# Find req about AB... Also pls tell which symm aur law u are using

Look at the symmetry of the circuit. The battery (suppose) is connected between points A and B.

From the symmetry current in AO will be same as current in OB

Similarly current CO equal to current in OD

The given circuit can be redrawn as

$R=4+4+\frac{1}{\left(\frac{1}{4}+\frac{1}{8}\right)}=8+\frac{1}{\left(\frac{2+1}{8}\right)}=8+\frac{8}{3}=\frac{32}{3}\phantom{\rule{0ex}{0ex}}\frac{1}{{R}_{AB}}=\frac{1}{4}+\frac{1}{8}+\frac{1}{\frac{32}{3}}=\frac{1}{4}+\frac{1}{8}+\frac{3}{32}=\frac{8+4+3}{32}=\frac{15}{32}\phantom{\rule{0ex}{0ex}}{R}_{AB}=\frac{32}{15}Ohm$

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