Đáp án:
\(\begin{array}{l}
a.\\
{R_{D1}} = 6\Omega \\
{R_{D2}} = 12\Omega \\
b.\\
{P_{d1}} = \frac{8}{3}W\\
{P_{d2}} = \frac{4}{3}W\\
c.\\
{A_{d1}} = \frac{4}{9}Wh\\
{A_{d2}} = \frac{2}{9}Wh
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{D1}} = \frac{{{U_{dm1}}^2}}{{{P_{dm1}}}} = \frac{{{6^2}}}{6} = 6\Omega \\
{R_{D2}} = \frac{{{U_{dm2}}^2}}{{{P_{dm2}}}} = \frac{{{6^2}}}{3} = 12\Omega \\
b.\\
R = {R_b} + \frac{{{R_{D2}}{R_{D1}}}}{{{R_{D2}}{R_{D1}}}} = 5 + \frac{{6.12}}{{6 + 12}} = 9\Omega \\
I = \frac{U}{R} = \frac{9}{9} = 1A\\
{I_{d1}}{R_{D1}} = {I_{d2}}{R_{D2}}\\
6{I_{d1}} = 12{I_{d2}}\\
{I_{d1}} + {I_{d2}} = I = 1\\
{I_{d1}} = \frac{2}{3}A\\
{I_{d2}} = \frac{1}{3}A\\
{P_{d1}} = {R_{d1}}.{I_{d1}}^2 = 6.{\frac{2}{3}^2} = \frac{8}{3}W\\
{P_{d2}} = {R_{d1}}.{I_{d1}}^2 = 12.{\frac{1}{3}^2} = \frac{4}{3}W\\
c.\\
{A_{d1}} = {P_{d1}}.t = \frac{8}{3}.\frac{1}{6} = \frac{4}{9}Wh\\
{A_{d2}} = {P_{d2}}.t = \frac{4}{3}.\frac{1}{6} = \frac{2}{9}Wh
\end{array}\)
