Bài 16:

a) Mạch gồm: R1 nt ((R2 nt R3) // (R4 nt R5)) nt R6

Ta có:

\(\begin{array}{l}
{R_{23}} = {R_2} + {R_3} = 8 + 8 = 16\Omega \\
{R_{45}} = {R_4} + {R_5} = 4 + 12 = 16\Omega \\
{R_{2345}} = \dfrac{{{R_{23}}.{R_{45}}}}{{{R_{23}} + {R_{45}}}} = \dfrac{{16.16}}{{16 + 16}} = 8\Omega \\
{R_{td}} = {R_1} + {R_{2345}} + {R_6} = 4 + 8 + 4 = 16\Omega 
\end{array}\)

b) Ta có:

\(\begin{array}{l}
I = \dfrac{U}{{{R_{td}}}} = \dfrac{{12}}{{16}} = 0,75A\\
 \Rightarrow {I_1} = {I_6} = I = 0,75A\\
{I_2} = {I_3} = {I_4} = {I_5} = \dfrac{I}{2} = \dfrac{{0,75}}{2} = 0,375A
\end{array}\)

c) Ta có:

\(\begin{array}{l}
{U_1} = {I_1}{R_1} = 0,75.4 = 3V\\
{U_2} = {I_2}{R_2} = 0,375.8 = 3V\\
{U_3} = {I_3}{R_3} = 0,375.8 = 3V\\
{U_4} = {I_4}{R_4} = 0,375.4 = 1,5V\\
{U_5} = {I_5}{R_5} = 0,375.12 = 4,5V\\
{U_6} = {I_6}{R_6} = 0,75.4 = 3V
\end{array}\)

d) Ta có:

\(\begin{array}{l}
{P_3} = {I_3}{U_3} = 0,375.3 = 1,125W\\
P = IU = 0,75.12 = 9W
\end{array}\)

e) Ta có:

\({U_{MN}} =  - {U_2} + {U_4} =  - 3 + 1,5 =  - 1,5V\)

f) Ta có:

\({U_{NB}} = {U_5} + {U_6} = 4,5 + 3 = 7,5V\)

Bài 17:

a) Mạch gồm: R8 nt (R7 // (R6 nt ((R1 nt R2) // R3 // (R4 nt R5))))

Ta có:

\(\begin{array}{l}
{R_{12}} = {R_1} + {R_2} = 3 + 5 = 8\Omega \\
{R_{45}} = {R_4} + {R_5} = 1 + 3 = 4\Omega \\
\dfrac{1}{{{R_{BD}}}} = \dfrac{1}{{{R_{12}}}} + \dfrac{1}{{{R_3}}} + \dfrac{1}{{{R_{45}}}} = \dfrac{1}{8} + \dfrac{1}{8} + \dfrac{1}{4} = \dfrac{1}{2}\\
 \Rightarrow {R_{BD}} = 2\Omega \\
{R_{CDB}} = {R_6} + {R_{BD}} = 10 + 2 = 12\Omega \\
{R_{CB}} = \dfrac{{{R_{CDB}}.{R_7}}}{{{R_{CDB}} + {R_7}}} = \dfrac{{12.6}}{{12 + 6}} = 4\Omega \\
{R_{td}} = {R_8} + {R_{CB}} = 6 + 4 = 10\Omega 
\end{array}\)

b) Ta có:

\(\begin{array}{l}
I = \dfrac{U}{{{R_{td}}}} = \dfrac{{24}}{{10}} = 2,4A\\
 \Rightarrow {I_8} = I = 2,4A\\
{I_7} = \dfrac{{{R_{CDB}}}}{{{R_{CDB}} + {R_7}}}.I = \dfrac{{12}}{{12 + 6}}.2,4 = 1,6A\\
{I_{CDB}} = I - {I_7} = 2,4 - 1,6 = 0,8A\\
 \Rightarrow {I_6} = {I_{DB}} = {I_{CDB}} = 0,8A\\
{U_{CDB}} = {I_{CDB}}.{R_{CDB}} = 0,8.12 = 9,6V\\
{I_1} = {I_2} = \dfrac{{{U_{CDB}}}}{{{R_{12}}}} = \dfrac{{9,6}}{8} = 1,2A\\
{I_3} = \dfrac{{{U_{CDB}}}}{{{R_3}}} = \dfrac{{9,6}}{8} = 1,2A\\
{I_4} = {I_5} = \dfrac{{{U_{CDB}}}}{{{R_{45}}}} = \dfrac{{9,6}}{4} = 2,4A
\end{array}\)

c) Ta có:

\(\begin{array}{l}
{U_{EF}} = {U_{ED}} + {U_{DF}} =  - {U_1} + {U_4}\\
 \Rightarrow {U_{EF}} =  - {I_1}{R_1} + {I_4}{R_4} =  - 1,2.3 + 2,4.1 =  - 1,2V
\end{array}\)