find the equivalent resistance between A and B.

Explanation

Method 1: it is a
wheatstone bridge, but it is
not balanced. There are no
series parallel connections.
But there are similar values
on input and output sides.
Here we see that even after
using symmetry, the circuit
does not reduce to series
parallel combiantion as in
previous examples. Therefore, we apply Kirchoff's voltage
law.
In loop 1 :
`V - 10(i-x) - 5x = 0 `
or `V - 10i+5x = 0 (i)`
In loop2 : `10(i - x) -5x - 5(2x-i) = 0`
or `10i -10x - 10x +5i -5x = 0`
or `15i -25x= 0`
or `x = (15)/(25)i 5x 3i (ii)`
Using (ii) and (i),
`V - 10i +3i = 0`
or `(V)/(I) = 7 Omega or R_(eq) = 7Omega`
Method 2: From shifted
symmetry, it is clear that the
current in the resistances
connected across nodes 1 and
2 will be same as the resistance
connected across nodes 5 and
4. Hence, potential differences
across these resistances will
be equal.
We can state the same for the resistances connected across
the nodes 6 and 5 and nodes 2 and 3. let us assingn the potentials
of points A and B to be V and 0, respectively, then we can
assign potentials of different nodes as shown in the figure. At
node 2
`((x-V))/(5) +[(x-V-x)]/(5) + ((x-0))/(10) = 0`
or` x = (4)/(7)V (i)`
At point A, using Kirchhoff's junction rule
`I=((V-x))/(5) +([V -(V -x)])/(10) = (2V -x)/(10) (ii)`
From (i) and (ii)
`(V)/(I) = 7Omega = R_(eq)`