Đáp án:

 bài 1

b.$i = 4\sqrt 2 \cos \left( {100\pi t + \frac{{3\pi }}{{10}}} \right)\left( A \right)$

${u_{MB}} = 320\cos \left( {100\pi t + \frac{{11\pi }}{{20}}} \right)\left( V \right)$

c.

$\begin{array}{l}
C = \frac{{{{10}^{ - 3}}}}{{4\pi }}\left( F \right)\\
{U_{MB\min }} = 200\sqrt 2 \left( V \right)
\end{array}$

bài 2:

$\begin{array}{l}
a.T = 2,{5.10^{ - 3}}s\\
f = 400\left( {Hz} \right)\\
b.{I_0} = 0,022\left( A \right)\\
{Q_0} = 8,{8.10^{ - 6}}\left( C \right)\\
c.q = 8,57,{10^{ - 6}}\left( C \right)
\end{array}$

Giải thích các bước giải:

 bài 1

$\begin{array}{l}
b.{Z_L} = L\omega  = \frac{{0,4}}{\pi }.100\pi  = 40\\
{Z_C} = \frac{1}{{C\omega }} = \frac{1}{{\frac{{{{10}^{ - 3}}}}{\pi }.100\pi }} = 10\\
Z = \sqrt {{{\left( {{Z_L} - {Z_C}} \right)}^2} + {{\left( {R + r} \right)}^2}}  = \sqrt {{{30}^2} + {{40}^2}}  = 50\\
{I_0} = \frac{{{U_{0AB}}}}{Z} = \frac{{200\sqrt 2 }}{{50}} = 4\sqrt 2 \\
\tan \varphi  = \frac{{{Z_L} - {Z_C}}}{{R + r}} = \frac{3}{4} \Rightarrow \varphi  = \frac{\pi }{5} = {\varphi _{{u_{AB}}}} - {\varphi _i} \Rightarrow {\varphi _i} = \frac{\pi }{2} - \frac{\pi }{5} = \frac{{3\pi }}{{10}}\\
 \Rightarrow i = 4\sqrt 2 \cos \left( {100\pi t + \frac{{3\pi }}{{10}}} \right)\left( A \right)\\
\tan {\varphi _{MB}} = \frac{{{Z_L}}}{{R + r}} = \frac{{40}}{{40}} = 1 \Rightarrow \varphi  = \frac{\pi }{4} = {\varphi _{{u_{MB}}}} - {\varphi _i} \Rightarrow {\varphi _{{u_{MB}}}} = \frac{\pi }{4} + \frac{{3\pi }}{{10}} = \frac{{11\pi }}{{20}}\\
{U_{0MB}} = {I_0}.{Z_{MB}} = {I_0}.\sqrt {{{\left( {r + R} \right)}^2} + Z_L^2}  = 4\sqrt 2 .\sqrt {{{40}^2} + {{40}^2}}  = 320\\
 \Rightarrow {u_{MB}} = 320\cos \left( {100\pi t + \frac{{11\pi }}{{20}}} \right)\left( V \right)\\
c,\\
{U_{MB}} = \frac{{{U_{AB}}}}{{\sqrt {{{\left( {{Z_L} - {Z_C}} \right)}^2} + {{\left( {R + r} \right)}^2}} }}.\sqrt {{{\left( {r + R} \right)}^2} + Z_L^2}  = \frac{{200}}{{\sqrt {{{\left( {40 - {Z_C}} \right)}^2} + {{40}^2}} }}.40\sqrt 2 \\
{U_{MB\min }} \Leftrightarrow \left( {40 - {Z_C}} \right) = 0 \Rightarrow \frac{1}{{C.100\pi }} = 40 \Rightarrow C = \frac{{{{10}^{ - 3}}}}{{4\pi }}\left( F \right)\\
{U_{MB\min }} = \frac{{200}}{{40}}.40\sqrt 2  = 200\sqrt 2 \left( V \right)
\end{array}$

bài 2

$\begin{array}{l}
a.\\
a.T = 2\pi \sqrt {LC}  = 2\pi \sqrt {{{4.10}^{ - 6}}{{.40.10}^{ - 3}}}  = 2,{5.10^{ - 3}}s\\
f = \frac{1}{T} = 400\left( {Hz} \right)\\
b.\\
{U_0} = {I_0}.{Z_C} = \frac{{{I_0}}}{{C\omega }} = \frac{{{I_0}}}{C}\sqrt {LC}  = {I_0}.\sqrt {\frac{L}{C}}  = 100{I_0}\\
{\left( {\frac{u}{{{U_0}}}} \right)^2} + {\left( {\frac{i}{{{I_o}}}} \right)^2} = 1\\
 \Rightarrow {\left( {\frac{2}{{100{I_0}}}} \right)^2} + {\left( {\frac{{{{10.10}^{ - 3}}}}{{{I_0}}}} \right)^2} = 1\\
 \Rightarrow {I_0} = 0,022\left( A \right)\\
{Q_0} = \frac{{{I_0}}}{\omega } = {I_0}\sqrt {LC}  = 8,{8.10^{ - 6}}\left( C \right)\\
c.\\
{\left( {\frac{q}{{{Q_0}}}} \right)^2} + {\left( {\frac{i}{{{I_0}}}} \right)^2} = 1\\
 \Rightarrow {\left( {\frac{q}{{8,{{8.10}^{ - 6}}}}} \right)^2} + {\left( {\frac{{{{5.10}^{ - 3}}}}{{0,022}}} \right)^2} = 1\\
 \Rightarrow q = 8,57,{10^{ - 6}}\left( C \right)
\end{array}$