Đáp án:

\(\begin{array}{l}
a.R = 3,2\Omega \\
b.\\
I = 7,5A\\
{I_1} = 4A\\
{I_2} = 2A\\
{I_3} = 1,5A\\
c.\\
P = 180W\\
{P_1} = 96W\\
{P_2} = 48W\\
{P_3} = 36W
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} = \dfrac{1}{6} + \dfrac{1}{{16}} + \dfrac{1}{{12}} = \dfrac{1}{{3,2}}\\
 \Rightarrow R = 3,2\Omega \\
b.\\
I = \dfrac{U}{R} = \dfrac{{24}}{{3,2}} = 7,5A\\
{U_1} = {U_2} = {U_3} = U = 24V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{24}}{6} = 4A\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{24}}{{12}} = 2A\\
{I_3} = \dfrac{{{U_3}}}{{{R_3}}} = \dfrac{{24}}{{16}} = 1,5A\\
c.\\
P = \dfrac{{{U^2}}}{R} = \dfrac{{{{24}^2}}}{{3,2}} = 180W\\
{P_1} = \dfrac{{U_1^2}}{{{R_1}}} = \dfrac{{{{24}^2}}}{6} = 96W\\
{P_2} = \dfrac{{U_2^2}}{{{R_2}}} = \dfrac{{{{24}^2}}}{{12}} = 48W\\
{P_3} = \dfrac{{U_3^2}}{{{R_3}}} = \dfrac{{{{24}^2}}}{{16}} = 36W
\end{array}\)