Giải giúp mình bài này với

Đáp án:

a) \(10\Omega \)

b) \(\dfrac{9}{{35}}A\)

c) \(\dfrac{3}{7}A\)

d) \(0,3A\)

Giải thích các bước giải:

a) Khi K1, K2 đều mở, mạch gồm: R4 nt R5 nt R6

Ta có:

\(\begin{array}{l}
{R_4} + {R_5} + {R_6} = \dfrac{U}{{{I_{{A_1}}}}}\\
 \Rightarrow {R_4} + 10 + 12 = \dfrac{{12}}{{\dfrac{3}{8}}}\\
 \Rightarrow {R_4} = 10\Omega 
\end{array}\)

b) Khi K1 đóng, K2 mở, mạch gồm: ((R3 nt R2 nt R1) //R4) nt R5 nt R6

Ta có:

\(\begin{array}{l}
{R_{123}} = {R_1} + {R_2} + {R_3} = 5 + 5 + 5 = 15\Omega \\
{R_{1234}} = \dfrac{{{R_{123}}.{R_4}}}{{{R_{123}} + {R_4}}} = \dfrac{{15.10}}{{15 + 10}} = 6\Omega \\
I = \dfrac{U}{{{R_{1234}} + {R_5} + {R_6}}} = \dfrac{{12}}{{6 + 10 + 12}} = \dfrac{3}{7}A\\
{I_{{A_2}}} = \dfrac{{{R_{123}}}}{{{R_{123}} + {R_4}}}.I = \dfrac{{15}}{{15 + 10}}.\dfrac{3}{7} = \dfrac{9}{{35}}A
\end{array}\)

c) Khi K1 mở, K2 đóng, mạch gồm: R4 nt R5 nt (R6 // (R9 nt R8 nt R7))

Ta có:

\(\begin{array}{l}
{R_{789}} = {R_7} + {R_8} + {R_9} = 8 + 8 + 8 = 24\Omega \\
{R_{6789}} = \dfrac{{{R_6}.{R_{789}}}}{{{R_6} + {R_{789}}}} = \dfrac{{12.24}}{{12 + 24}} = 8\Omega \\
I = \dfrac{U}{{{R_{6789}} + {R_4} + {R_5}}} = \dfrac{{12}}{{8 + 10 + 10}} = \dfrac{3}{7}A\\
{I_{{A_3}}} = I = \dfrac{3}{7}A
\end{array}\)

d) Khi K1, K2 cùng đóng, mạch gồm: (R4 // (R3 nt R2 nt R1)) nt R5 nt (R6 // (R9 nt R8 nt R7))

Ta có:

\(\begin{array}{l}
{R_{123}} = {R_1} + {R_2} + {R_3} = 5 + 5 + 5 = 15\Omega \\
{R_{1234}} = \dfrac{{{R_{123}}.{R_4}}}{{{R_{123}} + {R_4}}} = \dfrac{{15.10}}{{15 + 10}} = 6\Omega \\
{R_{789}} = {R_7} + {R_8} + {R_9} = 8 + 8 + 8 = 24\Omega \\
{R_{6789}} = \dfrac{{{R_6}.{R_{789}}}}{{{R_6} + {R_{789}}}} = \dfrac{{12.24}}{{12 + 24}} = 8\Omega \\
I = \dfrac{U}{{{R_{1234}} + {R_5} + {R_{6789}}}} = \dfrac{{12}}{{6 + 10 + 8}} = 0,5A\\
{I_{{A_4}}} = \dfrac{{{R_{123}}}}{{{R_{123}} + {R_4}}}.I = \dfrac{{15}}{{15 + 10}}.0,5 = 0,3A
\end{array}\)