Đáp án:
a) 15
b) \({I_1} = {I_2} = 0,4A;{I_3} = {I_4} = 0,2A\)
c) 0,7A
Giải thích các bước giải:
a)\(\begin{array}{l}
{{\rm{R}}_1}{\rm{//[}}{R_2}{\rm{nt}}({R_3}//{R_4}){\rm{]}}\\
{R_{34}} = \dfrac{{{R_3}{R_4}}}{{{R_3} + {R_4}}} = 10\\
{R_{234}} = {R_2} + {R_{34}} = 30\\
{R_{td}} = \dfrac{{{R_1}.{R_{234}}}}{{{R_1} + {R_{234}}}} = 15
\end{array}\)
b) \(\begin{array}{l}
{I_1} = \dfrac{U}{{{R_1}}} = \dfrac{U}{{30}}\\
{I_2} = \dfrac{U}{{{R_{td}}}} - {I_1} = \dfrac{U}{{15}} - \dfrac{U}{{30}} = \dfrac{U}{{30}}\\
{I_3} = {I_4} = \dfrac{{{I_2}}}{2} = \dfrac{U}{{60}}\\
{I_A} = {I_3} + {I_1} = \dfrac{U}{{20}} = 0,6 \Rightarrow U = 12V\\
\Rightarrow {I_1} = {I_2} = 0,4A;{I_3} = {I_4} = 0,2A
\end{array}\)
c) \({{\rm{R}}_1}{\rm{//[}}{R_2}{\rm{nt}}{R_3}{\rm{]}}\)
\(\begin{array}{l}
{{\rm{R}}_{23}} = {R_2} + {R_3} = 40\\
{R_{123}} = \dfrac{{{R_1}{R_{23}}}}{{{R_1} + {R_{23}}}} = \dfrac{{120}}{7}\\
{I_A} = I = \dfrac{U}{{{R_{123}}}} = 0,7A
\end{array}\)
