Question

[img]https://upload.cc/i1/2020/03/23/pnljZJ.jpg[/img]


the question is in the picture

R =4

Answer 1

computing

$R_1 =\frac{1}{\frac{1}{8}+\frac{1}{4}+\frac{1}{2}+\frac{1}{8}}=1 \Omega$

$R_2=\frac{1}{\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{\frac{1}{\frac{1}{3}+\frac{1}{6}}+8}}=\frac{40}{47}\Omega$

computing for equivalent circuit

$R_{eq}=\frac{1}{\frac{1}{10+1}+\frac{47}{40}}=\frac{440}{557}=0.851 \Omega$