Find the effective resistance between the points A and B in figure:

Hint: In order to calculate the effective resistance between A and B, we must be aware to calculate the effective resistance in series and parallel combination. There are many closed circuits around AB such as ACB, ADCB, AEDCB and AFEDCB. So, we need to combine all these resistances and find the resultant of all of them.

Complete step by step solution:
The effective resistance for series combination is given as:
\[{R_s} = {R_1} + {R_2} + ... + {R_n}\]
Where \[{R_s}\] is the resultant resistance for series combination and
\[{R_1},{R_2},{R_n}\] are the individual resistances.
Now, the resultant resistance for parallel combination is given as:
\[\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ... + \dfrac{1}{{{R_n}}}\]
Where, \[{R_p}\] is the resultant resistance for parallel combination.
Now, from the diagram it is clear that the resistors between AF and FE are connected in series. Hence, the effective resistance between them will be \[3\Omega + 3\Omega = 6\Omega \] .
Now, this \[6\Omega \] resistance will be in parallel with the \[6\Omega \] resistance between AE as:

\[\dfrac{1}{{{R_{AE}}}} = \dfrac{1}{6} + \dfrac{1}{6}\]

\[ \Rightarrow {R_{AE}} = 3\Omega \]
Now, the diagram will be as follows:

Now, resistors between AE and ED are in series, therefore the resultant will be given as:
\[{R_{AD}} = 3\Omega + 3\Omega \]
\[ \Rightarrow {R_{AD}} = 6\Omega \]
Now, this \[{R_{AD}} = 6\Omega \] will be in parallel with the middle resistor which has resistance of \[6\Omega \]. The resultant between AD will be given by parallel combination as
\[\dfrac{1}{{{R_{AD}}}} = \dfrac{1}{6} + \dfrac{1}{6}\]
\[ \Rightarrow {R_{AD}} = 3\Omega \]
The updated diagram will be as follows:

Now, we have resistances between AD and DC in series. Hence, the resultant of them will be:
\[3\Omega + 3\Omega = 6\Omega \] and this \[6\Omega \] is parallel to the \[6\Omega \] resistance in the center of figure.
Now, we have \[6\Omega \] and \[6\Omega \] resistances which are in parallel between AC, their resultant will be:
\[\dfrac{1}{{{R_{AC}}}} = \dfrac{1}{6} + \dfrac{1}{6}\]
\[ \Rightarrow {R_{AC}} = 3\Omega \]
So, the updated diagram will be as follows:

As we have to find the resistance between A and B, hence we will consider AC and CB in series which will be in parallel to lower AB:
So for AC and CB in series the resultant resistance is given as:
\[{R_{AB}} = 3\Omega + 3\Omega \]
\[ \Rightarrow {R_{AB}} = 6\Omega \]
This will be in parallel to lower AB hence the resultant will be:
\[\dfrac{1}{{{R_{AB}}}} = \dfrac{1}{6} + \dfrac{1}{3}\]
\[{R_{AB}} = \dfrac{{18}}{9}\]
\[ \Rightarrow {R_{AB}} = 2\Omega \]

Therefore, the effective resistance between AB is \[2\Omega \].

Note: We have to always remember that the effective resistance for series combination is given as:
\[{R_s} = {R_1} + {R_2} + ... + {R_n}\]
And the effective resistance for parallel combination is given as:
\[\dfrac{1}{{{R_p}}} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + ... + \dfrac{1}{{{R_n}}}\]
This problem can be solved easily if diagrams are drawn after calculating effective resistance between two points.