Đáp án:
$\begin{array}{l}
a.\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{S_2}}}{{{S_1}}}\\
b.\dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{S_1}}}{{{S_2}}}
\end{array}$
Giải thích các bước giải:
a. Khi mắc nt thì:
$\left\{ \begin{array}{l}
{Q_1} = {I^2}.{R_1}.t = {I^2}.\rho .\dfrac{l}{{{S_1}}}.t\\
{Q_2} = {I^2}.{R_2}.t = {I^2}.\rho .\dfrac{l}{{{S_2}}}.t
\end{array} \right. \Leftrightarrow \dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{{I^2}.\rho .\dfrac{l}{{{S_1}}}.t}}{{{I^2}.\rho .\dfrac{l}{{{S_2}}}.t}} = \dfrac{{{S_2}}}{{{S_1}}}$
b. Khi mắc // thì:
$\left\{ \begin{array}{l}
{Q_1} = \dfrac{{{U^2}}}{{{R_1}}}.t = \dfrac{{{U^2}}}{{\rho .\dfrac{l}{{{S_1}}}}}.t\\
{Q_2} = \dfrac{{{U^2}}}{{{R_2}}}.t = \dfrac{{{U^2}}}{{\rho .\dfrac{l}{{{S_2}}}}}.t
\end{array} \right. \Leftrightarrow \dfrac{{{Q_1}}}{{{Q_2}}} = \dfrac{{\dfrac{{{U^2}}}{{\rho .\dfrac{l}{{{S_1}}}}}.t}}{{\dfrac{{{U^2}}}{{\rho .\dfrac{l}{{{S_2}}}}}.t}} = \dfrac{{{S_1}}}{{{S_2}}}$
