Đáp án:
\(\begin{array}{l}
4.\\
a.\\
R = 23\Omega \\
b.\\
I = 0,5A\\
U = 11,5V\\
c.\\
{I_1} = {I_2} = {I_3} = 0,5A\\
{U_1} = 5V\\
{U_2} = I2,5V\\
{U_3} = 4V\\
d.\\
H = 95,83\% \\
5.\\
a.\\
R = 4\Omega \\
b.\\
I = 0,6A\\
c.\\
{Q_1} = 48J\\
d.\\
P = 1,8W\\
H = 80\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
4.\\
a.\\
R = {R_1} + {R_2} + {R_3} = 10 + 5 + 8 = 23\Omega \\
b.\\
I = \dfrac{E}{{R + r}} = \dfrac{{12}}{{23 + 1}} = 0,5A\\
U = RI = 23.0,5 = 11,5V\\
c.\\
{I_1} = {I_2} = {I_3} = I = 0,5A\\
{U_1} = {I_1}{R_1} = 0,5.10 = 5V\\
{U_2} = {I_2}{R_2} = 0,5.5 = 2,5V\\
{U_3} = {I_3}{R_3} = 0,5.8 = 4V\\
d.\\
H = \dfrac{R}{{R + r}} = \dfrac{{23}}{{23 + 1}} = 95,83\% \\
5.\\
a.\\
{R_{12}} = {R_1} + {R_2} = 1 + 5 = 6\Omega \\
R = \dfrac{{{R_{12}}{R_3}}}{{{R_{12}} + {R_3}}} = \dfrac{{6.12}}{{6 + 12}} = 4\Omega \\
b.\\
I = \dfrac{E}{{R + r}} = \dfrac{3}{{4 + 1}} = 0,6A\\
c.\\
{U_{12}} = U = {\rm{IR}} = 0,6.4 = 2,4V\\
{I_1} = \dfrac{{{U_{12}}}}{{{R_{12}}}} = \dfrac{{2,4}}{6} = 0,4A\\
{Q_1} = {R_1}I_1^2t = 1.0,{4^2}.5.60 = 48J\\
d.\\
P = EI = 3.0,6 = 1,8W\\
H = \dfrac{U}{E} = \dfrac{{2,4}}{3} = 80\%
\end{array}\)
