Đáp án:

\(\begin{array}{l}
a.\\
{U_{AB}} = 18V\\
b.\\
{I_{d1}} = 2A\\
{I_{d2}} = 3A\\
{I_R} = 1A\\
c.\\
{P_{d2}} = 36W\\
d.\\
P = 54W
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
{U_{AB}} = {U_{d1}} + {U_v} = 6 + 12 = 18V\\
b.\\
{R_{d1}} = \frac{{U_{dm1}^2}}{{{P_{dm1}}}} = \frac{{{6^2}}}{{12}} = 3\Omega \\
{R_{AC}} = \frac{{R{R_{d1}}}}{{R + {R_{d1}}}} = \frac{{6.3}}{{6 + 3}} = 2\Omega \\
{I_{d1}} = \frac{{{P_{dm1}}}}{{{U_{dm1}}}} = \frac{{12}}{6} = 2A\\
I = {I_{d2}} = \frac{{{U_{AC}}}}{{{R_{AC}}}} = \frac{{{U_{d1}}}}{{{R_{AC}}}} = \frac{6}{2} = 3A\\
{I_R} = I - {I_{d1}} = 3 - 2 = 1A\\
c.\\
{P_{d2}} = {U_V}.{I_{d2}} = 12.3 = 36W\\
d.\\
P = UI = 18.3 = 54W
\end{array}\)