Cho một đoạn mạch xoay chiều 

D

- Hướng dẫn giải

Đáp án D

Đặt \(k = {Z_L} - {Z_C}\)

+ Trong trường hợp 1:

\({P_1} = \frac{{{U^2}R}}{{{R^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}} = \frac{{{U^2}}}{{R + \frac{{{k^2}}}{R}}} \le \frac{{{U^2}}}{{2\left| k \right|}} = x\)

+ Trong trường hợp 2:

\({P_2} = \frac{{{U^2}\left( {R + r} \right)}}{{{{\left( {R + r} \right)}^2} + {{\left( {{Z_L} - {Z_C}} \right)}^2}}} = \frac{{{U^2}\left( {R + r} \right)}}{{{{\left( {R + r} \right)}^2} + {k^2}}}\)

Khi R=0 :

\({P_2} = \frac{{{U^2}r}}{{{r^2} + {k^2}}} = y\)

+ Từ đồ thị ta thấy, khi R=0,25r thì:

\(\begin{array}{l}{P_1} = {P_2} = 120W\\ \Rightarrow \left\{ \begin{array}{l}{P_1} = {P_2}\\{P_1} = 120W\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}\frac{{0,25r}}{{{{\left( {0,25r} \right)}^2} + {k^2}}} = \frac{{r + 0,25r}}{{{{\left( {r + 0,25r} \right)}^2} + {k^2}}}\\\frac{{{U^2}0,25r}}{{{{\left( {0,25r} \right)}^2} + {k^2}}} = 120\end{array} \right.\\ \Rightarrow \left\{ \begin{array}{l}{r^2} = 3,2{k^2}\\\frac{{{U^2}}}{{\left| k \right|}} = \frac{{720}}{{\sqrt 5 }}\end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l}x = \frac{{{U^2}}}{{2\left| k \right|}} = \frac{{360}}{{\sqrt 5 }}\\y = \frac{{{U^2}\sqrt {3,2} .\left| k \right|}}{{3,2{k^2} + {k^2}}} = \frac{{{U^2}}}{{\left| k \right|}}.\frac{{4\sqrt 5 }}{{21}} = \frac{{960}}{7}{\rm{W}}\end{array} \right.\\ \Rightarrow x + y = \frac{{360}}{{\sqrt 5 }} + \frac{{960}}{7} = 298,14{\rm{W}}\end{array}\)