Giúp em giải bài này với ạ

Đáp án:

a) \(\dfrac{{12}}{{21}}V\)

b) \(\dfrac{{24}}{{49}}{\rm{W}}\)

c) \(\dfrac{4}{{21}}V\)

d) \(85,7\% \)

Giải thích các bước giải:

a) Ta có: 

\(\begin{array}{l}
{R_{12}} = {R_1} + {R_2} = 2 + 4 = 6\Omega \\
{R_{34}} = {R_3} + {R_4} = 6 + 6 = 12\Omega \\
{R_{1234}} = \dfrac{{{R_{12}}.{R_{34}}}}{{{R_{12}} + {R_{34}}}} = \dfrac{{6.12}}{{6 + 12}} = 4\Omega \\
R = {R_{1234}} + {R_5} = 4 + 2 = 6\Omega \\
{I_5} = I = \dfrac{E}{{R + r}} = \dfrac{2}{{6 + 1}} = \dfrac{2}{7}A\\
{U_5} = {I_5}{R_5} = \dfrac{2}{7}.2 = \dfrac{4}{7}V\\
{I_1} = {I_2} = \dfrac{{{R_{34}}}}{{{R_{12}} + {R_{34}}}}.I = \dfrac{{12}}{{6 + 12}}.\dfrac{2}{7} = \dfrac{4}{{21}}A\\
{U_1} = {I_1}{R_1} = \dfrac{4}{{21}}.2 = \dfrac{8}{{21}}V\\
{U_2} = {I_2}{R_2} = \dfrac{4}{{21}}.4 = \dfrac{{16}}{{21}}V\\
{I_3} = {I_4} = I - {I_1} = \dfrac{2}{7} - \dfrac{4}{{21}} = \dfrac{2}{{21}}A\\
{U_3} = {I_3}{R_3} = \dfrac{2}{{21}}.6 = \dfrac{{12}}{{21}}V\\
{U_4} = {I_4}{R_4} = \dfrac{2}{{21}}.6 = \dfrac{{12}}{{21}}V
\end{array}\)

b) Ta có: 

\({P_N} = {I^2}R = {\left( {\dfrac{2}{7}} \right)^2}.6 = \dfrac{{24}}{{49}}{\rm{W}}\)

c) Ta có: 

\({U_{CD}} =  - {U_1} + {U_3} =  - \dfrac{8}{{21}} + \dfrac{{12}}{{21}} = \dfrac{4}{{21}}V\)

d) Hiệu suất là:

\(H = \dfrac{R}{{R + r}}.100\%  = \dfrac{6}{{6 + 1}}.100\%  = 85,7\% \)