Đáp án:
\(\begin{array}{l}
a.I = {I_1} = {I_2} = 0,8A\\
b.\\
{U_1} = 4V\\
{U_2} = 8V\\
c.\\
{l_1} = 58,824m\\
{R_1}' = \frac{5}{9}\Omega \\
d.\\
I = {I_1} = 1,44A\\
{I_2} = 0,48A\\
{I_3} = 0,96A
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
R = {R_1} + {R_2} = 5 + 10 = 15\Omega \\
I = {I_1} = {I_2} = \dfrac{U}{R} = \dfrac{{12}}{{15}} = 0,8A\\
b.\\
{U_1} = {I_1}{R_1} = 0,8.5 = 4V\\
{U_2} = U - {U_1} = 12 - 4 = 8V\\
c.\\
{R_1} = p\dfrac{{{l_1}}}{S}\\
\Rightarrow {l_1} = \dfrac{{{R_1}S}}{p} = \dfrac{{5.0,{{2.10}^{ - 6}}}}{{1,{{7.10}^{ - 8}}}} = 58,824m\\
{R_1}' = p\dfrac{{{l_1}'}}{{S'}} = 1,{7.10^{ - 8}}\dfrac{{\dfrac{{58,824}}{3}}}{{3.0,{{2.10}^{ - 6}}}} = \dfrac{5}{9}\Omega \\
d.\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{10.5}}{{10 + 5}} = \dfrac{{10}}{3}\Omega \\
R = {R_1} + {R_{23}} = 5 + \dfrac{{10}}{3} = \dfrac{{25}}{3}\Omega \\
I = {I_1} = \dfrac{U}{R} = \dfrac{{12}}{{\dfrac{{25}}{3}}} = 1,44A\\
{U_1} = {I_1}{R_1} = 1,44.5 = 7,2V\\
{U_2} = {U_3} = U - {U_1} = 12 - 7,2 = 4,8V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{4,8}}{{10}} = 0,48A\\
{I_3} = I - {I_2} = 1,44 - 0,48 = 0,96A
\end{array}\)