Đáp án:

\(\begin{array}{l}
a.\\
E = 14V\\
r = 3,4\Omega \\
b.\\
m = 0,384g\\
c.\\
{R_1} = 4,461\Omega 
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
E = {E_1} + {E_2} = 6 + 8 = 14V\\
r = {r_1} + {r_2} = 1,7 + 1,7 = 3,4\Omega \\
b.\\
{R_{d1}} = \frac{{U_{dm1}^2}}{{{P_{dm}}}} = \frac{{{6^2}}}{6} = 6\Omega \\
R = \frac{{({R_1} + {R_{d1}}){R_2}}}{{{R_1} + {R_{d1}} + {R_2}}} = \frac{{(3 + 6).6}}{{3 + 6 + 6}} = 3,6\Omega \\
I = \frac{E}{{R + r}} = \frac{{14}}{{3,6 + 3,4}} = 2A = {I_1} + {I_2}(1)\\
{I_1}({R_1} + {R_{d1}}) = {I_2}R\\
{I_1}(6 + 3) = 6{I_2}\\
{I_2} = 1,2A\\
m = \frac{{A{I_2}t}}{{96500n}} = \frac{{64.1,2.965}}{{96500.2}} = 0,384g\\
c.\\
{R_{d2}} = \frac{{U_{dm2}^2}}{{{P_{dm2}}}} = \frac{{{8^2}}}{8} = 8\Omega \\
R = \frac{{({R_1} + {R_{d1}}){R_{d2}}}}{{{R_1} + {R_{d1}} + {R_{d2}}}} = \frac{{({R_1} + 6).8}}{{{R_1} + 6 + 8}} = \frac{{({R_1} + 6).8}}{{{R_1} + 14}}\Omega \\
I = \frac{E}{{R + r}} = \frac{{14}}{{3,4 + \frac{{({R_1} + 6).8}}{{{R_1} + 14}}}}\\
{I_1} = \frac{{{U_{d2}}}}{{{R_1} + {R_{d1}}}} = \frac{8}{{{R_1} + 6}}\\
{I_1} = I - {I_2}\\
\frac{8}{{{R_1} + 6}} = \frac{{14}}{{3,4 + \frac{{({R_1} + 6).8}}{{{R_1} + 14}}}} - 1\\
{R_1} = 4,461\Omega 
\end{array}\)