Đáp án:
a,$\begin{array}{l}
{\xi _n} = {\xi _1} + {\xi _2}\\
{r_b} = {r_1} + {r_2}
\end{array}$
b.$I = 0,86\left( A \right)$ ${U_{AB}} = 16,34\left( V \right)$
${R_1} = 2\left( \Omega \right)$
Giải thích các bước giải:
$\begin{array}{l}
a.{\xi _n} = {\xi _1} + {\xi _2}\\
{r_b} = {r_1} + {r_2}\\
b.{R_1}nt\left( {{R_2}//{R_3}} \right)\\
{R_{23}} = \frac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \frac{{12.4}}{{12 + 4}} = 3\\
{R_{123}} = {R_1} + {R_{23}} = 16 + 3 = 19\\
I = \frac{{{\xi _b}}}{{{R_{123}} + {r_b}}} = \frac{{3 + 9}}{{1 + 4 + 19}} = 0,86\left( A \right)
\end{array}$
${U_{AB}} = I.{R_{123}} = 0,86.19 = 16,34\left( V \right)$
công suất mạch ngoài
${P_n} = {I^2}.{R_{123}} = {\left( {\frac{{{\xi _b}}}{{{R_{123}} + {r_b}}}} \right)^2}.{R_{123}} = \frac{{\xi _b^2}}{{{R_{123}} + \frac{{r_b^2}}{{{R_{123}}}} + 2{r_b}}}$
để ${P_{nmax}}$ ta có:
${R_{123}} + \frac{{r_b^2}}{{{R_{123}}}} \ge 2{r_b} \Leftrightarrow {R_{123}} = \frac{{r_b^2}}{{{R_{123}}}} \Rightarrow {r_b} = {R_{123}} = {R_1} + 3 \Rightarrow 5 = {R_1} + 3 \Rightarrow {R_1} = 2\left( \Omega \right)$
