Lời giải:Đáp án C
Công suất tiêu thụ R là: \(P={{I}^{2}}R=\frac{{{\text{E}}^{2}}R}{{{(R+r)}^{2}}}\)
\({{P}_{1}}={{P}_{2}}\Rightarrow \frac{{{\text{E}}^{2}}{{R}_{1}}}{{{\left( {{R}_{1}}+r \right)}^{2}}}=\frac{{{\text{E}}^{2}}{{R}_{2}}}{{{\left( {{R}_{2}}+r \right)}^{2}}}\Leftrightarrow {{R}_{1}}{{\left( {{R}_{2}}+r \right)}^{2}}={{R}_{2}}{{\left( {{R}_{1}}+r \right)}^{2}}\Rightarrow r=\sqrt{{{R}_{1}}.{{R}_{2}}}\)
\(\left\{ \begin{align}
& {{R}_{1}}=2(\Omega ) \\
& {{R}_{2}}=8(\Omega ) \\
\end{align} \right.\Rightarrow r=4(\Omega )\)
Mặt khác: \({{P}_{\max }}=\frac{{{\text{E}}^{2}}}{4r}=\frac{{{20}^{2}}}{4.4}=25(W)\)
