Đáp án:

\(\dfrac{9}{4}\)

Giải thích các bước giải:

* Khi mắc đèn 1, ta có:

\(\begin{array}{l}
{I_1} = \dfrac{U}{{{R_1} + R}}\\
{P_1} = U{I_1} = \dfrac{{{U^2}}}{{{R_1} + R}}\\
{P_1}' = I_1^2R = \dfrac{{{U^2}R}}{{{{\left( {{R_1} + R} \right)}^2}}}
\end{array}\)

* Khi mắc đèn 2, ta có:

\(\begin{array}{l}
{I_2} = \dfrac{U}{{{R_2} + R}}\\
{P_2} = U{I_2} = \dfrac{{{U^2}}}{{{R_2} + R}}\\
 \Rightarrow \dfrac{{{P_1}}}{{{P_2}}} = \dfrac{{\dfrac{{{U^2}}}{{{R_1} + R}}}}{{\dfrac{{{U^2}}}{{{R_2} + R}}}} = \dfrac{{{R_2} + R}}{{{R_1} + R}} = \dfrac{{12}}{8}\\
{P_2}' = I_2^2R = \dfrac{{{U^2}R}}{{{{\left( {{R_2} + R} \right)}^2}}}\\
 \Rightarrow \dfrac{{{P_1}'}}{{{P_2}'}} = \dfrac{{\dfrac{{{U^2}R}}{{{{\left( {{R_1} + R} \right)}^2}}}}}{{\dfrac{{{U^2}R}}{{{{\left( {{R_2} + R} \right)}^2}}}}} = {\left( {\dfrac{{{R_2} + R}}{{{R_1} + R}}} \right)^2} = {\left( {\dfrac{{12}}{8}} \right)^2} = \dfrac{9}{4}
\end{array}\)