Solved 1 Answer
See More Answers for FREE
Enhance your learning with StudyX
Receive support from our dedicated community users and experts
See up to 20 answers per week for free
Experience reliable customer service
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 ...