Đáp án B

\(P={{I}^{2}}R=\dfrac{{{U}^{2}}}{{{R}^{2}}+{{\left( {{Z}_{L}}-{{Z}_{C}} \right)}^{2}}}\to {{P}_{\max }}=\dfrac{{{U}^{2}}}{2R}\leftrightarrow R=\left| {{Z}_{L}}-{{Z}_{C}} \right|\)
Từ đồ thị: \(R=40\Omega \leftrightarrow {{P}_{{{1}_{\max }}}}=\dfrac{{{U}_{1}}^{2}}{2R}=250\to {{U}_{1}}=100\sqrt{2}\left( V \right)\)
Có 2 giá trị của R là \({{R}_{1}}=20\Omega ;{{R}_{2}}=80\Omega \) cho cùng công suất P1 = 200W
\(\dfrac{20{{U}_{1}}^{2}}{{{20}^{2}}+{{\left( {{Z}_{{{L}_{1}}}}-{{Z}_{{{C}_{1}}}} \right)}^{2}}}=\dfrac{80{{U}_{1}}^{2}}{{{80}^{2}}+{{\left( {{Z}_{{{L}_{1}}}}-{{Z}_{{{C}_{1}}}} \right)}^{2}}}\to \left| {{Z}_{{{L}_{1}}}}-{{Z}_{{{C}_{1}}}} \right|=40\Omega \)
Có 2 giá trị của R là \({{R}_{1}}=80\Omega ;{{R}_{2}}=180\Omega \) cho cùng công suất P2 = 200W \(\dfrac{80{{U}_{2}}^{2}}{{{20}^{2}}+{{\left( {{Z}_{{{L}_{2}}}}-{{Z}_{{{C}_{2}}}} \right)}^{2}}}=\dfrac{180{{U}_{1}}^{2}}{{{180}^{2}}+{{\left( {{Z}_{{{L}_{2}}}}-{{Z}_{{{C}_{2}}}} \right)}^{2}}}\to \left| {{Z}_{{{L}_{2}}}}-{{Z}_{{{C}_{2}}}} \right|=120\Omega\)
\(\begin{align} {{P}_{1}}={{P}_{2}}\to \dfrac{20{{U}_{1}}^{2}}{{{20}^{2}}+{{\left( {{Z}_{{{L}_{1}}}}-{{Z}_{{{C}_{1}}}} \right)}^{2}}}=\dfrac{80{{U}_{2}}^{2}}{{{80}^{2}}+{{\left( {{Z}_{{{L}_{2}}}}-{{Z}_{{{C}_{2}}}} \right)}^{2}}} \\ \to \dfrac{20.{{\left( 100\sqrt{2} \right)}^{2}}}{{{20}^{2}}+{{40}^{2}}}=\dfrac{80{{U}_{2}}^{2}}{{{80}^{2}}+{{120}^{2}}}\to {{U}_{2}}=20\sqrt{130}\left( V \right)\to {{P}_{2}}=130W \\ \end{align}\)