Đáp án:
\(\begin{array}{l}
a.R = 12\Omega \\
b.\\
I = 1A\\
{I_1} = 0,6A\\
{I_2} = 0,4A\\
c.{l_1} = 5m
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
R = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{20.30}}{{20 + 30}} = 12\Omega \\
b.\\
I = \dfrac{U}{R} = \dfrac{{12}}{{12}} = 1A\\
U = {U_1} = {U_2} = 12V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{12}}{{20}} = 0,6A\\
{I_2} = I - {I_1} = 1 - 0,6 = 0,4A\\
c.\\
{R_1} = {p_1}\dfrac{{{l_1}}}{{{S_1}}}\\
\Rightarrow {l_1} = \dfrac{{{R_1}{S_1}}}{{{p_1}}} = \dfrac{{20.0,{{1.10}^{ - 6}}}}{{0,{{4.10}^{ - 6}}}} = 5m
\end{array}\)