Đáp án:
\(\begin{array}{l}
1.\\
R = 8\Omega \\
{I_1} = 0,5A\\
{U_1} = 1,5V\\
{U_2} = {U_3} = 2,5V\\
{I_2} = \frac{1}{3}A\\
{I_3} = \frac{1}{6}A\\
2.\\
R = 4\Omega \\
{U_3} = 12V\\
{I_3} = 1A\\
{I_1} = {I_2}2A\\
{U_1} = 4V\\
{U_2} = 8V
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1.\\
{R_{23}} = \dfrac{{{R_2}{R_3}}}{{{R_2} + {R_3}}} = \dfrac{{7,5.15}}{{7,5 + 15}} = 5\Omega \\
R = {R_1} + {R_{23}} = 3 + 5 = 8\Omega \\
I = {I_1} = \dfrac{U}{R} = \dfrac{4}{8} = 0,5A\\
{U_1} = {I_1}{R_1} = 0,5.3 = 1,5V\\
{U_2} = {U_3} = U - {U_1} = 4 - 1,5 = 2,5V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{{2,5}}{{7,5}} = \dfrac{1}{3}A\\
{I_3} = I - {I_2} = 0,5 - \dfrac{1}{3} = \dfrac{1}{6}A\\
2.\\
{R_{12}} = {R_1} + {R_2} = 2 + 4 = 6\Omega \\
R = \dfrac{{{R_{12}}{R_3}}}{{{R_{12}}{R_3}}} = \dfrac{{6.12}}{{6 + 12}} = 4\Omega \\
{U_{12}} = {U_3} = U = 12V\\
{I_3} = \dfrac{{{U_3}}}{{{R_3}}} = \dfrac{{12}}{{12}} = 1A\\
{I_1} = {I_2} = \dfrac{{{U_{12}}}}{{{R_{12}}}} = \dfrac{{12}}{6} = 2A\\
{U_1} = {I_1}{R_1} = 2.2 = 4V\\
{U_2} = U - {U_1} = 12 - 4 = 8V
\end{array}\)
