Đáp án:

a) \(100\sqrt 2 \left( V \right)\) ; \(100\sqrt 2 \left( V \right)\)

b) \(108,{43^o}\)

c) \( - {45^o}\)

Giải thích các bước giải:

a) Ta có:

\(\begin{array}{l}
\dfrac{{{U_L}}}{{{U_{RC}}}} = \dfrac{{{Z_L}}}{{\sqrt {{R^2} + Z_C^2} }} \Rightarrow \dfrac{{200\sqrt 2 }}{{200}} = \dfrac{{{Z_L}}}{{\sqrt {{R^2} + Z_C^2} }}\\
 \Rightarrow \sqrt 2  = \dfrac{{{Z_L}}}{{\sqrt {{R^2} + Z_C^2} }} \Rightarrow 2\left( {{R^2} + Z_C^2} \right) = Z_L^2\\
\dfrac{{{U_L}}}{U} = \dfrac{{{Z_L}}}{Z} \Rightarrow \dfrac{{200\sqrt 2 }}{{200}} = \dfrac{{{Z_L}}}{{\sqrt {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} }}\\
 \Rightarrow 2\left( {{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}} \right) = Z_L^2\\
 \Rightarrow 2\left( {{R^2} + Z_L^2 - 2{Z_L}{Z_C} + Z_C^2} \right) = Z_L^2\\
 \Rightarrow Z_L^2 + 2\left( {Z_L^2 - 2{Z_L}{Z_C}} \right) = Z_L^2\\
 \Rightarrow {Z_L} = 2{Z_C}\\
 \Rightarrow {U_L} = 2{U_C}\\
 \Rightarrow {U_C} = \dfrac{{{U_L}}}{2} = \dfrac{{200\sqrt 2 }}{2} = 100\sqrt 2 \left( V \right)\\
2\left( {{R^2} + Z_C^2} \right) = Z_L^2 \Rightarrow 2\left( {{R^2} + Z_C^2} \right) = 4Z_C^2\\
 \Rightarrow R = {Z_C} \Rightarrow {U_R} = {U_C} = 100\sqrt 2 \left( V \right)
\end{array}\)

b) Ta có:

\(\begin{array}{l}
\tan {\varphi _{AN}} = \dfrac{{{Z_L}}}{R} = \dfrac{{2{Z_C}}}{{{Z_C}}} = 2 \Rightarrow {\varphi _{AN}} = 63,{43^o}\\
\tan {\varphi _{MB}} =  - \dfrac{{{Z_C}}}{R} =  - 1 \Rightarrow {\varphi _{MB}} =  - {45^o}
\end{array}\)

Độ lệch pha giữa u AN và u MB là:

\(\Delta \varphi  = {\varphi _{AN}} - {\varphi _{MB}} = 108,{43^o}\)

c) Ta có:

\(\begin{array}{l}
{\varphi _{NB}} =  - {90^o}\\
{\varphi _{MB}} =  - {45^o}
\end{array}\)

Độ lệch pha giữa u NB và u MB là:

\(\Delta \varphi  = {\varphi _{NB}} - {\varphi _{MB}} =  - {45^o}\)