Đáp án:
\(\begin{array}{l}
1.\\
a.\\
{E_b} = 40V\\
{r_b} = 20\Omega \\
b.\\
{E_b} = 2V\\
{r_b} = \dfrac{1}{{20}}\Omega \\
2.\\
a.\\
{E_b} = 6V\\
{r_b} = 1,5\Omega \\
b.\\
{E_b} = 3V\\
{r_b} = \dfrac{1}{3}\Omega
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1.\\
a.\\
{E_b} = 20E = 20.2 = 40V\\
{r_b} = 20r = 20.1 = 20\Omega \\
b.\\
{E_b} = E = 2V\\
{r_b} = \dfrac{r}{{20}} = \dfrac{1}{{20}}\Omega \\
2.\\
a.\\
{E_b} = {E_1} + {E_2} = 3 + 3 = 6V\\
{r_b} = {r_1} + {r_2} = 0,5 + 1 = 1,5\Omega \\
b.\\
{E_b} = {E_1} = {E_2} = 3V\\
{r_b} = \dfrac{{{r_1}{r_2}}}{{{r_1} + {r_2}}} = \dfrac{{1.0,5}}{{1 + 0,5}} = \dfrac{1}{3}\Omega
\end{array}\)
