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Step 1Please see the answer belowStep 2The above figure ean be redrawion asThe equivalent resistanee between A and O isR_{1}=(3 R+3 R) 11(3 R)=(6 R) 11(3 R)=\frac{(6 R)(3 R)}{6 R+3 R}=2 RLimilarly the equivalent resistance between O and B isR_{2}=(5 R+5 R) 11(5 R)=(10 R) 1(5 R)=\frac{(10 R)(5 R)}{10 R+5 R}=\frac{50 R}{15}=\frac{10 R}{3}N O W. The equivalent resistanee between A and B is,R_{e q}=R_{1}+R_{2}=2 R+\frac{10 R}{3}=\frac{6 R+10 R}{3}=\frac{16 R}{3}As, R=18 \Omega\therefore R_{e q}=\frac{16}{3} \times(18 \Omega)=96 \Omega ...