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B

- Hướng dẫn giải

Phương pháp giải:

Giải chi tiết:

Không mất tính tổng quát, giả sử phương trình của \({u_C} = {U_{0C}}.\cos \omega t\,\,\left( V \right)\), độ lệch pha giữa i và u_cd là φ với \(\left( {0 < \varphi  < \dfrac{\pi }{2}} \right)\)

→ Phương trình: \({u_{cd}} = {U_{0cd}}.\cos \left( {\omega t + \dfrac{\pi }{2}} \right)\,\,\left( V \right)\,\,\,\left( 1 \right)\)

Ta có:

\(\begin{array}{l}{u_{cd}} = {U_{0cd}}.\left[ {\cos \omega t.\cos \left( {\dfrac{\pi }{2} + \varphi } \right) - \sin \omega t.\sin \left( {\dfrac{\pi }{2} + \varphi } \right)} \right]\,\\\,\,\,\,\,\,\,\, = {U_{0cd}}.\left( { - \sin \omega t} \right).\cos \varphi  - {U_{0cd}}.\cos \omega t.\sin \omega t\,\,\,\,\left( 2 \right)\end{array}\)

Từ (1) và (2) ta có: \(\left\{ \begin{array}{l}\cos \omega t = \dfrac{{{u_C}}}{{{U_{0C}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 3 \right)\\\cos \varphi .\sin \omega t + \dfrac{{{u_C}}}{{{U_{0C}}}}.\sin \varphi  = \dfrac{{{u_{cd}}}}{{{U_{0cd}}}}\,\,\,\,\,\left( 4 \right)\end{array} \right.\)

Thay (3) vào (4) có:

\(\cos \varphi .\sin \omega t + \dfrac{{{u_C}}}{{{U_{0C}}}}.\sin \varphi  =  - \dfrac{{{u_{cd}}}}{{{U_{0cd}}}} \Rightarrow \sin \omega t = \dfrac{{ - \dfrac{{{u_{cd}}}}{{{U_{0cd}}}} - \dfrac{{{u_C}}}{{{U_{0C}}}}.\sin \varphi }}{{\cos \varphi }}\,\,\,\,\,\,\,\,\,\,\left( 5 \right)\)

Từ (3) và (5) \({\left( {\dfrac{{{u_C}}}{{{U_{0C}}}}} \right)^2} + {\left( {\dfrac{{\dfrac{{{u_{cd}}}}{{{U_{0cd}}}} - \dfrac{{{u_C}}}{{{U_{0C}}}}.\sin \varphi }}{{\cos \varphi }}} \right)^2} = 1\)

Nhìn vào đồ thị: \(\left\{ \begin{array}{l}{u_C} = 1\\{u_{cd}} =  - 2a\end{array} \right.;\left\{ \begin{array}{l}{u_C} = 2\\{u_{cd}} =  - 2a\end{array} \right.;\left\{ \begin{array}{l}{u_C} = 2\\{u_{cd}} =  - a\end{array} \right.;\left\{ \begin{array}{l}{u_C} =  - 1\\{u_{cd}} = 2a\end{array} \right.\)

\( \Rightarrow \left\{ \begin{array}{l}{\left( {\dfrac{1}{{{U_{0C}}}}} \right)^2} + \left( {\dfrac{{\sin \varphi }}{{{U_{0C}}.\cos \varphi }} - \dfrac{{2a}}{{{U_{0cd}}.\cos \varphi }}} \right) = 1\,\,\,\,\left( 6 \right)\\{\left( {\dfrac{2}{{{U_{0C}}}}} \right)^2} + \left( {\dfrac{{2\sin \varphi }}{{{U_{0C}}.\cos \varphi }} - \dfrac{{2a}}{{{U_{0cd}}.\cos \varphi }}} \right) = 1\,\,\,\,\left( 7 \right)\\{\left( {\dfrac{2}{{{U_{0C}}}}} \right)^2} + \left( {\dfrac{{2\sin \varphi }}{{{U_{0C}}.\cos \varphi }} - \dfrac{a}{{{U_{0cd}}.\cos \varphi }}} \right) = 1\,\,\,\,\,\left( 8 \right)\end{array} \right.\)

Từ (7) và (8) ta có: \(\dfrac{{3{a^2}}}{{U_{0cd}^2.{{\cos }^2}\varphi }} - \dfrac{{4a.\sin \varphi }}{{{U_{0C}}.{U_{0cd}}.{{\cos }^2}\varphi }} = 0\)

Nhận thấy \(\left\{ \begin{array}{l}\cos \varphi  \ne 0\\a \ne 0\end{array} \right. \Rightarrow \dfrac{{3a}}{{{U_{0cd}}}} = \dfrac{{4\sin \varphi }}{{{U_{0C}}}}\,\,\,\,\,\left( 9 \right)\)

Từ (6) và (8) ta có:

\(\begin{array}{l}\dfrac{3}{{U_{0C}^2}} + \dfrac{{3.{{\sin }^2}\varphi }}{{U_{0C}^2.{{\cos }^2}\varphi }} = \dfrac{{3.{a^2}}}{{U_{0cd}^2.{{\cos }^2}\varphi }} \Rightarrow \left( {1 + {{\tan }^2}\varphi } \right).\dfrac{1}{{U_{0C}^2}} = \dfrac{1}{{{{\cos }^2}\varphi }}.\dfrac{{{a^2}}}{{U_{0cd}^2}}\\ \Rightarrow \dfrac{1}{{U_{0C}^2}} = \dfrac{{{a^2}}}{{U_{0cd}^2}} \Rightarrow {U_{0cd}} = a{U_{0C}}\,\,\,\,\,\,\left( {10} \right)\end{array}\)

Từ (9) và (10) ta có:

\(a = \dfrac{{3a}}{{4\sin \varphi }} \Rightarrow \sin \varphi  = \dfrac{3}{4};\left( {a \ne 0} \right) \Rightarrow \left[ \begin{array}{l}\varphi  = \arcsin \dfrac{3}{4} + k2\pi \\\varphi  = \pi  - \arcsin \dfrac{3}{4} + k2\pi \end{array} \right.;k \in Z\)

Do \(0 < \varphi  < \dfrac{\pi }{2} \Rightarrow \varphi  = \arcsin \dfrac{3}{4}\)

Góc lệch giữa u_cd và u_C  bằng \(\dfrac{\pi }{2} + \arcsin \dfrac{3}{4} \approx 2,42rad\)

Chọn B