Đáp án:
\(\begin{align}
& 10) \\
& \text{a)W}={{3.10}^{-5}}J \\
& b){{I}_{0}}=0,05A \\
& 11) \\
& a)i=0,05A \\
& b)i=0,02A \\
\end{align}\)
Giải thích các bước giải:
Bài 10:
a) năng lượng điện từ:
\(\begin{align}
& \text{W}={{\text{W}}_{C}}+{{\text{W}}_{L}} \\
& =\frac{1}{2}.C.{{u}^{2}}+\frac{1}{2}.L.{{i}^{2}} \\
& =\frac{1}{2}.0,{{4.10}^{-6}}{{.10}^{2}}+\frac{1}{2}.0,2.{{({{10.10}^{-3}})}^{2}} \\
& ={{3.10}^{-5}}J \\
\end{align}\)
b)Cường độ dòng điện:
\(\begin{align}
& \text{W}=\dfrac{1}{2}.L.I_{0}^{2} \\
& \Rightarrow {{I}_{0}}=\sqrt{\frac{2W}{L}}=\sqrt{\frac{2.7,{{5.10}^{-6}}}{{{6.10}^{-3}}}}=0,05A \\
\end{align}\)
Bài 11:
$\begin{align}
& {{I}_{0}}=0,1A \\
& {{\text{W}}_{C}}=2{{W}_{L}} \\
\end{align}$
mà:
\(\begin{align}
& \text{W}={{\text{W}}_{C}}+{{\text{W}}_{L}}=4{{W}_{L}} \\
& \Leftrightarrow \dfrac{1}{2}.L.I_{0}^{2}=4.\frac{1}{2}.L.{{i}^{2}} \\
& \Rightarrow i=\sqrt{\frac{0,{{1}^{2}}}{4}}=0,05A \\
\end{align}\)
b) \(\begin{align}
& {{\text{W}}_{C}}={{\text{W}}_{L}}\Rightarrow \text{W}=2{{W}_{L}} \\
& \Leftrightarrow {{2.10}^{-6}}=2\frac{1}{2}{{.5.10}^{-3}}.{{i}^{2}} \\
& \Rightarrow i=0,02A \\
\end{align}\)
