Đáp án:
\(\begin{array}{l}
a.\\
R = 8\Omega \\
b.\\
{I_2} = 0,75A\\
{U_2} = 3V\\
{U_1} = {U_3} = 3V\\
{I_3} = 0,5A\\
{I_1} = 0,25A\\
c.\\
{R_3} = 2,4\Omega
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
{R_{13}} = \frac{{{R_1}{R_3}}}{{{R_1} + {R_3}}} = \frac{{6.12}}{{6 + 12}} = 4\Omega \\
R = {R_2} + {R_{13}} = 4 + 4 = 8\Omega \\
b.\\
I = {I_2} = \frac{U}{R} = \frac{6}{8} = 0,75A\\
{U_2} = {I_2}{R_2} = 0,75.4 = 3V\\
{U_1} = {U_3} = U - {U_2} = 6 - 3 = 3V\\
{I_3} = \frac{{{U_3}}}{{{R_3}}} = \frac{3}{6} = 0,5A\\
{I_1} = I - {I_3} = 0,75 - 0,5 = 0,25A\\
c.\\
I = 0,75 + 0,25 = 1A\\
R = \frac{U}{I} = \frac{6}{1} = 6\Omega \\
{R_{13}} = R - {R_2} = 6 - 4 = 2\Omega \\
\frac{1}{{{R_{13}}}} = \frac{1}{{{R_1}}} + \frac{1}{{{R_3}}}\\
\frac{1}{2} = \frac{1}{{12}} + \frac{1}{{{R_3}}}\\
{R_3} = 2,4\Omega
\end{array}\)