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
–
Ricky Escobar
·
3 years, 10 months ago

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
–
Muhammed Shemeer K
·
3 years, 10 months ago

symertrycity will help u
calculate the some point as same if distance from source and sink are same
it will make easier
–
Dhurjati Das
·
3 years, 10 months ago

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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.
–
Soham Mullick
·
3 years, 10 months ago

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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.
–
Aditya Karekatte
·
3 years, 10 months ago

@Arushit Mudgal
–
i m in 10th standard, and according to cbse 10 physics syllabi, there is nothing as kirchoff law
–
Aßhĩмanyu Singh
·
3 years, 10 months ago

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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.
–
Mahathir Ahmad
·
3 years, 10 months ago

## Comments

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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 – Ricky Escobar · 3 years, 10 months ago

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– Ritvik Choudhary · 3 years, 9 months ago

hey ricky is there any other method ????Log in to reply

– Aßhĩмanyu Singh · 3 years, 9 months ago

abe mujhe to ye method jyaada easy lag rha haiLog in to reply

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 – Muhammed Shemeer K · 3 years, 10 months ago

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– Shubham Srivastava · 3 years, 10 months ago

What will be the current in 3 ohm resistance?Log in to reply

– Ricky Escobar · 3 years, 10 months ago

\(\frac{1}{9} \ \mathrm{A} \approx 0.1111 \ \mathrm{A}\)Log in to reply

1ohm – Ashutosh Giri · 3 years ago

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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 – Dhurjati Das · 3 years, 10 months ago

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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. – Soham Mullick · 3 years, 10 months ago

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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. – Aditya Karekatte · 3 years, 10 months ago

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– Aßhĩмanyu Singh · 3 years, 10 months ago

plzzz can u give me step by step solution to thisLog in to reply

– Arushit Mudgal · 3 years, 10 months ago

FIrst Reply.. u are in which standard and do you know Kirchoff LawLog in to reply

– Aßhĩмanyu Singh · 3 years, 10 months ago

i m in 10th standard, and according to cbse 10 physics syllabi, there is nothing as kirchoff lawLog in to reply

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. – Mahathir Ahmad · 3 years, 10 months ago

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– Aßhĩмanyu Singh · 3 years, 10 months ago

although i know abot v , i and parallel and series circuitsLog in to reply

– Aßhĩмanyu Singh · 3 years, 10 months ago

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

u can use star & delta techniques that will simplify it – Amgad Ahmed · 3 years, 10 months ago

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sory where do you get that question – Gonna Sing · 3 years, 10 months ago

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plzzz anyone – Aßhĩмanyu Singh · 3 years, 10 months ago

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mantap, dapet soal darimananih? – Putu Andika · 3 years, 10 months ago

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– Ridhan Fadhilah · 3 years, 10 months ago

that is indonesian language. iya mantap nihLog in to reply

– Aditya Karekatte · 3 years, 10 months ago

English, please?Log in to reply

– Putu Andika · 3 years, 10 months ago

okay, i'llLog in to reply

is it 3/8? – Mahathir Ahmad · 3 years, 10 months ago

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From SOmewhere i got the hint that it's ans is 4/3 – Aßhĩмanyu Singh · 3 years, 10 months ago

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R=3 DUE TO SYMMETRY OF THE CIRCUIT – Satyam Suman · 3 years, 10 months ago

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– Aßhĩмanyu Singh · 3 years, 10 months ago

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