Đáp án:
\(\begin{array}{l}
a.{I_2} = 0,72A\\
b.{R_3} = 120\Omega
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{60.30}}{{60 + 30}} = 20\Omega \\
R = {R_1} + {R_{23}} = 5 + 20 = 25\Omega \\
I = \dfrac{U}{R} = \dfrac{{54}}{{25}} = 2,16A\\
{U_2} = {U_3}\\
\Rightarrow {I_2}{R_2} = {I_3}{R_3}\\
\left. \begin{array}{l}
\Rightarrow 60{I_2} = 30{I_3}\\
{I_2} + {I_3} = I = 2,16
\end{array} \right\} \Rightarrow {I_2} = 0,72A\\
b.\\
{U_2} = {I_2}{R_2} = 0,8.60 = 48V\\
{U_1} + {U_2} = U \Rightarrow {U_1} = U - {U_2} = 54 - 48 = 6V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{6}{5} = 1,2A\\
{I_3} = I - {I_2} = 1,2 - 0,8 = 0,4A\\
{R_3} = \dfrac{{{U_3}}}{{{I_3}}} = \dfrac{{48}}{{0,4}} = 120\Omega
\end{array}\)
