The correct answer is \(\frac{299R}{330} \approx .906R \).

Using Pranav A.'s method of folding the circuit, and then applying a Delta-Y transformation (http://en.wikipedia.org/wiki/Y-%CE%94transform), I was able to reduce the circuit to a 13-resistor,10-node circuit(http://i.imgur.com/OQ0t9f8.png). From there, you can use the Node-Voltage method (http://en.wikipedia.org/wiki/Nodalanalysis), along with assuming a test voltage applied across A and B, to solve the circuit, although it requires using a rather large system of equations. If anyone knows of a way to further reduce the circuit from the 13-resistor state so that you don't have to get such a large system please tell me.
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Ricky Escobar
·
4 years ago

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@Ricky Escobar
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I too had the same problem. I had to apply the Y-delta transform again and again but it wasn't ending up with nice values so I gave up.
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Pranav Arora
·
4 years ago

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IF ANYONE OF YOU FIND THIS PROBLEM IN A BOOK,DON'T THINK I HAVE COPIED IT....I CREATED IT MYSELF....
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Raja Metronetizen
·
4 years ago

I am not sure about it but first step is obviously folding the circuit about AB. This step retains only the half portion of circuit with all the resistors equal to R/2 except the middle line of four resistors which still have R resistance. The second step involves removing "fake" nodes but I still need to figure out those nodes.
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Pranav Arora
·
4 years ago

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@Pranav Arora
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there is a great theory to compute the equivalent resistance of such big complicated circuits of resistance....it opines that if any system of resistance is symmetrical about its two opposite point here the two points "A" and "B" then all the points equidistant from this two terminal would be at same potential ......i knew this fact or theorem but i don't know how to apply it here.......i think it would have to be applied here.....try it....
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Raja Metronetizen
·
4 years ago

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Delta Y transformation is the only way(known to me), and you will definitely need a good calculator!!
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Shourya Pandey
·
4 years ago

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well really...... current had to do lot of hard work!!!!!
please clearly mention your problem, the question is not clearly visible.
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Diksha Verma
·
4 years ago

@Sreehari Vp
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What are the values of the resistors...are they all same or different
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Mukesh G
·
4 years ago

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@Mukesh G
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you have not seen the whole problem..... i have written it in the right corner........ however, each and every resistor have equivalent resistance=R.......
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Raja Metronetizen
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4 years ago

Consider each resistor 'R'
Fold the circuit along AB. When resistors overlap, consider it a parallel combination and replace it by 'R/2'.

So, except the 4 resistors in the AB line, you get half the circuit where each resistor is 'R/2'. Now, draw a line of symmetry along the perpendicular bisector of AB. There are two nodes on this line, the mid point of AB and the midpoint of the bottom line. Remove these connections as follows : for the top:

something like ---------
^

and for the bottom, something like : v

Now use the normal laws and solve the problem.
–
Kashyap Mahler
·
4 years ago

## Comments

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TopNewestThe correct answer is \(\frac{299R}{330} \approx .906R \).

Using Pranav A.'s method of folding the circuit, and then applying a Delta-Y transformation (http://en.wikipedia.org/wiki/Y-%CE%94

transform), I was able to reduce the circuit to a 13-resistor,10-node circuit(http://i.imgur.com/OQ0t9f8.png). From there, you can use the Node-Voltage method (http://en.wikipedia.org/wiki/Nodalanalysis), along with assuming a test voltage applied across A and B, to solve the circuit, although it requires using a rather large system of equations. If anyone knows of a way to further reduce the circuit from the 13-resistor state so that you don't have to get such a large system please tell me. – Ricky Escobar · 4 years agoLog in to reply

– Pranav Arora · 4 years ago

I too had the same problem. I had to apply the Y-delta transform again and again but it wasn't ending up with nice values so I gave up.Log in to reply

IF ANYONE OF YOU FIND THIS PROBLEM IN A BOOK,DON'T THINK I HAVE COPIED IT....I CREATED IT MYSELF.... – Raja Metronetizen · 4 years ago

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– Sreehari Vp · 4 years ago

Do you know the answer?Log in to reply

– Raja Metronetizen · 4 years ago

no, sorry i don't know it..but if anyone describes me how to approach, i could understand....Log in to reply

Thanks Ricky E. !! ... You've been brilliant. – Sung Moo Hong · 4 years ago

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– Shourya Pandey · 4 years ago

Is that a pun on the word brilliant? lol! :PLog in to reply

I am not sure about it but first step is obviously folding the circuit about AB. This step retains only the half portion of circuit with all the resistors equal to R/2 except the middle line of four resistors which still have R resistance. The second step involves removing "fake" nodes but I still need to figure out those nodes. – Pranav Arora · 4 years ago

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– Raja Metronetizen · 4 years ago

there is a great theory to compute the equivalent resistance of such big complicated circuits of resistance....it opines that if any system of resistance is symmetrical about its two opposite point here the two points "A" and "B" then all the points equidistant from this two terminal would be at same potential ......i knew this fact or theorem but i don't know how to apply it here.......i think it would have to be applied here.....try it....Log in to reply

Delta Y transformation is the only way(known to me), and you will definitely need a good calculator!! – Shourya Pandey · 4 years ago

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well really...... current had to do lot of hard work!!!!! please clearly mention your problem, the question is not clearly visible. – Diksha Verma · 4 years ago

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Turns out to be 'R' (I think so) – Kashyap Mahler · 4 years ago

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Image Link I simulated this problem. these resistors have 21/19R ohms. I think It is hard to solve by hand. – Sung Moo Hong · 4 years ago

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Very complicated...... – Sreehari Vp · 4 years ago

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– Mukesh G · 4 years ago

What are the values of the resistors...are they all same or differentLog in to reply

– Raja Metronetizen · 4 years ago

you have not seen the whole problem..... i have written it in the right corner........ however, each and every resistor have equivalent resistance=R.......Log in to reply

I think the answer is 52R/45. – Rafael Saboya · 4 years ago

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Consider each resistor 'R' Fold the circuit along AB. When resistors overlap, consider it a parallel combination and replace it by 'R/2'.

So, except the 4 resistors in the AB line, you get half the circuit where each resistor is 'R/2'. Now, draw a line of symmetry along the perpendicular bisector of AB. There are two nodes on this line, the mid point of AB and the midpoint of the bottom line. Remove these connections as follows : for the top:

something like --------- ^

and for the bottom, something like :

vNow use the normal laws and solve the problem. – Kashyap Mahler · 4 years ago

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– Kaustubh Olpadkar · 3 years, 9 months ago

What's your answer??Log in to reply

first we change the delta connection into star connection and now we solve it easy series and parallal combination – Attain k Gupta · 4 years ago

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– Raja Metronetizen · 4 years ago

would you please elaborate it....?...i'm unaware of these connections.....Log in to reply

– Ricky Escobar · 4 years ago

http://en.wikipedia.org/wiki/Y-%CE%94_transformLog in to reply