Đáp án:
$\begin{align}
& a)Z=80\Omega ;{{I}_{0}}=1,5\sqrt{2}A;i=1,5\sqrt{2}cos(100\pi t-\frac{\pi }{6})A \\
& b){{u}_{R}}=60\sqrt{6}cos(100\pi t-\frac{\pi }{6})V \\
& {{u}_{L}}=120\sqrt{2}cos(100\pi t+\frac{\pi }{3})V \\
& {{u}_{C}}=60\sqrt{2}cos(100\pi t-\frac{2}{3}\pi )V \\
\end{align}$
Giải thích các bước giải:
$R=40\sqrt{3}\Omega ;L=\frac{4}{5}\pi H;C=\frac{{{10}^{-3}}}{4\pi }F;u=120\sqrt{2}cos100\pi t$
a) tổng trở:$\begin{align}
& {{Z}_{L}}=\omega .L=100\pi .\frac{4}{5\pi }=80\Omega \\
& {{Z}_{C}}=\frac{1}{\omega .C}=\frac{1}{100\pi .\frac{{{10}^{-3}}}{4\pi }}=40\Omega \\
& \Rightarrow Z=\sqrt{{{R}^{2}}+{{({{Z}_{L}}-{{Z}_{C}})}^{2}}}=\sqrt{{{(40\sqrt{3})}^{2}}+{{(80-40)}^{2}}}=80\Omega \\
\end{align}$
Cường độ cực đại: ${{I}_{0}}=\frac{{{U}_{0}}}{Z}=\frac{120\sqrt{2}}{80}=1,5\sqrt{2}A$
Độ lệch pha: $\begin{align}
& \tan \varphi =\frac{{{Z}_{L}}-{{Z}_{C}}}{R}=\frac{40}{40\sqrt{3}}\Rightarrow \varphi =\frac{\pi }{6} \\
& {{\varphi }_{i}}={{\varphi }_{u}}-\varphi =-\frac{\pi }{6}rad \\
\end{align}$
biểu thức: $i=1,5\sqrt{2}cos(100\pi t-\frac{\pi }{6})$
b) hiệu điện thế
$\begin{align}
& {{U}_{0R}}={{I}_{0}}.R=1,5\sqrt{2}.40\sqrt{3}=60\sqrt{6}V \\
& {{U}_{0L}}={{I}_{0}}.{{Z}_{L}}=1,5\sqrt{2}.80=120\sqrt{2}V \\
& {{U}_{0C}}={{I}_{0}}.{{Z}_{C}}=1,5\sqrt{2}.40=60\sqrt{2}V \\
\end{align}$
Pha:
$\begin{align}
& {{\varphi }_{uL}}={{\varphi }_{i}}+\frac{\pi }{2}=\frac{\pi }{3}rad \\
& {{\varphi }_{uC}}={{\varphi }_{i}}-\frac{\pi }{2}=-\frac{2\pi }{3}rad \\
& {{\varphi }_{uR}}={{\varphi }_{i}}=-\frac{\pi }{6}rad \\
\end{align}$
Phương trình:
$\begin{align}
& {{u}_{R}}=60\sqrt{6}cos(100\pi t-\frac{\pi }{6})V \\
& {{u}_{L}}=120\sqrt{2}cos(100\pi t+\frac{\pi }{3})V \\
& {{u}_{C}}=60\sqrt{2}cos(100\pi t-\frac{2}{3}\pi )V \\
\end{align}$
