Đáp án:
a.
\(\begin{array}{l}
{R_4} = 6\Omega \\
{R_2} = 12\Omega
\end{array}\)
b.\({I_A} = 0,667A\)
c. \({I_A} = 0A\)
Giải thích các bước giải:
a.
khi \({K_1}\) đóng, \({K_2}\) mở, vôn kế chỉ 0,5A
\(({R_1}//{R_3})nt{R_2}\)
\(\begin{array}{l}
{I_A} = {I_3} = 0,5A\\
{U_{AN}} = {I_3}.{R_3} = 0,5.6 = 3V\\
{I_1} = \frac{{{U_{AN}}}}{{{R_1}}} = \frac{3}{{12}} = 0,25A\\
I = {I_1} + {I_3} = 0,25 + 0,5 = 0,75A\\
{U_{MB}} = U - {U_{AN}} = 12 - 3 = 9V\\
{R_2} = \frac{{{U_{MB}}}}{I} = \frac{9}{{0,75}} = 12\Omega
\end{array}\)
khi \({K_2}\) đóng, \({K_1}\) mở, vôn kế chỉ 0,4A
\({R_3}nt({R_2}//{R_4})\)
\(\begin{array}{l}
{I_2} = {I_A} = 0,4A\\
{U_{MB}} = {I_2}.{R_2} = 0,4.12 = 4,8V\\
{U_{AN}} = U - {U_{MB}} = 12 - 4,8 = 7,2V\\
I = \frac{{{U_{AN}}}}{{{R_3}}} = \frac{{7,2}}{6} = 1,2A\\
{I_4} = I - {I_2} = 1,2 - 0,4 = 0,8A\\
{R_4} = \frac{{{U_{MB}}}}{{{I_4}}} = \frac{{4,8}}{{0,8}} = 6\Omega
\end{array}\)
b.
\({R_3}nt{R_2}\)
\(\begin{array}{l}
{R_{23}} = {R_2} + {R_3} = 12 + 6 = 18\Omega \\
{I_A} = I = \frac{U}{{{R_{23}}}} = \frac{{12}}{{18}} = 0,667A
\end{array}\)
c.
\(\begin{array}{l}
{R_{13}} = \frac{{{R_1}{R_3}}}{{{R_1} + {R_3}}} = \frac{{12.6}}{{12 + 6}} = 4\Omega \\
{R_{24}} = \frac{{{R_2}{R_4}}}{{{R_2} + {R_4}}} = \frac{{12.6}}{{12 + 6}} = 4\Omega \\
R = {R_{13}} + {R_{24}} = 4 + 4 = 8\Omega \\
I = \frac{U}{R} = \frac{{12}}{8} = 1,5A
\end{array}\)
\(\begin{array}{l}
{I_3}{R_3} = {I_1}{R_1}\\
6{I_3} = 12{I_1}\\
I = {I_1} + {I_3} = 1,5A\\
\Rightarrow {I_3} = 1A\\
{I_2}{R_2} = {I_4}{R_4}\\
12{I_2} = 6{I_4}\\
\Rightarrow {I_4} = 1A
\end{array}\)
ta có:
\(\begin{array}{l}
{I_3} = {I_A} + {I_4}\\
1 = {I_A} + 1\\
\Rightarrow {I_A} = 0A
\end{array}\)
