Đáp án:

1. \(\dfrac{{2995}}{{751}}V\)

2. \(\dfrac{{913}}{{2260}}A\)

Giải thích các bước giải:

1. Giả sử I5 có chiều từ trên xuống dưới.

Ta có:

\(\begin{array}{l}
{U_{23}} = {U_{21}} + {U_{13}}\\
 \Rightarrow {U_5} =  - {U_1} + {U_2}\\
 \Rightarrow {I_5}{R_5} =  - {I_1}{R_1} + {I_2}{R_2}\\
 \Rightarrow 30{I_5} =  - 21{I_1} + 42{I_2}\\
 \Rightarrow  - 21{I_1} + 42{I_2} - 30{I_5} = 0\\
 \Rightarrow 7{I_1} - 14{I_2} + 10{I_5} = 0\\
{U_{23}} = {U_{24}} + {U_{43}}\\
 \Rightarrow {U_5} = {U_3} - {U_7} - {U_4}\\
 \Rightarrow {I_5}{R_5} = {I_3}{R_3} - {I_7}{R_7} - {I_4}{R_4}\\
 \Rightarrow 30{I_5} = 20\left( {{I_1} - {I_5}} \right) - \left( {42 + 20} \right)\left( {{I_2} + {I_5}} \right)\\
 \Rightarrow 30{I_5} = 20\left( {{I_1} - {I_5}} \right) - 62\left( {{I_2} + {I_5}} \right)\\
 \Rightarrow 20{I_1} - 62{I_2} - 112{I_5} = 0\\
{U_1} + {U_3} + {U_6} = U\\
 \Rightarrow {I_1}{R_1} + {I_3}{R_3} + {I_6}{R_6} = 33\\
 \Rightarrow 21{I_1} + 20\left( {{I_1} - {I_5}} \right) + 2\left( {{I_3} + {I_7}} \right) = 33\\
 \Rightarrow 41{I_1} - 20{I_5} + 2\left( {{I_1} - {I_5} + {I_2} + {I_5}} \right) = 33\\
 \Rightarrow 43{I_1} + 2{I_2} - 20{I_5} = 33\\
 \Rightarrow \left\{ \begin{array}{l}
{I_1} = \dfrac{{547}}{{751}}A\\
{I_2} = \dfrac{{246}}{{751}}A\\
{I_5} =  - \dfrac{{77}}{{1502}}A
\end{array} \right.
\end{array}\)

Vậy I5 có chiều từ dưới lên trên.

Ta có:

\(\begin{array}{l}
{U_5} = {I_5}{R_5} = \dfrac{{77}}{{1502}}.30 = \dfrac{{1155}}{{751}}V\\
{U_4} = {I_4}{R_4} = \left( {{I_2} - {I_5}} \right){R_4} = \dfrac{{4150}}{{751}}V\\
{U_V} =  - {U_5} + {U_4} = \dfrac{{2995}}{{751}}V
\end{array}\)

2. Mạch gồm: (R1 // (R2 nt (R5 // R4))) nt (R3 // R7) nt R6

Ta có:

\(\begin{array}{l}
{R_{45}} = \dfrac{{{R_4}.{R_5}}}{{{R_4} + {R_5}}} = \dfrac{{20.30}}{{20 + 30}} = 12\Omega \\
{R_{245}} = {R_2} + {R_{45}} = 42 + 12 = 54\Omega \\
{R_{1245}} = \dfrac{{{R_1}.{R_{245}}}}{{{R_1} + {R_{245}}}} = \dfrac{{21.54}}{{21 + 54}} = 15,12\Omega \\
{R_{37}} = \dfrac{{{R_3}{R_7}}}{{{R_3} + {R_7}}} = \dfrac{{20.20}}{{20 + 20}} = 10\Omega \\
{R_{td}} = {R_{1245}} + {R_{37}} + {R_6} = 15,12 + 10 + 2 = 27,12\Omega \\
I = \dfrac{U}{{{R_{td}}}} = \dfrac{{33}}{{27,12}} = \dfrac{{275}}{{226}}A\\
{I_{245}} = \dfrac{{{R_1}}}{{{R_1} + {R_{245}}}}.I = \dfrac{{21}}{{21 + 54}}.\dfrac{{275}}{{226}} = \dfrac{{77}}{{226}}A\\
{I_4} = \dfrac{{{R_5}}}{{{R_4} + {R_5}}}.{I_{245}} = \dfrac{{30}}{{20 + 30}}.\dfrac{{77}}{{226}} = \dfrac{{231}}{{1130}}A\\
{I_7} = \dfrac{{{R_3}}}{{{R_3} + {R_7}}}.I = \dfrac{1}{2}.\dfrac{{275}}{{226}} = \dfrac{{275}}{{452}}A\\
{I_A} = {I_7} - {I_4} = \dfrac{{275}}{{452}} - \dfrac{{231}}{{1130}} = \dfrac{{913}}{{2260}}A
\end{array}\)