Đáp án:
\(\begin{array}{l}
13.\\
a.R = 3\Omega \\
b.\\
{I_1} = 0,5A\\
{I_2} = 0,25A\\
{I_3} = 0,25A\\
14.\\
a.R = 30\Omega \\
b.\\
I = {I_2} = {I_3} = {I_1} = \dfrac{5}{6}A
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
13.\\
a.\\
\dfrac{1}{R} = \dfrac{1}{{{R_1}}} + \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} = \dfrac{1}{6} + \dfrac{1}{{12}} + \dfrac{1}{{12}} = \dfrac{1}{2}\\
\Rightarrow R = 3\Omega \\
b.\\
{U_2} = {U_3} = {U_1} = {R_1}{I_1} = 6.0,5 = 3V\\
{I_2} = \dfrac{{{U_2}}}{{{R_2}}} = \dfrac{3}{{12}} = 0,25A\\
{I_3} = \dfrac{{{U_3}}}{{{R_3}}} = \dfrac{3}{{12}} = 0,25A\\
14.\\
a.\\
R = {R_1} + {R_2} + {R_3} = 6 + 12 + 12 = 30\Omega \\
b.\\
I = {I_2} = {I_3} = {I_1} = \dfrac{{{U_1}}}{{{R_1}}} = \dfrac{5}{6}A
\end{array}\)
