Đáp án:
\(\begin{align}
& {{I}_{2}}=0,3A \\
& {{I}_{3}}=0,2A \\
\end{align}\)
Giải thích các bước giải:
Mạch tương đương
\(\left[ {{R}_{1}}nt({{R}_{2}}//{{R}_{3}}) \right]//{{R}_{4}}\)
a> điện trở tương đương
\({{R}_{td}}=\frac{U}{I}=\frac{36}{2}=18\Omega \)
Điện trở :
\(\begin{align}
& \dfrac{1}{{{R}_{td}}}=\dfrac{1}{{{R}_{123}}}+\dfrac{1}{{{R}_{4}}}\Rightarrow \frac{1}{{{R}_{123}}}=\dfrac{1}{18}-\dfrac{1}{24}=\dfrac{1}{72} \\
& {{R}_{123}}={{R}_{1}}+{{R}_{23}}\Rightarrow {{R}_{23}}=72-2.24=24\Omega \\
& \dfrac{1}{{{R}_{23}}}=\dfrac{1}{{{R}_{2}}}+\dfrac{1}{{{R}_{3}}}\Rightarrow \dfrac{1}{{{R}_{3}}}=\dfrac{1}{24}-\dfrac{1}{40}=\dfrac{1}{60} \\
& \Rightarrow {{R}_{3}}=60\Omega \\
\end{align}\)
b>
\(\begin{align}
& {{U}_{4}}={{U}_{123}}=U=36V\Rightarrow {{I}_{123}}=\dfrac{U}{{{R}_{123}}}=\dfrac{36}{72}=0,5A \\
& {{I}_{23}}={{I}_{1}}\Rightarrow {{U}_{23}}={{I}_{23}}.{{R}_{23}}=0,5.24=12V \\
& {{U}_{23}}={{U}_{2}}={{U}_{3}} \\
& \Rightarrow {{I}_{2}}=\dfrac{{{U}_{2}}}{{{R}_{2}}}=\dfrac{12}{40}=0,3A \\
& \Rightarrow {{I}_{3}}=\dfrac{{{U}_{3}}}{{{R}_{3}}}=\dfrac{12}{60}=0,2A \\
\end{align}\)
