Đáp án:

 A

Giải thích các bước giải:

Câu 30: 
\(u=120\sqrt{2}\text{cos200t(V);i=}\sqrt{2}\text{cos(200t+}\dfrac{\pi }{6})A;{{R}_{0}}=40\sqrt{3}\Omega ;{{Z}_{L}}=\omega .L=200.0,2=40\Omega \)

Tổng trở: 
\(Z=\dfrac{{{U}_{0}}}{{{I}_{0}}}=\dfrac{120\sqrt{2}}{\sqrt{2}}=120\Omega \)
Ta có độ lệch pha: 
\(\begin{array}{*{35}{l}}
   \text{ }\!\!~\!\!\text{ } & \tan \varphi =\dfrac{{{Z}_{L}}-{{Z}_{C}}}{{{R}_{0}}+R}  \\
   \text{ }\!\!~\!\!\text{  }\!\!~\!\!\text{ } & \Leftrightarrow \tan (-\dfrac{\pi }{6})=\dfrac{40-{{Z}_{C}}}{40\sqrt{3}+R}=-\frac{\sqrt{3}}{3}  \\
   \text{ }\!\!~\!\!\text{  }\!\!~\!\!\text{ } & {{Z}_{C}}=\frac{(40\sqrt{3}+R)\sqrt{3}}{3}+40(1)  \\
   \text{ }\!\!~\!\!\text{ } & {}  \\
\end{array}\)

tổng trở: 
\(\begin{align}
  & {{Z}^{2}}={{(R+{{R}_{0}})}^{2}}+{{({{Z}_{L}}-{{Z}_{C}})}^{2}} \\ 
 & {{120}^{2}}={{(40\sqrt{3}+{{R}_{0}})}^{2}}+{{(40-{{Z}_{C}})}^{2}} \\ 
 & {{120}^{2}}={{(40\sqrt{3}+{{R}_{0}})}^{2}}+{{\left[ 40-(\dfrac{(40\sqrt{3}+R)\sqrt{3}}{3}+40) \right]}^{2}} \\ 
 & R=20\sqrt{3}\Omega  \\ 
\end{align}\)

Dung kháng: 
\(\begin{align}
  & {{Z}_{C}}=\dfrac{(40\sqrt{3}+20\sqrt{3})\sqrt{3}}{3}+40=100\Omega  \\ 
 & {{Z}_{C}}=\dfrac{1}{\omega .C}\Rightarrow C=\dfrac{1}{100.200}={{50.10}^{-6}}F=50\mu F \\ 
\end{align}\)