Đáp án:
\(\begin{array}{l}
a.{R_2} = 12\Omega \\
b.{R_3} = 24\Omega \\
c.{I_A} = 0,375A
\end{array}\)
Giải thích các bước giải:
a.
\(\begin{array}{l}
a.\\
{R_1}nt{R_2}\\
R = \dfrac{U}{I} = \dfrac{{12}}{{0,6}} = 20\Omega \\
{R_1} + {R_2} = R\\
\Rightarrow {R_2} = R - {R_1} = 20 - 8 = 12\Omega \\
b.\\
({R_2}//{R_3})nt{R_1}\\
R = \dfrac{U}{I} = \dfrac{{12}}{{0,75}} = 16\Omega \\
R = {R_1} + {R_{23}}\\
\Rightarrow {R_{23}} = R - {R_1} = 16 - 8 = 8\Omega \\
\dfrac{1}{{{R_{23}}}} = \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}\\
\Rightarrow \dfrac{1}{8} = \dfrac{1}{{12}} + \dfrac{1}{{{R_3}}}\\
\Rightarrow {R_3} = 24\Omega \\
c.\\
{R_1}nt{R_3}\\
R = {R_1} + {R_3} = 8 + 24 = 32\Omega \\
{I_A} = I = \dfrac{U}{R} = \dfrac{{12}}{{32}} = 0,375A
\end{array}\)