Đáp án:

 \(484\Omega \)

Giải thích các bước giải:

a) Ta có: 

\(\begin{array}{l}
{I_1} = \dfrac{{{P_1}}}{{{U_1}}} = \dfrac{{100}}{{220}} = \dfrac{5}{{11}}A\\
{I_2} = \dfrac{{{P_2}}}{{{U_2}}} = \dfrac{{25}}{{220}} = \dfrac{5}{{44}}A\\
{R_1} = \dfrac{{{U_1}}}{{{I_1}}} = \dfrac{{220}}{{\dfrac{5}{{11}}}} = 484\Omega \\
{R_2} = \dfrac{{{U_2}}}{{{I_2}}} = \dfrac{{220}}{{\dfrac{5}{{44}}}} = 1936\Omega 
\end{array}\)

b) Ta có: 

\(\begin{array}{l}
I = \dfrac{U}{{{R_1} + {R_2}}} = \dfrac{{220}}{{484 + 1936}} = \dfrac{1}{{11}}A\\
{P_1} = {I^2}{R_1} = {\left( {\dfrac{1}{{11}}} \right)^2}.484 = 4W\\
{P_2} = {I^2}{R_2} = {\left( {\dfrac{1}{{11}}} \right)^2}.1936 = 16W
\end{array}\)