Đáp án:
\[\begin{array}{l}
a.{R_{td}} = 6\left( \Omega \right)\\
b.\\
{I_d} = 0,6\left( A \right)\\
{I_R} = 0,4\left( A \right)\\
c.{R_2} = 6\left( \Omega \right)
\end{array}\]
Giải thích các bước giải:
\(\begin{array}{l}
11.\\
a.{R_{td}} = \frac{{{R_d}R}}{{{R_d} + R}} = \frac{{10.15}}{{10 + 15}} = 6\left( \Omega \right)\\
b.R//{R_d} \Rightarrow U = {U_d} = {U_R}\\
{I_d} = \frac{U}{{{R_d}}} = \frac{6}{{10}} = 0,6\left( A \right)\\
{I_R} = \frac{U}{R} = \frac{6}{{15}} = 0,4\left( A \right)\\
c.R{'_{td}} = \frac{U}{I} = \frac{6}{{0,5}} = 12 > {R_{td}}\\
\Rightarrow {R_2}nt\left( {R//{R_d}} \right)\\
R{'_{td}} = {R_2} + {R_{td}}\\
\Rightarrow 12 = {R_2} + 6\\
\Rightarrow {R_2} = 6\left( \Omega \right)
\end{array}\)
