Đáp án:
\(\begin{array}{l}
a.\\
{E_b} = 18V\\
{r_b} = 2\Omega \\
b.\\
R = 10\Omega \\
I = 1,5A\\
{P_{E1}} = 18W\\
{P_{E2}} = 9W\\
{P_1} = 9{\rm{W}}\\
{P_2} = 13,5W\\
c.{R_x} = 3\Omega
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{E_b} = {E_1} + {E_2} = 12 + 6 = 18V\\
{r_b} = {r_1} + {r_2} = 1 + 1 = 2\Omega \\
b.\\
R = {R_1} + {R_2} = 4 + 6 = 10\Omega \\
I = \dfrac{{{E_b}}}{{R + {r_b}}} = \dfrac{{18}}{{10 + 2}} = 1,5A\\
{P_{E1}} = {E_1}I = 12.1,5 = 18W\\
{P_{E2}} = {E_2}I = 6.1,5 = 9W\\
{P_1} = {R_1}{I^2} = 4.1,{5^2} = 9{\rm{W}}\\
{P_2} = {R_2}{I^2} = 6.1,{5^2} = 13,5W\\
c.\\
I' = \dfrac{{{E_b}}}{{R' + {r_b}}}\\
\Rightarrow 2 = \dfrac{{18}}{{R' + 2}}\\
\Rightarrow R' = 7\Omega \\
R' = {R_1} + {R_x}\\
\Rightarrow {R_x} = R' - {R_1} = 7 - 4 = 3\Omega
\end{array}\)
