Find the equivalent resistance between A & B of the resistor’s network. Each value of the resistor is R.

The correct option is 4R/5.

Let us find the equivalent resistance between terminals A and B by utilizing the symmetry of the circuit network shown in the image.

Looking at the provided circuit diagram, we observe a symmetric network of 12 resistors, each of value R, connected between terminal A on the left and terminal B on the right.

Let us label the key nodes in the network as follows:
- A: The input terminal node on the far left.
- B: The output terminal node on the far right.
- O: The central intersection node.
- C: The upper-left junction node connected to A, O, and E.
- D: The lower-left junction node connected to A, O, and F.
- E: The upper-right junction node connected to B, O, and C.
- F: The lower-right junction node connected to B, O, and D.

To analyze the circuit, we apply a potential difference of 1 V across the terminals by setting the potentials:
$V_A=1V$
$V_B=0V$

Due to the top-bottom symmetry of the circuit, the potentials at the symmetric upper and lower nodes must be equal:
$V_C=V_D$
$V_E=V_F$

Due to the left-right symmetry and the applied potentials, the potential at the central node O is exactly halfway between the terminal potentials:
$V_O=0.5V$

Furthermore, the potential distribution must satisfy the relation:
$V_E=1-V_C$

Now, let us write Kirchhoff's Current Law (KCL) at node C. The sum of currents leaving node C through the three connected resistors must be zero:

$\frac{V_C-V_A}{R}+\frac{V_C-V_E}{R}+\frac{V_C-V_O}{R}=0$

Multiplying the entire equation by R and substituting the known potential values:

$(V_C-1)+(V_C-(1-V_C\left)\right)+(V_C-0.5)=0$

Simplify this equation to solve for $V_C$:

$3V_C-1.5-1+V_C=0$

$4V_C-2.5=0$

$V_C=\frac{2.5}{4}=\frac{5}{8}V$

Therefore, we have:

$V_C=V_D=\frac{5}{8}V$

Now, let us calculate the total current I entering the network at node A:

$I=\frac{V_A-V_C}{R}+\frac{V_A-V_D}{R}+\frac{V_A-V_O}{R}$

Substitute the values of the potentials:

$I=\frac{1-\frac{5}{8}}{R}+\frac{1-\frac{5}{8}}{R}+\frac{1-0.5}{R}$

$I=\frac{3}{8R}+\frac{3}{8R}+\frac{1}{2R}$

$I=\frac{3}{8R}+\frac{3}{8R}+\frac{4}{8R}=\frac{10}{8R}=\frac{5}{4R}$

Finally, we apply Ohm's law to the equivalent network to determine the equivalent resistance $R_{eq}$:

$R_{eq}=\frac{V_A-V_B}{I}=\frac{1}{\frac{5}{4R}}=\frac{4R}{5}$