Câu 4: 

Ta có: 

\(\begin{array}{l}
{U_{BA}} = {E_1}{d_1} = {4.10^4}.0,08 = 3200V\\
 \Rightarrow {V_B} - {V_A} = 3200 \Rightarrow {V_B} = 3200V\\
{U_{BC}} = {E_2}{d_2} = {5.10^4}.0,1 = 5000V\\
 \Rightarrow {V_B} - {V_C} = 5000 \Rightarrow {V_C} =  - 1800V
\end{array}\)

Câu 5: 

Ta có: 

\(\begin{array}{l}
{R_{nt}} = {R_1} + {R_2} = 90\Omega \\
{R_{ss}} = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = 20\\
 \Rightarrow \dfrac{{{R_1}{R_2}}}{{90}} = 20 \Rightarrow {R_1}{R_2} = 1800\\
 \Rightarrow {R_1}\left( {90 - {R_1}} \right) = 1800\\
 \Rightarrow \left[ \begin{array}{l}
{R_1} = 60\Omega \\
{R_1} = 30\Omega 
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
{R_2} = 30\Omega \\
{R_2} = 60\Omega 
\end{array} \right.
\end{array}\)

Bài 6: 

a) Ta có: 

\(\begin{array}{l}
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{6.3}}{{6 + 3}} = 2\Omega \\
R = {R_1} + {R_{23}} = 3 + 2 = 5\Omega 
\end{array}\)

b) Ta có: 

\({I_A} = \dfrac{E}{{R + r}} = \dfrac{6}{{5 + 1}} = 1A\)