Đáp án:
\(\begin{array}{l}
a.\\
R = 12\Omega \\
b.\\
U = 72V
\end{array}\)
Giải thích các bước giải:
\((({R_4}nt{R_5}nt{R_6})//{R_2}//{R_3})nt{R_1}\)
\(\begin{array}{l}
a.\\
{R_{456}} = {R_4} + {R_5} + {R_6} = 2 + 2 + 2 = 6\Omega \\
\dfrac{1}{{{R_{23456}}}} = \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}} + \dfrac{1}{{{R_{456}}}} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{1}{2}\\
\Rightarrow {R_{23456}} = 2\Omega \\
R = {R_1} + {R_{23456}} = 10 + 2 = 12\Omega \\
b.\\
{I_4} = {I_{456}} = 2A\\
{U_{456}} = {U_{23456}} = {I_{456}}{R_{456}} = 2.6 = 12V\\
I = {I_{23456}} = \dfrac{{{U_{23456}}}}{{{R_{23456}}}} = \dfrac{{12}}{2} = 6A\\
U = {\rm{IR}} = 6.12 = 72V
\end{array}\)
