Đáp án:
\(\begin{array}{l}
a.\\
R = 15,04\Omega \\
b.\\
{I_4} = 2,327A\\
{I_3} = 1,6757A\\
{I_1} = {I_2} = 0,6513A
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{12}} = {R_1} + {R_2} = 15 + 3 = 18\Omega \\
{R_{123}} = \frac{{{R_{12}}{R_3}}}{{{R_{12}} + {R_3}}} = \frac{{18.7}}{{18 + 7}} = 5,04\Omega \\
R = {R_{123}} + {R_4} = 10 + 5,04 = 15,04\Omega \\
b.\\
{I_4} = I = \frac{U}{R} = \frac{{35}}{{15,04}} = 2,327A\\
{U_{CB}} = U - {U_{AC}} = U - {I_4}{R_4} = 35 - 2,327.10 = 11,73V\\
{I_3} = \frac{{{U_{CB}}}}{{{R_3}}} = \frac{{11,73}}{7} = 1,6757A\\
{I_1} = {I_2} = I - {I_3} = 2,327 - 1,6757 = 0,6513A
\end{array}\)
