Câu 39: D

Ta có: 

\(\begin{array}{l}
{R_{13}} = {R_1} + {R_3} = 3 + 3 = 6\Omega \\
{R_{24}} = {R_2} + {R_4} = 3 + 6 = 9\Omega \\
{R_{AB}} = \dfrac{{{R_{13}}.{R_{24}}}}{{{R_{13}} + {R_{24}}}} = \dfrac{{6.9}}{{6 + 9}} = 3,6\Omega \\
I = \dfrac{E}{{{R_{AB}} + r}} = \dfrac{{7,8}}{{3,6 + 0,4}} = 1,95A\\
{I_1} = \dfrac{{{R_{24}}}}{{{R_{13}} + {R_{24}}}}.I = \dfrac{9}{{6 + 9}}.1,95 = 1,17A\\
{I_2} = I - {I_1} = 1,95 - 1,17 = 0,78A\\
{U_1} = {I_1}{R_1} = 1,95.3 = 5,85V\\
{U_2} = {I_2}{R_2} = 0,78.3 = 2,34V\\
{U_{MN}} =  - {U_1} + {U_2} =  - 5,85 + 2,34 =  - 3,51V
\end{array}\)

Câu 40: A

Ta có: 

\(\begin{array}{l}
P = {I^2}R = {\left( {\dfrac{E}{{R + r}}} \right)^2}R\\
 \Rightarrow 16 = {\left( {\dfrac{{12}}{{R + 2}}} \right)^2}.R\\
 \Rightarrow \left( \begin{array}{l}
{R_1} = 4\Omega \\
{R_2} = 1\Omega 
\end{array} \right.\\
{H_1} = \dfrac{{{R_1}}}{{{R_1} + r}}.100\%  = \dfrac{4}{{4 + 2}}.100\%  = 66,67\% \\
{H_2} = \dfrac{{{R_2}}}{{{R_2} + r}}.100\%  = \dfrac{1}{{1 + 2}}.100\%  = 33,33\% 
\end{array}\)