Đáp án:
\(\begin{array}{l}
+ {R_1}nt{R_2}nt{R_3}\\
R = 50\Omega \\
+ {R_1}nt({R_2}//{R_3})\\
R = 19,375\Omega \\
+ {R_2}nt({R_1}//{R_3})\\
R = \dfrac{{155}}{7}\Omega \\
+ {R_3}nt({R_1}//{R_2})\\
R = 26\Omega \\
+ {R_1}//{R_2}//{R_3}\\
R = \dfrac{{150}}{{31}}\Omega
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
+ {R_1}nt{R_2}nt{R_3}\\
R = {R_1} + {R_2} + {R_3} = 10 + 15 + 25 = 50\Omega \\
+ {R_1}nt({R_2}//{R_3})\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{15.25}}{{15 + 25}} = 9,375\Omega \\
R = {R_1} + {R_{23}} = 10 + 9,375 = 19,375\Omega \\
+ {R_2}nt({R_1}//{R_3})\\
{R_{13}} = \dfrac{{{R_1}{R_3}}}{{{R_1} + {R_3}}} = \dfrac{{10.25}}{{10 + 25}} = \dfrac{{50}}{7}\Omega \\
R = {R_2} + {R_{13}} = 15 + \dfrac{{50}}{7} = \dfrac{{155}}{7}\Omega \\
+ {R_3}nt({R_1}//{R_2})\\
{R_{12}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{10.15}}{{10 + 15}} = 6\Omega \\
R = {R_3} + {R_{12}} = 20 + 6 = 26\Omega \\
+ {R_1}//{R_2}//{R_3}\\
\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} = \dfrac{1}{{10}} + \dfrac{1}{{15}} + \dfrac{1}{{25}} = \dfrac{1}{{\dfrac{{150}}{{31}}}}\\
\Rightarrow R = \dfrac{{150}}{{31}}\Omega
\end{array}\)
