Đáp án:

a) \(r = 6\Omega \)

b) \(5\sqrt {10} \left( V \right)\)

Giải thích các bước giải:

a) Ta có:

\(\begin{array}{l}
{P_1} = {P_2}\\
 \Rightarrow I_1^2{R_1} = I_2^2{R_2}\\
 \Rightarrow {\left( {\dfrac{U}{{{R_1} + r}}} \right)^2}{R_1} = {\left( {\dfrac{U}{{{R_2} + r}}} \right)^2}{R_2}\\
 \Rightarrow \dfrac{{{R_1}}}{{{{\left( {{R_1} + r} \right)}^2}}} = \dfrac{{{R_2}}}{{{{\left( {{R_2} + r} \right)}^2}}}\\
 \Rightarrow \dfrac{4}{{{{\left( {r + 4} \right)}^2}}} = \dfrac{9}{{{{\left( {r + 9} \right)}^2}}}\\
 \Rightarrow r = 6\Omega 
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
{P_1} = I_1^2{R_1} = {\left( {\dfrac{U}{{{R_1} + r}}} \right)^2}{R_1} = 10\\
 \Rightarrow {\left( {\dfrac{U}{{4 + 6}}} \right)^2}.4 = 10\\
 \Rightarrow U = 5\sqrt {10} \left( V \right)
\end{array}\)