geomineq
Member
Vấn đề.
Thay đổi $\omega $ để ${{U}_{RL}}$ max
Các cao thủ kiểm tra giúp;;);;)
Ta có
$\begin{align}
& {{U}_{RL}}=\dfrac{U\sqrt{{{R}^{2}}+Z_{L}^{2}}}{\sqrt{{{R}^{2}}+{{\left({{Z}_{L}}-{{Z}_{C}} \right)}^{2}}}}=U\sqrt{\dfrac{{{R}^{2}}+{{L}^{2}}{{\omega }^{2}}}{{{R}^{2}}+{{\left(L\omega -\dfrac{1}{C\omega } \right)}^{2}}}} \\
& =U\sqrt{\dfrac{{{R}^{2}}+{{L}^{2}}{{\omega }^{2}}}{{{L}^{2}}{{\omega }^{2}}+\left({{R}^{2}}-\dfrac{2L}{C} \right)+\dfrac{1}{{{C}^{2}}{{\omega }^{2}}}}} \\
\end{align}$
Đặt ${{\omega }^{2}}=x$ thì ${{U}_{RL}}=U.\sqrt{\dfrac{{{R}^{2}}+{{L}^{2}}x}{{{L}^{2}}x+\left({{R}^{2}}-\dfrac{2L}{C} \right)+\dfrac{1}{{{C}^{2}}x}}}=U.\sqrt{\dfrac{{{L}^{2}}{{x}^{2}}+{{R}^{2}}x}{{{L}^{2}}{{x}^{2}}+\left({{R}^{2}}-\dfrac{2L}{C} \right)x+\dfrac{1}{{{C}^{2}}}}}$
Xét $y=\dfrac{{{L}^{2}}{{x}^{2}}+{{R}^{2}}x}{{{L}^{2}}{{x}^{2}}+\left({{R}^{2}}-\dfrac{2L}{C} \right)x+\dfrac{1}{{{C}^{2}}}}$ thì
$\begin{align}
& y'=\dfrac{\left(2{{L}^{2}}x+{{R}^{2}} \right)\left({{L}^{2}}{{x}^{2}}+\left( {{R}^{2}}-\dfrac{2L}{C} \right)x+\dfrac{1}{{{C}^{2}}} \right)}{{{\left({{L}^{2}}{{x}^{2}}+\left( {{R}^{2}}-\dfrac{2L}{C} \right)x+\dfrac{1}{{{C}^{2}}} \right)}^{2}}} \\
& -\dfrac{\left(2{{L}^{2}}x+\left( {{R}^{2}}-\dfrac{2L}{C} \right) \right)\left({{L}^{2}}{{x}^{2}}+{{R}^{2}}x \right)}{{{\left({{L}^{2}}{{x}^{2}}+\left( {{R}^{2}}-\dfrac{2L}{C} \right)x+\dfrac{1}{{{C}^{2}}} \right)}^{2}}} \\
& =\dfrac{-\dfrac{2{{L}^{3}}}{C}{{x}^{2}}+\dfrac{2{{L}^{2}}}{{{C}^{2}}}x+\dfrac{{{R}^{2}}}{{{C}^{2}}}}{{{\left({{L}^{2}}{{x}^{2}}+\left( {{R}^{2}}-\dfrac{2L}{C} \right)x+\dfrac{1}{{{C}^{2}}} \right)}^{2}}} \\
\end{align}$
$\begin{align}
& y'=0\Leftrightarrow -\dfrac{2{{L}^{3}}}{C}{{x}^{2}}+\dfrac{2{{L}^{2}}}{{{C}^{2}}}x+\dfrac{{{R}^{2}}}{{{C}^{2}}}=0 \\
& \Leftrightarrow x=\dfrac{1}{2LC}\pm \sqrt{\dfrac{1}{4{{L}^{2}}{{C}^{2}}}+\dfrac{{{R}^{2}}}{2{{L}^{3}}C}} \\
\end{align}$
$x={{\omega }^{2}}>0$ nên ${{\omega }_{RL}}=\sqrt{x}=\sqrt{\dfrac{1}{2LC}+\sqrt{\dfrac{1}{4{{L}^{2}}{{C}^{2}}}+\dfrac{{{R}^{2}}}{2{{L}^{3}}C}}}$ .
$0<{{\omega }_{CH}}=\dfrac{1}{\sqrt{LC}}<{{\omega }_{RL}}$ .
Thay đổi $\omega $ để ${{U}_{RL}}$ max
Các cao thủ kiểm tra giúp;;);;)
Ta có
$\begin{align}
& {{U}_{RL}}=\dfrac{U\sqrt{{{R}^{2}}+Z_{L}^{2}}}{\sqrt{{{R}^{2}}+{{\left({{Z}_{L}}-{{Z}_{C}} \right)}^{2}}}}=U\sqrt{\dfrac{{{R}^{2}}+{{L}^{2}}{{\omega }^{2}}}{{{R}^{2}}+{{\left(L\omega -\dfrac{1}{C\omega } \right)}^{2}}}} \\
& =U\sqrt{\dfrac{{{R}^{2}}+{{L}^{2}}{{\omega }^{2}}}{{{L}^{2}}{{\omega }^{2}}+\left({{R}^{2}}-\dfrac{2L}{C} \right)+\dfrac{1}{{{C}^{2}}{{\omega }^{2}}}}} \\
\end{align}$
Đặt ${{\omega }^{2}}=x$ thì ${{U}_{RL}}=U.\sqrt{\dfrac{{{R}^{2}}+{{L}^{2}}x}{{{L}^{2}}x+\left({{R}^{2}}-\dfrac{2L}{C} \right)+\dfrac{1}{{{C}^{2}}x}}}=U.\sqrt{\dfrac{{{L}^{2}}{{x}^{2}}+{{R}^{2}}x}{{{L}^{2}}{{x}^{2}}+\left({{R}^{2}}-\dfrac{2L}{C} \right)x+\dfrac{1}{{{C}^{2}}}}}$
Xét $y=\dfrac{{{L}^{2}}{{x}^{2}}+{{R}^{2}}x}{{{L}^{2}}{{x}^{2}}+\left({{R}^{2}}-\dfrac{2L}{C} \right)x+\dfrac{1}{{{C}^{2}}}}$ thì
$\begin{align}
& y'=\dfrac{\left(2{{L}^{2}}x+{{R}^{2}} \right)\left({{L}^{2}}{{x}^{2}}+\left( {{R}^{2}}-\dfrac{2L}{C} \right)x+\dfrac{1}{{{C}^{2}}} \right)}{{{\left({{L}^{2}}{{x}^{2}}+\left( {{R}^{2}}-\dfrac{2L}{C} \right)x+\dfrac{1}{{{C}^{2}}} \right)}^{2}}} \\
& -\dfrac{\left(2{{L}^{2}}x+\left( {{R}^{2}}-\dfrac{2L}{C} \right) \right)\left({{L}^{2}}{{x}^{2}}+{{R}^{2}}x \right)}{{{\left({{L}^{2}}{{x}^{2}}+\left( {{R}^{2}}-\dfrac{2L}{C} \right)x+\dfrac{1}{{{C}^{2}}} \right)}^{2}}} \\
& =\dfrac{-\dfrac{2{{L}^{3}}}{C}{{x}^{2}}+\dfrac{2{{L}^{2}}}{{{C}^{2}}}x+\dfrac{{{R}^{2}}}{{{C}^{2}}}}{{{\left({{L}^{2}}{{x}^{2}}+\left( {{R}^{2}}-\dfrac{2L}{C} \right)x+\dfrac{1}{{{C}^{2}}} \right)}^{2}}} \\
\end{align}$
$\begin{align}
& y'=0\Leftrightarrow -\dfrac{2{{L}^{3}}}{C}{{x}^{2}}+\dfrac{2{{L}^{2}}}{{{C}^{2}}}x+\dfrac{{{R}^{2}}}{{{C}^{2}}}=0 \\
& \Leftrightarrow x=\dfrac{1}{2LC}\pm \sqrt{\dfrac{1}{4{{L}^{2}}{{C}^{2}}}+\dfrac{{{R}^{2}}}{2{{L}^{3}}C}} \\
\end{align}$
$x={{\omega }^{2}}>0$ nên ${{\omega }_{RL}}=\sqrt{x}=\sqrt{\dfrac{1}{2LC}+\sqrt{\dfrac{1}{4{{L}^{2}}{{C}^{2}}}+\dfrac{{{R}^{2}}}{2{{L}^{3}}C}}}$ .
$0<{{\omega }_{CH}}=\dfrac{1}{\sqrt{LC}}<{{\omega }_{RL}}$ .