Calculate equivalent resistance

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Vibhor
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Homework Statement



Circle of triangular grid wires with centre point labelled O, outer circumference drawn as a separate circular wire, showing a triangular lattice of equal-resistance segments.


Homework Equations

The Attempt at a Solution



In the attached image the outer circular wire is represented as N . We have to find the equivalent resistance between O and N . Consider a battery of voltage V between O and N . From O there are six similar paths OA ,OB,OC,OD,OE,OF . Now , All six points would be equipotential . Consider any point , say A . No current would flow between A and B . From A there are three paths AX, AN,AY . AX and AY are symmetric w.r.t to AN so current would flow from A to X and then from X to N . Similarly current would flow from A to Y to N . Current would flow directly from A to N .I have drawn the currents in attached image .

Resistance between A and N would be 2r || r || 2r = r/2 .Resistance between O and N via A would be r+r/2 = 3r/2 . Now there are six similar parallel paths from O to N .Hence equivalent resistance would be r/4 .

Is that correct ? Is the reasoning also valid ?

Many Thanks
 
Attachments
  • Circle of wire containing a triangular lattice of equal wire segments, with centre point labelled O; the outer circle is the circumference wire.
    question.PNG
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  • Triangular lattice of equal resistors inside a circular outer wire, with points labelled O at centre, A to F on the inner ring, and N, X and Y on the outer circle.
    label.PNG
    10.2 KB · Views: 560
  • Triangular lattice network of resistors inside a circular outer wire labelled N, with centre node O, nodes labelled A to F and X and Y on the rim, and green arrows showing assumed current directions from A toward X, Y and N.
    current.PNG
    10.6 KB · Views: 599
Vibhor said:

Homework Statement



View attachment 211076

Homework Equations

The Attempt at a Solution



In the attached image the outer circular wire is represented as N . We have to find the equivalent resistance between O and N . Consider a battery of voltage V between O and N . From O there are six similar paths OA ,OB,OC,OD,OE,OF . Now , All six points would be equipotential . Consider any point , say A . No current would flow between A and B . From A there are three paths AX, AN,AY . AX and AY are symmetric w.r.t to AN so current would flow from A to X and then from X to N . Similarly current would flow from A to Y to N . Current would flow directly from A to N .I have drawn the currents in attached image .

Resistance between A and N would be 2r || r || 2r = r/2 .Resistance between O and N via A would be r+r/2 = 3r/2 . Now there are six similar parallel paths from O to N .Hence equivalent resistance would be r/4 .

Is that correct ? Is the reasoning also valid ?

Many Thanks
That all looks right to me.
 
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haruspex said:
That all looks right to me.

Thanks for confirming .

But official key states r/6 as the answer :eek: .
 
Vibhor said:
Thanks for confirming .

But official key states r/6 as the answer :eek: .
That is clearly wrong. We could connect ABCDEF with a zero resistance, and the effective resistance from O to there would be r/6. There must be something to add to get to N.
 
haruspex said:
That is clearly wrong. We could connect ABCDEF with a zero resistance, and the effective resistance from O to there would be r/6. There must be something to add to get to N.

Right .

Sorry , I had read answer of some other question in the key o:) .

Answer key states answer as r/3 .Solution says there would be no current in AX , XN , AY, YN wires . Current would flow from O to A and from A to N . Resistance between O and N would be 2r . Hence equivalent resistance is r/3 .
 
Vibhor said:
Solution says there would be no current in AX , XN , AY, YN wires .
Again, clearly false.
 
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