Đáp án:
\(\begin{array}{l}
{R_2} = \frac{{35}}{{12}}\Omega \\
{R_1} = \frac{{99}}{{425}}\Omega
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
{U_{d1}} = {U_{d2}} + {U_2}\\
6 = 2,5 + {U_2}\\
{U_2} = 3,5V\\
{I_2} = {I_{d2}} = \frac{{{P_{dm2}}}}{{{U_{dm2}}}} = \frac{3}{{2,5}} = 1,2A\\
{R_2} = \frac{{{U_2}}}{{{I_2}}} = \frac{{3,5}}{{1,2}} = \frac{{35}}{{12}}\Omega \\
I = {I_1} = {I_{d1}} + {I_{d2}} = \frac{{{P_{dm1}}}}{{{U_{dm1}}}} + {I_{d2}} = \frac{3}{6} + 1,2 = 1,7A\\
{R_{d1}} = \frac{{{U_{d1}}}}{{{I_{d1}}}} = \frac{6}{{0,5}} = 12\Omega \\
{R_{d2}} = \frac{{{U_{d2}}}}{{{I_{d2}}}} = \frac{{2,5}}{{1,2}} = \frac{{25}}{{12}}\Omega \\
I = \frac{E}{{R + r}}\\
1,7 = \frac{{6,6}}{{R + 0,12}}\\
R = \frac{{1599}}{{425}}\Omega \\
{R_{2d1d2}} = \frac{{{R_{d1}}({R_{d2}} + {R_2})}}{{{R_{d1}} + {R_{d2}} + {R_2}}} = \frac{{12(\frac{{25}}{{12}} + \frac{{35}}{{12}})}}{{12 + \frac{{25}}{{12}} + \frac{{35}}{{12}}}} = \frac{{60}}{{17}}\Omega \\
R = {R_{2d1d2}} + {R_1}\\
\frac{{1599}}{{425}} = \frac{{60}}{{17}} + {R_1}\\
{R_1} = \frac{{99}}{{425}}\Omega
\end{array}\)
