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Đáp án:

 \(\dfrac{{27}}{{14}}\Omega \)

Giải thích các bước giải:

Mạch gồm: ((R4 nt (R3 // (R1 nt R2))) // R6 // (R8 nt R7)) nt R5

Ta có:

\(\begin{array}{l}
{R_{12}} = {R_1} + {R_2} = 1 + 5 = 6\Omega \\
{R_{123}} = \dfrac{{{R_{12}}.{R_3}}}{{{R_{12}} + {R_3}}} = \dfrac{{6.3}}{{6 + 3}} = 2\Omega \\
{R_{1234}} = {R_{123}} + {R_4} = 2 + 3 = 5\Omega \\
{R_{78}} = {R_7} + {R_8} = 5 + 1 = 6\Omega \\
\dfrac{1}{{{R_{1234678}}}} = \dfrac{1}{{{R_{1234}}}} + \dfrac{1}{{{R_6}}} + \dfrac{1}{{{R_{78}}}}\\
 \Rightarrow \dfrac{1}{{{R_{1234678}}}} = \dfrac{1}{5} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{7}{{10}}\\
 \Rightarrow {R_{1234678}} = \dfrac{{10}}{7}\Omega \\
{R_{td}} = {R_{1234678}} + {R_5} = \dfrac{{10}}{7} + 0,5 = \dfrac{{27}}{{14}}\Omega 
\end{array}\)