Đáp án:
\(\begin{array}{l}
a.\\
R = \frac{{24}}{{19}}\Omega \\
I = \frac{{456}}{{67}}A\\
U = \frac{{576}}{{67}}V\\
b.\\
{I_1} = \frac{{384}}{{67}}A\\
{I_2} = \frac{{48}}{{67}}A\\
{I_3} = \frac{{24}}{{67}}A\\
c.{m_{Ag}} = 1,0824g
\end{array}\)
\(\begin{array}{l}
d.\\
{P_{ng}} = \dfrac{{5472}}{{67}}{\rm{W}}\\
{H_{ng}} = 71,64\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{6.12}}{{6 + 12}} = 4\Omega \\
{R_{23b}} = {R_{23}} + {R_b} = 4 + 4 = 8\Omega \\
R = \dfrac{{{R_1}{R_{23b}}}}{{{R_1} + {R_{23b}}}} = \dfrac{{1,5.8}}{{1,5 + 8}} = \dfrac{{24}}{{19}}\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{{12}}{{\dfrac{{24}}{{19}} + 0,5}} = \dfrac{{456}}{{67}}A\\
U = {\rm{IR}} - \dfrac{{456}}{{67}}.\dfrac{{24}}{{19}} = \dfrac{{576}}{{67}}V\\
b.\\
{U_1} = {U_{23b}} = U = \dfrac{{576}}{{67}}V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{\dfrac{{576}}{{67}}}}{{1,5}} = \dfrac{{384}}{{67}}A\\
{I_b} = {I_{23}} = I - {I_1} = \dfrac{{456}}{{67}} - \dfrac{{384}}{{67}} = \dfrac{{72}}{{67}}A\\
{U_2} = {U_3} = {U_{23}} = {I_{23}}{R_{23}} = \dfrac{{72}}{{67}}.4 = \dfrac{{288}}{{67}}V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{\dfrac{{288}}{{67}}}}{6} = \dfrac{{48}}{{67}}A\\
{I_3} = {I_{23}} - {I_3} = \dfrac{{72}}{{67}} - \dfrac{{48}}{{67}} = \dfrac{{24}}{{67}}A\\
c.\\
{m_{Ag}} = \dfrac{{A{I_b}t}}{{Fn}} = \dfrac{{108.\frac{{72}}{{67}}.900}}{{96500.1}} = 1,0824g
\end{array}\)
\(\begin{array}{l}
d.\\
{P_{ng}} = EI = 12.\dfrac{{456}}{{67}} = \dfrac{{5472}}{{67}}{\rm{W}}\\
{H_{ng}} = \dfrac{U}{E} = \dfrac{{\dfrac{{576}}{{67}}}}{{12}} = 71,64\%
\end{array}\)
