All resistances in figure are 1 $$\Omega$$ each. The value of current '$$I$$' is $$\frac{a}{5}$$ A. The value of $$a$$ is ______.

All resistors are 1Ω.

Start at the top-right:

Two 1Ω in parallel:

R$$=\frac{1}{2}$$

In series with the resistor before it:

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

Now this is in parallel with the lower identical 32\frac3223​ branch:

$$R=\frac{\frac{3}{2}}{2}=\frac{3}{4}$$

Now add the resistor to its left in series:

$$R=1+\frac{3}{4}=\frac{7}{4}$$

By symmetry, bottom half also gives

$$\frac{7}{4}$$

Now these two are in parallel:

$$R=\frac{\frac{7}{4}}{2}=\frac{7}{8}$$

Finally add the first leftmost series resistor:

$$R_{eq}=1+\frac{7}{8}=\frac{15}{8}$$

which is

$$1.875Ω$$

Now

$$I=\frac{V}{R}=\frac{3}{15/8}$$

$$=\frac{8}{5}$$

Given

$$I=\frac{a}{5}$$

so

$$\frac{a}{5}=\frac{8}{5}$$

$$a=8$$