R Woessss

Using 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: https://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

plzzz anyone

sory where do you get that question

u can use star & delta techniques that will simplify it

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.

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

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.

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

1ohm

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

is it 3/8?

mantap, dapet soal darimananih?

R=3 DUE TO SYMMETRY OF THE CIRCUIT