Idnetical resistor each of resistance r are connected as shown in figure. Calculate equivalent resistance between A and E.

Explanation

Let the potentials of nodes A and B be V and O, respecitvely,
,
from symmetry, it is clear that the current in the resistance
AB and AH will be same. Similaryl , the current in resistances
DE and FE will be same hence potential differnce across
these resistances will be same.
Let the potentials of point D and F be x, then the potentials
of nodes B and H will be (V - x). the potential difference
across the resistances BH and DF is zero. the net resistacne
can be ginven as `R_O(eq) = V//I`
Nodal equation at A is
`1 =((V-(V-x))/(r ) + (V- (V-x))/(r )) = (2x)/(r )`
Nodal equation at B is
`((V -x)- V)/(r ) = (V -x) - (x)/((2//3r)) +((V-x) -x)/(r )`
From (i) and (ii),
`I = (2)/(r )((5)/(12)V) or (V)/(I) = (6)/(I) r =R_(eq)`
Method:2 if the battery of emf V is connected across A and E, cirucit beomes as shown in fig there fore currents in AB, AH, DE, and FE are identical (I_1 say)
Current is BD and FE are indential to each other `(I_2, say)`.
Currents i BF and HD are idnetical to each other `(I_3 say)` no
currents flows through BH and DF hecen currents throgh
differents resistors will be as shown in fign applying Kirchhoff's voltage law at junction B, we get
`I_1 = I_2 +I_3 (i)` `
Applying Kirchhoff's voltage law on mesh BFHB, we get
`I_3r - (2)/(3)r I_2 = 0 or 3I_3 = 2I_2 (ii)`
Now appling Kirchhoff's voltage law on mesh AHFEJKA,
we get
`I_1r +I_2((2)/(3)r) + I_1r- V = 0 (iii)`
From Eqs. (i) and (ii) we get `I_2 = 1.5I_3 and I_1 = 2.5I_3`
Substituting these values in Eqs. (iii) we get,
`6I_3r= V or (V)/(I_3) = 6r`
Total current drawn from battery is `I = 2I_1 =5I_3` Therefore,
equivalent resistance is
`(V)/(5I_3) = (1)/(5)xx6r = 1.2r`