Đáp án:
\(\begin{array}{l}
a.\\
{E_b} = 9V\\
{r_b} = 1,5\Omega \\
b.\\
R = 3\Omega \\
I = 2A\\
U = 6V\\
c.\\
{I_1} = 2A\\
{U_1} = 4V\\
{U_2} = {U_3} = 2V\\
{I_2} = 1A\\
{I_3} = 1A\\
d.{Q_1} = 480J
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{E_b} = 3E = 3.3 = 9V\\
r' = 3r = 3.1 = 3\Omega \\
{r_b} = \dfrac{{r'}}{2} = \dfrac{3}{2} = 1,5\Omega \\
b.\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{2.2}}{{2 + 2}} = 1\Omega \\
R = {R_1} + {R_{23}} = 2 + 1 = 3\Omega \\
I = \dfrac{{{E_b}}}{{R + {r_b}}} = \dfrac{9}{{3 + 1,5}} = 2A\\
U = {\rm{IR}} = 2.3 = 6V\\
c.\\
{I_1} = I = 2A\\
{U_1} = {I_1}{R_1} = 2.2 = 4V\\
{U_2} = {U_3} = U - {U_1} = 6 - 4 = 2V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{2}{2} = 1A\\
{I_3} = I - {I_2} = 2 - 1 = 1A\\
d.\\
{Q_1} = {R_1}{I_1}{t^2} = {2.2^2}.60 = 480J
\end{array}\)
