Đáp án:
\(\begin{array}{l}
a.\\
U = 72V\\
P = 129,6W\\
b.\\
{I_1} = 1,2A\\
{I_2} = 0,6A\\
c.P' = 28,8W
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
R = \dfrac{{{R_1}{R_2}}}{{{R_1} + {R_2}}} = \dfrac{{60.120}}{{60 + 120}} = 40\Omega \\
U = {\rm{IR}} = 1,8.40 = 72V\\
P = UI = 72.1,8 = 129,6W\\
b.\\
{U_1} = {U_2} = U = 72V\\
{I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{72}}{{60}} = 1,2A\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{72}}{{120}} = 0,6A\\
c.\\
R' = {R_1} + {R_2} = 60 + 120 = 180\Omega \\
P' = \dfrac{{{U^2}}}{{R'}} = \dfrac{{{{72}^2}}}{{180}} = 28,8W
\end{array}\)