Bài 1: 

Ta có: 

\(\begin{array}{l}
{I_{dm}} = \dfrac{{{P_{dm}}}}{{{U_{dm}}}} = \dfrac{5}{{14}}A\\
U = {U_{dm}} + {I_{dm}}.R = 14 + \dfrac{5}{{14}}.10 = 17,57V
\end{array}\)

Bài 2: B

* Khi chỉ dùng R1 thì: \(Q = \dfrac{{{U^2}}}{{{R_1}}}{t_1}\)

* Khi dùng R1 // R2 thì: 

\(\begin{array}{l}
Q = \dfrac{{{U^2}}}{{{R_{12}}}}.{t_2}\\
 \Rightarrow \dfrac{{{t_1}}}{{{R_1}}} = \dfrac{{{t_2}}}{{{R_{12}}}} \Rightarrow \dfrac{{{R_1}}}{{\dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}}}} = \dfrac{{{t_1}}}{{{t_2}}}\\
 \Rightarrow \dfrac{{{R_1} + {R_2}}}{{{R_2}}} = \dfrac{{{t_1}}}{{{t_2}}}\\
 \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} + 1 = \dfrac{6}{4} \Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = 0,5
\end{array}\)

* Khi chỉ dùng R2 thì:

\(\begin{array}{l}
Q = \dfrac{{{U^2}}}{{{R_2}}}{t_3} \Rightarrow \dfrac{{{t_1}}}{{{R_1}}} = \dfrac{{{t_3}}}{{{R_2}}}\\
 \Rightarrow {t_3} = \dfrac{{{R_2}}}{{{R_1}}}.{t_1} = 2{t_1} = 12ph
\end{array}\)