Đáp án:

\(\begin{array}{l}
a.\\
{U_V} = 6V\\
{I_A} = 1A\\
b.\\
R = 3\Omega \\
{P_R} = 3W\\
c.A = 16200J = 4,{5.10^{ - 3}}kWh\\
d.{U_V} = 4V
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\\
{U_V} = {U_{dm}} = 6V\\
{I_A} = I = {I_{dm}} = \dfrac{{{P_{dm}}}}{{{U_{dm}}}} = \dfrac{6}{6} = 1A\\
b.\\
{U_R} = U - {U_V} = 9 - 6 = 3V\\
R = \dfrac{{{U_R}}}{I} = \dfrac{3}{1} = 3\Omega \\
{P_R} = R{I^2} = {3.1^2} = 3W\\
c.\\
A = UIt = 9.1.30.60 = 16200J = 4,{5.10^{ - 3}}kWh\\
d.\\
R' = \dfrac{{{R_{\max }}}}{2} = \dfrac{{15}}{2} = 7,5\Omega \\
{R_d} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{6^2}}}{6} = 6\Omega \\
{R_{td}} = R' + {R_d} = 7,5 + 6 = 13,5\Omega \\
I' = \dfrac{U}{{{R_{td}}}} = \dfrac{9}{{13,5}} = \dfrac{2}{3}A\\
{U_V} = I'.{R_d} = \dfrac{2}{3}.6 = 4V
\end{array}\)