Đáp án:
\(R = r\)
Giải thích các bước giải:
* K mở:
Ta có:
\(\begin{array}{l}
{R_{23}} = {R_2} + {R_3} = 2R\\
{R_{AB}} = \dfrac{{{R_1}{R_{23}}}}{{{R_1} + {R_{23}}}} = \dfrac{{R.2R}}{{3R}} = \dfrac{2}{3}R\\
{R_N} = {R_{AB}} + {R_4} = \dfrac{2}{3}R + R = \dfrac{5}{3}R\\
I = \dfrac{E}{{{R_N} + r}} = \dfrac{E}{{\dfrac{5}{3}R + r}}\\
{P_1} = {I^2}.{R_N} = \dfrac{{{E^2}}}{{{{\left( {\dfrac{5}{3}R + r} \right)}^2}}}.\dfrac{5}{3}R
\end{array}\)
* K đóng:
\(\begin{array}{l}
{R_{34}} = \dfrac{{{R^2}}}{{2R}} = \dfrac{R}{2}\\
{R_{134}} = R + \dfrac{R}{2} = \dfrac{3}{2}R\\
{R_N} = \dfrac{{\dfrac{3}{2}R.R}}{{\dfrac{3}{2}R + R}} = \dfrac{3}{5}R\\
I = \dfrac{E}{{{R_N} + r}} = \dfrac{E}{{\dfrac{3}{5}R + r}}\\
{P_2} = {I^2}.{R_N} = \dfrac{{{E^2}}}{{{{\left( {\dfrac{3}{5}R + r} \right)}^2}}}.\dfrac{3}{5}R
\end{array}\)
* Đề công suất là như nhau:
\(\begin{array}{l}
{P_1} = {P_2}\\
\Rightarrow \dfrac{{{E^2}}}{{{{\left( {\dfrac{5}{3}R + r} \right)}^2}}}.\dfrac{5}{3}R = \dfrac{{{E^2}}}{{{{\left( {\dfrac{3}{5}R + r} \right)}^2}}}.\dfrac{3}{5}R\\
\Rightarrow \dfrac{3}{5}.{\left( {\dfrac{5}{3}R + r} \right)^2} = \dfrac{5}{3}.{\left( {\dfrac{3}{5}R + r} \right)^2}\\
\Rightarrow \dfrac{3}{5}\left( {\dfrac{5}{3}R + r} \right) = \dfrac{3}{5}R + r\\
\Rightarrow R + \dfrac{3}{5}r = \dfrac{3}{5}R + r\\
\Rightarrow \dfrac{2}{5}R = \dfrac{2}{5}r\\
\Rightarrow R = r
\end{array}\)
