Đáp án:
\(\begin{array}{l}
6.D\\
7.D\\
{R_1} = \dfrac{{U_1^2}}{{{P_1}}} = \dfrac{{{{120}^2}}}{{75}} = 192\Omega \\
{R_2} = \dfrac{{U_2^2}}{{{P_2}}} = \dfrac{{{{120}^2}}}{{45}} = 320\Omega \\
\Rightarrow {R_2} > {R_1}\\
8.C\\
\dfrac{{R'}}{R} = \dfrac{{p\dfrac{{l'}}{{S'}}}}{{p\dfrac{l}{S}}} = \dfrac{{l'S}}{{lS'}} = \dfrac{{\frac{l}{4}.S}}{{l.4S}} = \dfrac{1}{{16}}\\
9.A.0,6A\\
R = {R_1} + {R_2} = 100 + 150 = 250\Omega \\
I = \dfrac{U}{R} = \dfrac{{150}}{{250}} = 0,6A\\
10.B.0,5A\\
U = {U_1} = {U_2} = {U_3} = {U_4} = {U_5} = 10V\\
{I_1} = {I_2} = {I_3} = {I_4} = {I_5} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{{10}}{{20}} = 0,5A
\end{array}\)
