Đáp án:

 a,$8\left( \Omega  \right)$

b.$\begin{array}{l}
1,8\left( A \right) = {I_4}\\
{U_4} = 10,8\left( V \right)\\
{U_1} = 3,6\left( V \right)\\
{I_1} = 1,2\left( A \right)\\
{I_2} = {I_3} = 0,6\left( A \right)\\
{U_2} = 1,8\left( V \right)\\
{U_3} = 1,8\left( V \right)
\end{array}$

c,${U_{CD}} = 12,6\left( V \right)$

d.$\begin{array}{l}
{U_{NG}} = 14,4\left( V \right)\\
H = 80\% 
\end{array}$

Giải thích các bước giải:

$\begin{array}{l}
a.\left( {\left( {{R_2}nt{R_3}} \right)//{R_1}} \right)nt{R_4}\\
{R_{23}} = {R_2} + {R_3} = 3 + 3 = 6\\
{R_{123}} = \frac{{{R_{23}}.{R_1}}}{{{R_{23}} + {R_1}}} = \frac{{6.3}}{{6 + 3}} = 2\\
{R_{1234}} = {R_{123}} + {R_4} = 2 + 6 = 8\left( \Omega  \right)
\end{array}$

b.$\begin{array}{l}
I = \frac{{{E_b}}}{{{r_b} + {R_{1234}}}} = \frac{{9.2}}{{1.2 + 8}} = 1,8\left( A \right) = {I_4}\\
{U_4} = {I_4}.{R_4} = 1,8.6 = 10,8\left( V \right)\\
{U_{123}} = {U_1} = {U_{23}} = I.{R_{123}} = 1,8.2 = 3,6\left( V \right)\\
{I_1} = \frac{{{U_1}}}{{{R_1}}} = \frac{{3,6}}{3} = 1,2\left( A \right)\\
{I_2} = {I_3} = \frac{{{U_{23}}}}{{{R_{23}}}} = \frac{{3,6}}{6} = 0,6\left( A \right)\\
{U_2} = {I_2}.{R_2} = 0,6.3 = 1,8\left( V \right)\\
{U_3} = {I_3}.{R_3} = 0,6.3 = 1,8\left( V \right)
\end{array}$

c,${U_{CD}} = {U_3} + {U_4} = 12,6\left( V \right)$

d.$\begin{array}{l}
{U_{NG}} = {U_{1234}} = I.{R_{1234}} = 1,8.8 = 14,4\left( V \right)\\
H = \frac{{{R_{1234}}}}{{{R_{1234}} + {r_b}}} = \frac{8}{{8 + 2.1}} = 80\% 
\end{array}$