The following simulation shows that when a \(5 \ \mathrm{V}\) voltage is applied across the circuit, a \(5 \ \mathrm{A}\) current results, which implies the equivalent resistance of the circuit is \(1 \ \Omega\), verifying that \(R=\frac{3}{2} \ \Omega\) is correct: http://imgur.com/WoS0Yn7

Look at the circuit, It is symmetric for any value of R. We can proceed to result in many ways including Kirchoff's law , One of the easiest method is given here . It can be only applied for a symmetric circuit(caution).
Consider a 1 A current is flowing through it , as the circuit is symmetric 1/3 A will flow through each 1 ohm .Also 1/3 A will spit to 3 ohm and R, then finally 1/3 A will flow through last 1 ohm (1/3 A through each 1 ohm). Thus we get over all resistance as 1/3+(3 parallel to R)/3+1/3 =1 ,(3 parallel to R)/3=1/3 ,Thus R=3/2 .
If any explanation needed, then please specify

hey abhimanyu u can try to unferstand this. as points A and B are given name other 6 points as C,D,....H. Then try to figure out if a current say I starts from point A u can see the just next 3 corners are all connected by a 1 ohm resistor. so the current will be divided equally. then try to figure out those corners and the wires connecting then are they equal? try to find out the symmetry. U also can find from back ward also. after starting from A u try to picturise if the current I enters b and the next three corners of B are all equal resistance wire connected so the current would have come all equally from all those 3 corners try to ratioanalise ur thinking this way. i hope u will solve the remaining part urself.

There is a similar question in haliday,resnick and krane volume-2 physics book.Basically,here,you have to notice the symmetry and the points with the same potential.Through it,you can rearrange this circuit and solve it.Hopefully if i have not done any mistake in calculation the answer is 3/8.

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TopNewestUsing the methods outlined on this page: http://www.rfcafe.com/miscellany/factoids/kirts-cogitations-256.htm, you can set up the following equation:

\((\frac{1}{1 \ \Omega}+\frac{1}{1 \ \Omega}+\frac{1}{1 \ \Omega})^{-1} + (\frac{1}{3 \ \Omega}+\frac{1}{3 \ \Omega}+\frac{1}{3 \ \Omega}+\frac{1}{R}+\frac{1}{R}+\frac{1}{R})^{-1} +(\frac{1}{1 \ \Omega}+\frac{1}{1 \ \Omega}+\frac{1}{1 \ \Omega})^{-1} = 1 \ \Omega\).

Simplifying,

\((\frac{1}{3} \ \Omega) + (\frac{1}{1 \ \Omega} + \frac{3}{R})^{-1} + (\frac{1}{3} \ \Omega) = 1 \ \Omega\)

\((\frac{1}{1 \ \Omega} + \frac{3}{R})^{-1} = \frac{1}{3} \ \Omega\)

\(\frac{1}{1 \ \Omega}+\frac{3}{R}=\frac{3}{1 \ \Omega}\)

\(\frac{3}{R}=\frac{2}{1 \ \Omega}\)

\(R=\frac{3}{2} \ \Omega\).

The following simulation shows that when a \(5 \ \mathrm{V}\) voltage is applied across the circuit, a \(5 \ \mathrm{A}\) current results, which implies the equivalent resistance of the circuit is \(1 \ \Omega\), verifying that \(R=\frac{3}{2} \ \Omega\) is correct: http://imgur.com/WoS0Yn7

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hey ricky is there any other method ????

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abe mujhe to ye method jyaada easy lag rha hai

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Look at the circuit, It is symmetric for any value of R. We can proceed to result in many ways including Kirchoff's law , One of the easiest method is given here . It can be only applied for a symmetric circuit(caution). Consider a 1 A current is flowing through it , as the circuit is symmetric 1/3 A will flow through each 1 ohm .Also 1/3 A will spit to 3 ohm and R, then finally 1/3 A will flow through last 1 ohm (1/3 A through each 1 ohm). Thus we get over all resistance as 1/3+(3 parallel to R)/3+1/3 =1 ,(3 parallel to R)/3=1/3 ,Thus R=3/2 . If any explanation needed, then please specify

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What will be the current in 3 ohm resistance?

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\(\frac{1}{9} \ \mathrm{A} \approx 0.1111 \ \mathrm{A}\)

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1ohm

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symertrycity will help u calculate the some point as same if distance from source and sink are same it will make easier

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hey abhimanyu u can try to unferstand this. as points A and B are given name other 6 points as C,D,....H. Then try to figure out if a current say I starts from point A u can see the just next 3 corners are all connected by a 1 ohm resistor. so the current will be divided equally. then try to figure out those corners and the wires connecting then are they equal? try to find out the symmetry. U also can find from back ward also. after starting from A u try to picturise if the current I enters b and the next three corners of B are all equal resistance wire connected so the current would have come all equally from all those 3 corners try to ratioanalise ur thinking this way. i hope u will solve the remaining part urself.

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Try applying Kirchoff's Law. By Conservation of Charge, the current entering at A should leave at B, right? So start there.

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plzzz can u give me step by step solution to this

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FIrst Reply.. u are in which standard and do you know Kirchoff Law

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There is a similar question in haliday,resnick and krane volume-2 physics book.Basically,here,you have to notice the symmetry and the points with the same potential.Through it,you can rearrange this circuit and solve it.Hopefully if i have not done any mistake in calculation the answer is 3/8.

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although i know abot v , i and parallel and series circuits

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hey, actually i am in class 10th and we have not covered this topic yet.

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u can use star & delta techniques that will simplify it

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sory where do you get that question

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plzzz anyone

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mantap, dapet soal darimananih?

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that is indonesian language. iya mantap nih

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English, please?

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okay, i'll

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is it 3/8?

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From SOmewhere i got the hint that it's ans is 4/3

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R=3 DUE TO SYMMETRY OF THE CIRCUIT

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what do you mean, can't it be solved with kirchoff's law. well i dont know about it

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