keep picking loops until you have as much variables as you have equations (avoiding those that will duplicate existing equations).
Thank you for all the help! :smile:
I think i can do it now, this is the worst question i have ever done on finding the equivalent resistance. Is their any easier way to do this? Any trick?
 
  • #20
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Ah well, usually it starts looking as daunting.
And then you simply set up all equations using as many symmetries as you can.
And then it turns out you can solve it.
Afterwards, you'll see a couple of symmetries or some such that you missed.
But by then it doesn't matter anymore since you have the solution (assuming that you do).
 
  • #21
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Ah well, usually it starts looking as daunting.
And then you simply set up all equations using as many symmetries as you can.
And then it turns out you can solve it.
Afterwards, you'll see a couple of symmetries or some such that you missed.
But by then it doesn't matter anymore since you have the solution (assuming that you do).
Most of the times i refrain myself from using the Kirchoff's laws to find the equivalent resistance. I never thought that i will have to use Kirchoff's laws here. I kind of get bored by making equations and selecting loops (I need to do that in my school exams as i can't use tricks in those exams. :tongue2: ). I will see if i could find an alternative way to solve it.

Thanks once again! :smile:
 
  • #22
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Let me know if you find an alternative solution! ;)

However, when you can't find equivalent resistances using regular parallel or series replacements, you will be stuck with Kirchhoff's laws.
 
  • #23
ehild
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There is a trick: If you find two points in a circuit which are at the same potential you can connect them with a wire, so the two points become a single node.
See the attachment. When connecting a battery between A and B, both E and F are "at the middle" so at the same potential. connect E and F and fold the circuit so E and F become a single node: all pairs (A,F), (C,F), D,F), (B,F) are connected with two resistors in parallel. The orange lines represent R resistance, the blue represent R/2 resistance. The resistor in the middle can be halved, making a new node G. G is at the same potential as the node (EF). You can draw a short between G and (EF), and you have all parallel and series resistors between A and B at the end.

ehild
 

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  • #24
SammyS
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attachment.php?attachmentid=49405&d=1343554966.jpg



Using what ehild gave you, now you can make use of Δ - Y transforms.

Change the ACE and BFD Δs to Ys & you can use series/parallel analysis !
 
  • #25
ehild
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Using what ehild gave you, now you can make use of Δ - Y transforms.

Change the ACE and BFD Δs to Ys & you can use series/parallel analysis !
It is not needed, Sammy. The node EF can be connected to point G and all resistors are either series of parallel.

(I hate Δ - Y transform:tongue2:)

ehild.
 

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