Đáp án:
\(\begin{array}{l}
a.R = 24\Omega \\
b.\\
I = 5A\\
{I_1} = 2A\\
{I_2} = 3A\\
c.{S_2} = 9,{204.10^{ - 10}}{m^2}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
R = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{60.40}}{{60 + 40}} = 24\Omega \\
b.\\
I = \dfrac{U}{R} = \dfrac{{120}}{{24}} = 5A\\
{U_1} = {U_2} = U = 120V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{120}}{{60}} = 2A\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{120}}{{40}} = 3A\\
c.\\
l = N\pi \dfrac{{{d^2}}}{4} = 150\pi \dfrac{{0,{{025}^2}}}{4} = \dfrac{3}{{128}}\pi m\\
{R_2} = \rho \dfrac{l}{{{S_2}}} \Rightarrow {S_2} = \rho \dfrac{l}{{{R_2}}} = 0,{5.10^{ - 6}}\dfrac{{\dfrac{3}{{128}}\pi }}{{40}} = 9,{204.10^{ - 10}}{m^2}
\end{array}\)
