Đáp án:
\(\begin{array}{l}
a.m = 0,64g\\
b.{I_A} = 5A\\
c.E = 10V\\
d.H = 80\%
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{I_d} = {I_2} = {I_{dm}} = \dfrac{{{P_{dm}}}}{{{U_{dm}}}} = \dfrac{3}{3} = 1A\\
m = \dfrac{{A{I_2}t}}{{Fn}} = \dfrac{{64.1.1930}}{{96500.2}} = 0,64g\\
b.\\
{R_d} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{3^2}}}{3} = 3\Omega \\
{R_{2d}} = {R_2} + {R_d} = 3 + 5 = 8\Omega \\
{U_1} = U = {U_{2d}} = {I_2}.{R_{2d}} = 1.8 = 8V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{8}{2} = 4A\\
{I_A} = I = {I_1} + {I_2} = 1 + 4 = 5A\\
c.\\
R = \dfrac{{{R_1}{R_{2d}}}}{{{R_1} + {R_{2d}}}} = \dfrac{{2.8}}{{2 + 8}} = 1,6\Omega \\
E = I(R + r) = 5(1,6 + 0,4) = 10V\\
d.\\
H = \dfrac{U}{E} = \dfrac{8}{{10}} = 80\%
\end{array}\)