Điện tích tức thời $q={{q}_{0}}cos(\omega t+\varphi )$
Hiệu điện thế (điện áp) tức thời $u=\dfrac{q}{C}=\dfrac{{{q}_{0}}}{C}c\text{os}(\omega t+\varphi )={{U}_{0}}c\text{os}(\omega t+\varphi )$
Dòng điện tức thời $i=q=-\omega {{q}_{0}}sin(\omega t+\varphi )={{I}_{0}}cos(\omega t+\varphi +\dfrac{\pi }{2})$
$\left\{ \begin{align}& \frac{{{i}^{2}}}{I_{0}^{2}}+\frac{{{q}^{2}}}{q_{0}^{2}}=1 \\ & \frac{{{i}^{2}}}{I_{0}^{2}}+\frac{{{u}^{2}}}{U_{0}^{2}}=1 \\ \end{align} \right.$
${\left( {\dfrac{u}{{{U_o}}}} \right)^2} + {\left( {\dfrac{i}{{{I_o}}}} \right)^2} = 1{\left( {\dfrac{u}{{{U_o}}}} \right)^2} + {\left( {\dfrac{{{I_0}\dfrac{{\sqrt 3 }}{2}}}{{{I_o}}}} \right)^2} = 1 \to u = \dfrac{{{U_0}}}{2}.$
${{\left( \dfrac{u}{{{U}_{o}}} \right)}^{2}}+{{\left( \dfrac{i}{{{I}_{o}}} \right)}^{2}}=1\xrightarrow{{{I}_{0}}={{U}_{0}}\sqrt{\dfrac{C}{L}}}{{\left( \dfrac{u}{{{U}_{o}}} \right)}^{2}}+{{\left( \dfrac{i\sqrt{L}}{{{U}_{0}}\sqrt{C}} \right)}^{2}}=1\to {{i}^{2}}=\dfrac{C}{L}(U_{0}^{2}-{{u}^{2}})$
Thay số: ${{i}^{2}}=\dfrac{C}{L}(U_{0}^{2}-{{u}^{2}})=\dfrac{{{9.10}^{-9}}}{{{4.10}^{-3}}}\left( {{5}^{2}}-{{3}^{2}} \right)=3,{{6.10}^{-5}}\Rightarrow i={{6.10}^{-3}}A=6\text{ }mA$.
\({{\left( \dfrac{q}{{{q}_{0}}} \right)}^{2}}+{{\left( \dfrac{i}{{{I}_{0}}} \right)}^{2}}=1\to {{\left( \dfrac{q}{{{q}_{0}}} \right)}^{2}}+{{\left( \dfrac{i}{\omega .{{q}_{0}}} \right)}^{2}}=1\)
Thay số \(\xrightarrow{{}}{{\left( \dfrac{{{10}^{-7}}}{{{2.10}^{-7}}} \right)}^{2}}+{{\left( \dfrac{i}{{{2.10}^{-7}}{{.2.10}^{4}}} \right)}^{2}}=1\to i=2\sqrt{3}\text{ }mA\).
\({{\left( \dfrac{u}{{{U}_{0}}} \right)}^{2}}+{{\left( \dfrac{i}{{{I}_{0}}} \right)}^{2}}=1\xrightarrow{{{I}_{0}}={{U}_{0}}\sqrt{\dfrac{C}{L}}}{{\left( \dfrac{u}{{{U}_{0}}} \right)}^{2}}+{{\left( \dfrac{i}{{{U}_{0}}\sqrt{\dfrac{C}{L}}} \right)}^{2}}=1\to {{u}^{2}}+\dfrac{L{{i}^{2}}}{C}=U_{0}^{2}\)
\(\xrightarrow{\text{ }}{{4}^{2}}+\dfrac{0,1.0,{{02}^{2}}}{{{10.10}^{-6}}}=U_{0}^{2}\to {{U}_{0}}=2\sqrt{5}\text{ V}\text{.}\)
${\left( {\dfrac{u}{{{U_o}}}} \right)^2} + {\left( {\dfrac{i}{{{I_o}}}} \right)^2} = 1{\left( {\dfrac{u}{{{U_o}}}} \right)^2} + {\left( {\dfrac{{i\sqrt L }}{{{U_0}\sqrt C }}} \right)^2} = 1 \to {i^2} = \dfrac{C}{L}(U_0^2 - {u^2})$
${\left( {\dfrac{q}{{{q_o}}}} \right)^2} + {\left( {\dfrac{i}{{{I_o}}}} \right)^2} = 1{\left( {\dfrac{q}{{{q_o}}}} \right)^2} + {\left( {\dfrac{{0,5{I_0}}}{{{I_o}}}} \right)^2} = 1 \to q = \dfrac{{\sqrt 3 }}{2}{q_0}$
${{\left( \dfrac{q}{{{q}_{o}}} \right)}^{2}}+{{\left( \dfrac{i}{{{I}_{o}}} \right)}^{2}}=1\to {{\left( \dfrac{q}{{{q}_{o}}} \right)}^{2}}+{{\left( \dfrac{i}{\omega {{q}_{0}}} \right)}^{2}}=1$
$\xrightarrow[\text{ }]{{{q}_{0}}=\text{ }{{4.10}^{-12}}C;\text{ }q\text{ }=\text{ }{{2.10}^{-12}}\text{C; }\omega \text{ = 1}{{\text{0}}^{7}}\text{rad/s}}i=2\sqrt{3}{{.10}^{-5}}A$.
\({\left( {\dfrac{u}{{{U_o}}}} \right)^2} + {\left( {\dfrac{i}{{{I_o}}}} \right)^2} = 1 \to {\left( {\dfrac{{\dfrac{1}{2}{U_0}}}{{{U_o}}}} \right)^2} + {\left( {\dfrac{i}{{{I_o}}}} \right)^2} = 1 \to i = {I_o}\dfrac{{\sqrt 3 }}{2} = \left( {{U_0}\sqrt {\dfrac{C}{L}} } \right)\dfrac{{\sqrt 3 }}{2}\)
\({{\left( \dfrac{q}{{{q}_{o}}} \right)}^{2}}+{{\left( \dfrac{i}{{{I}_{o}}} \right)}^{2}}=1\to {{\left( \dfrac{q}{{{q}_{o}}} \right)}^{2}}+{{\left( \dfrac{i}{\omega {{q}_{o}}} \right)}^{2}}=1\to {{\left( \dfrac{q}{{{q}_{o}}} \right)}^{2}}+{{\left( \dfrac{i\sqrt{LC}}{{{q}_{o}}} \right)}^{2}}=1\to {{q}_{0}}=\sqrt{{{q}^{2}}+LC.{{i}^{2}}}\)
\(\xrightarrow{\text{ }}\text{ }{{q}_{0}}=\sqrt{{{\left( {{2.10}^{-8}} \right)}^{2}}+0,{{5.10}^{-6}}{{.6.10}^{-6}}.{{\left( {{20.10}^{-3}} \right)}^{2}}}={{4.10}^{-8}}C.\)
\({{\left( \dfrac{q}{{{q}_{o}}} \right)}^{2}}+{{\left( \dfrac{i}{{{I}_{o}}} \right)}^{2}}=1\to {{\left( \dfrac{q}{{{q}_{o}}} \right)}^{2}}+{{\left( \dfrac{i}{\omega {{q}_{0}}} \right)}^{2}}=1\)
\(\xrightarrow[\text{ }]{{{q}_{0}}=\text{ }{{10}^{-9}}C;\text{ i }=\text{ 6}{{.10}^{-6}}\text{C; }\omega \text{ = 1}{{\text{0}}^{4}}\text{rad/s}}\text{ }q={{8.10}^{-10}}C\).
Khi dòng điện trong mạch là i thì điện tích trên tụ có độ lớn là: $q=\dfrac{1}{\omega }\sqrt{I_{0}^{2}-{{i}^{2}}}$
Vậy $\dfrac{{{q}_{1}}}{{{q}_{2}}}=\dfrac{{{\omega }_{2}}}{{{\omega }_{1}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}=0,5$.
\({\left( {\dfrac{u}{{{U_0}}}} \right)^2} + {\left( {\dfrac{i}{{{I_0}}}} \right)^2} = 1{\left( {\dfrac{u}{{{I_0}\sqrt {\dfrac{L}{C}} }}} \right)^2} + {\left( {\dfrac{i}{{{I_0}}}} \right)^2} = 1 \to \dfrac{{C{u^2}}}{L} + {i^2} = I_0^2\)
Thay số \(\dfrac{{{{10.10}^{ - 6}}{{.8}^2}}}{{0,1}} + 0,{06^2} = I_0^2 \to {I_0} = 0,1{\rm{ A}}{\rm{.}}\)