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: 

\({P_1} = {P_2} = 120W \Rightarrow \left\{ \begin{array}{l} {P_1} = {P_2}\\ {P_1} = 120W \end{array} \right.\)

\( \Rightarrow \left\{ \begin{array}{l} \frac{{0,25r}}{{{{\left( {0,25r} \right)}^2} + {k^2}}} = \frac{{r + 0,25r}}{{{{\left( {r + 0,25r} \right)}^2} + {k^2}}}\\ \frac{{{U^2}.0,25r}}{{{{\left( {0,25r} \right)}^2} + {k^2}}} = 120 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} {r^2} = 3,2{k^2}\\ \frac{{{U^2}}}{{\left| k \right|}} = \frac{{720}}{{\sqrt 5 }} \end{array} \right.\)

+ Từ đó ta có:

\(\left\{ \begin{array}{l} x = \frac{{{U^2}}}{{2\left| k \right|}} = \frac{{360}}{{\sqrt 5 }}\\ y = \frac{{{U^2}.\sqrt {3,2} .\left| k \right|}}{{3,2{k^2} + {k^2}}} = \frac{{{U^2}}}{{\left| k \right|}}.\frac{{4\sqrt 5 }}{{21}} = \frac{{960}}{7}W \end{array} \right. \Rightarrow x + y = \frac{{360}}{{\sqrt 5 }} + \frac{{960}}{7} \approx 298,14\;W\)