Đáp án:
\(\begin{array}{l}
a.\\
R = 6\Omega \\
I = 0,5A\\
b.{R_3} = 2\Omega
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a.\\
R = {R_1} + {R_2} = 2 + 4 = 6\Omega \\
I = \dfrac{U}{R} = \dfrac{3}{6} = 0,5A\\
b.\\
I' = {I_1} = 0,75A\\
R' = \dfrac{U}{{I'}} = \dfrac{3}{{0,75}} = 4\Omega \\
R' = {R_1} + {R_{23}}\\
\Rightarrow {R_{23}} = R' - {R_1} = 4 - 2 = 2\Omega \\
\dfrac{1}{{{R_{23}}}} = \dfrac{1}{{{R_2}}} + \dfrac{1}{{{R_3}}}\\
\Rightarrow \frac{1}{{{R_3}}} = \dfrac{1}{{{R_{23}}}} - \dfrac{1}{{{R_2}}} = \dfrac{1}{2} - \dfrac{1}{4} = \dfrac{1}{2}\\
\Rightarrow {R_3} = 2\Omega
\end{array}\)