Đáp án:

 \(R = 1\Omega \)

Giải thích các bước giải:

a) K đóng, mạch gồm: R1 // ((Rac // Rbc) nt R2)

Ta có: 

\(\begin{array}{l}
{R_{AC}} = {R_{BC}} = 6\Omega \\
{R_{ABC}} = \dfrac{{{R_{AC}}.{R_{BC}}}}{{{R_{AC}} + {R_{BC}}}} = \dfrac{{6.6}}{{6 + 6}} = 3\Omega \\
{R_{2AB}} = {R_2} + {R_{ABC}} = 3 + 3 = 6\Omega \\
R = \dfrac{{{R_1}{R_{2AB}}}}{{{R_1} + {R_{2AB}}}} = \dfrac{{3.6}}{{3 + 6}} = 2\Omega \\
I = \dfrac{E}{{R + r}} = \dfrac{8}{{2 + 2}} = 2A\\
{I_2} = \dfrac{{{R_1}}}{{{R_1} + {R_{2AB}}}}.I = \dfrac{3}{{3 + 6}}.2 = \dfrac{2}{3}A\\
{I_{AC}} = \dfrac{{{R_{BC}}}}{{{R_{AC}} + {R_{BC}}}}.{I_2} = \dfrac{6}{{6 + 6}}.\dfrac{2}{3} = \dfrac{1}{3}A\\
{I_A} = I - {I_{AC}} = 2 - \dfrac{1}{3} = \dfrac{5}{3}A\\
m = \dfrac{{{I_2}At}}{{nF}} = \dfrac{{\dfrac{2}{3}.64.\left( {16.60 + 5} \right)}}{{2.96500}} = \dfrac{{16}}{{75}}g
\end{array}\)

b) K mở, mạch gồm: Rac nt (R2 // (Rcb nt R1))

Ta có:

\(\begin{array}{l}
{R_{AC}} = x \Rightarrow {R_{BC}} = R - x\\
{R_{1BC}} = {R_1} + {R_{BC}} = 3 + R - x\\
{R_{12BC}} = \dfrac{{{R_{1BC}}.{R_2}}}{{{R_{1BC}} + {R_2}}} = \dfrac{{3\left( {3 + R - x} \right)}}{{3 + 3 + R - x}}\\
 \Rightarrow {R_{12BC}} = \dfrac{{3\left( {3 + R - x} \right)}}{{R - x + 6}}\\
I = \dfrac{E}{{{R_{12BC}} + {R_{AC}} + r}} = \dfrac{8}{{\dfrac{{3\left( {3 + R - x} \right)}}{{R - x + 6}} + x + 2}}\\
 \Rightarrow I = \dfrac{{8\left( {R - x + 6} \right)}}{{3\left( {3 + R - x} \right) + \left( {x + 2} \right)\left( {R - x + 6} \right)}}\\
 \Rightarrow I = \dfrac{{8\left( {R - x + 6} \right)}}{{ - {x^2} + x + xR + 21 + 5R}}\\
 \Rightarrow I = \dfrac{{8\left( {R - x + 6} \right)}}{{ - {x^2} + x\left( {R + 1} \right) + 21 + 5R}}\\
{I_1} = \dfrac{{{R_2}}}{{{R_{1BC}} + {R_2}}}.I = \dfrac{3}{{R - x + 6}}.\dfrac{{8\left( {R - x + 6} \right)}}{{ - {x^2} + x\left( {R + 1} \right) + 21 + 5R}}\\
 \Rightarrow {I_1} = \dfrac{{24}}{{ - {x^2} + x\left( {R + 1} \right) + 21 + 5R}}\\
 \Rightarrow {I_1} = \dfrac{{24}}{{ - {x^2} + 2.x.\dfrac{{R + 1}}{2} - {{\left( {\dfrac{{R + 1}}{2}} \right)}^2} + {{\left( {\dfrac{{R + 1}}{2}} \right)}^2} + 21 + 5R}}\\
 \Rightarrow {I_1} = \dfrac{{24}}{{{{\left( {\dfrac{{R + 1}}{2}} \right)}^2} + 21 + 5R - {{\left( {x - \dfrac{{R + 1}}{2}} \right)}^2}}}\\
{\left( {x - \dfrac{{R + 1}}{2}} \right)^2} \ge 0 \Rightarrow {I_1} \ge \dfrac{{24}}{{{{\left( {\dfrac{{R + 1}}{2}} \right)}^2} + 21 + 5R}}
\end{array}\)

Dấu = xảy ra khi: 

\(x = \dfrac{{R + 1}}{2} = 1 \Rightarrow R = 1\Omega \)