Đáp án:

\(\begin{array}{l}
a.\\
{E_b} = 15V\\
{r_b} = 5,1\Omega \\
b.\\
I = 2A\\
{I_1} = 1,2A\\
{I_2} = 0,8A\\
c.\\
{P_1} = 10W\\
{P_2} = 12W\\
{P_3} = 8W
\end{array}\) 

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
{E_b} = {E_1} + {E_2} + {E_3} = 5 + 6 + 4 = 15V\\
r = {r_1} + {r_2} + {r_3} = 1,5 + 1,6 + 2 = 5,1\Omega \\
b.\\
R = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{4.6}}{{4 + 6}} = 2,4\Omega \\
I = \dfrac{{{E_b}}}{{R + {r_b}}} = \dfrac{{15}}{{2,4 + 5,1}} = 2A\\
{U_1} = {U_2} = U = RI = 2,4.2 = 4,8V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{4,8}}{4} = 1,2A\\
{I_2} = I - {I_1} = 2 - 1,2 = 0,8A\\
c.\\
{P_1} = {E_1}I = 5.2 = 10W\\
{P_2} = {E_2}I = 6.2 = 12W\\
{P_3} = {E_3}I = 4.2 = 8W
\end{array}\)