Đáp án:
\(\begin{array}{l}
a.\\
{U_1} = 10V\\
{I_1} = 1,25A\\
{I_2} = {I_3} = 1,25A\\
{U_2} = 3,75V\\
{U_3} = 6,25V\\
b.\\
P = 25W\\
H = 50\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{23}} = {R_2} + {R_3} = 3 + 8 = 8\Omega \\
R = \dfrac{{{R_1}{R_{23}}}}{{{R_1} + {R_{23}}}} = \dfrac{{8.8}}{{8 + 8}} = 4\Omega \\
E = {E_1} + {E_2} = 10 + 10 = 20V\\
r = {r_1} + {r_2} = 2 + 2 = 4\Omega \\
I = \frac{E}{{R + r}} = \dfrac{{20}}{{4 + 4}} = 2,5A\\
{U_{23}} = {U_1} = U = {\rm{IR}} = 2,5.4 = 10V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{10}}{8} = 1,25A\\
{I_2} = {I_3} = {I_{23}} = I - {I_1} = 2,5 - 1,25 = 1,25A\\
{U_2} = {I_2}{R_2} = 1,25.3 = 3,75V\\
{U_3} = {I_3}{R_3} = 1,25.5 = 6,25V\\
b.\\
P = R{I^2} = 4.2,{5^2} = 25W\\
H = \dfrac{U}{E} = \dfrac{{10}}{{20}} = 50\%
\end{array}\)
