We are all made of stars

Evaluating the equivalent resistance of one side of the star's pentagon (AECDFA), we get $\frac{1}{6} + \frac{1}{12} = \frac{1}{Side}\Rightarrow AE = AF = EC = FD = 4 \Omega$ .

Because $AC = AE + EC$ and $AD = AF + FD$ , we get $AC = AD = 8 \Omega$ .

We now are left with a square-like circuit ACBDA, with a connection between C and D. This connection's resistance can be despised because no electric current goes through it (Wheatstone Bridge). Because $AB = AC + CB = AD + DB$ , we finally get $\frac{1}{8+6} + \frac{1}{8+6} = \frac{1}{Final}\Rightarrow Final = 7 \Omega$ .

it's easier to work it out from point B to point A

from point B we can see that the current is split into two part, to point C and point D.

point C to E is a parallel circuit, as well as point E to A

point D to F and F to A is identical with point C to E and E to A so we can use the same method to calculate them.

there's no current flowing through the resistor in between C-D, so we don't have to calculate that.

calculating the resistor value:

point A-E = E-C = A- F= F-D

1/6+1/12 = 1/R

R= 4 ohms

point A-C = A-D

R= 4+4 = 8 ohms

parallel circuit A-B

1/14 +1/14=1/R

Rsum= 7 ohms

Between points ( A , E ), ( E, C ), ( A , F ) and ( F , D ) two resistors in series are parallel withanother resistor.

\Rightarrow net resistance between A and E

$= \frac {(R+R) \cdot R}{(R+R) + R}$

( this is because net resistance in parallel $= \frac{R_{1} \cdot R_{2}}{R_{1} + R_{2}}$ )

$= \frac {2R}{3}$

Similarly, net resistance between ( E, C ), ( A , F ) and ( F , D ) $= \frac{2R}{3}$ so, total resistance between A and C & A and D $= \frac {4}{3}$

Now, redrawing the diagram and a close look will reveal the formation of 'Weatsone-bridge'.

Hence no current flows in wire $\overline{CD}$ . Therefore, resistance in branch \overline{AC} is in series with $\overline{CB}$ . So, net resistance in \overline{ACB} is $\frac {4R}{3} + R = \frac {7R}{3}$ .

the net resistance between A and B is $\frac {7R}{6} = 7 \Omega$ .

due to junction separation at c nd d the equivalent resistance is 4R/3 THEN apply whestone bridge at junction abcd so that cd connection is removed then ac nd cb are in series similarly ad nd bd r in series so that the ev resistance is 7R/3 THEN the two r in parallal so answer is 7R/6=7

The "outer" resistors along CE, EA, AF, and FD have a total resistance of 12 ohms. In parallel with each 6-ohm resistor along the pentagon, the equivalent resistance between each of the four pairs is 4 ohms. AC and AD then have equivalent resistances of 8 ohms (2 four-ohm resistors in series). ACD can then undergo a delta-wye transformation. This results in the 8-8-4 delta being equivalent to a 3.2-1.6-1.6 wye. Each 1.6-ohm resistor is then in series with a 6-ohm resistor (either BC or BD). The parallel pair of 7.6-ohm resistors is then equivalent to 3.8 ohms. Combined in series with the last 3.2-ohm resistor from the delta-wye transformation yields the final result of 7 ohms.

Hint : Break the star into simple circuits

Resistance between D & F consisting of 3 resistance in which two are in series and one in parallel with the other two = (2/3) R Similarly resistance between F& A, A &E, E & C are also the same i.e (2/3) R

This means resistance between CE = resistance between EA =resistance between AF =resistance between FD = (2/3) R

Now resistance between CA = resistance between AD = (4/3) R

Now the star has reduced into a balanced Wheatstone Bridge. Thus no current will flow through the wire CD, hence the wire could be removed.

Resistance in wire AFDB = Resistance in wire AECB = (7/3) R

Hence net resistance between A & B = (1/2) * (7/3) R = (7/6) R Substituting value of R = 6ohm we get net resistance between A & B as 7ohm

Note that the resistances between C and E, D and F, F and A, and E and A are a simple section of two parallel branches, with one having two of the resistors in series and the other having one resistor. Thus the total resistance between the mentioned nodes becomes 4 ohms.

Next, the path from C to A and the path from D to A becomes a series circuit with two 4-ohm resistors. The total resistance along these paths is 8 ohms.

We now arrive at a unbalanced Wheastone bridge-like configuration, with the top 3 resistors in a triangle with side lengths 6 ohms and the bottom two parts of the unbalanced bridge being of 8 ohms. Applying a delta-wye transformation to the top three resistors, we find that the resistance of each of the resistors in the wye configuration is 2 ohms.

Now, the resistors between B and A have a 2 ohm resistor connected in series with a parallel network, whose branches each have resistance 10 ohms. Thus the total resistance of the parallel part is 5 ohms, and in series with the 2 ohm resistor we arrive at the fact that the resistance between B and A is 7 ohms.

Not a solution, but a hint!! After you get to the point where you can see a triangle ADC of resistances, Use WHEATSTONE'S METER BRIDGE CONCEPT to (i) either rule out the resistance between C & D or (ii) Short circuit the points C & D. Then carry on with your simple calculations!!