Giúp mình vs mọi người ơi !

Đáp án:

a) \(81,4\% \)

b) \(\dfrac{{108}}{7}\Omega \)

Giải thích các bước giải:

a) Ta có: 

\(\begin{array}{l}
{E_b} = 2E = 2.12 = 24V\\
{r_b} = 2r = 2.1 = 2\Omega \\
{R_d} = \dfrac{{U_{dm}^2}}{{{P_{dm}}}} = \dfrac{{{{12}^2}}}{6} = 24\Omega \\
{R_{dR}} = {R_d} + R = 24 + 8 = 32\Omega \\
{R_N} = \dfrac{{{R_1}{R_{dR}}}}{{{R_1} + {R_{dR}}}} = \dfrac{{12.32}}{{12 + 32}} = \dfrac{{96}}{{11}}\Omega \\
I = \dfrac{{{E_b}}}{{{R_N} + {r_b}}} = \dfrac{{24}}{{\dfrac{{96}}{{11}} + 2}} = \dfrac{{132}}{{59}}A\\
{I_R} = \dfrac{{{R_1}}}{{{R_1} + {R_{dR}}}}.I = \dfrac{{12}}{{12 + 32}}.\dfrac{{132}}{{59}} = \dfrac{{36}}{{59}}A\\
{I_{{R_1}}} = I - {I_R} = \dfrac{{132}}{{59}} - \dfrac{{36}}{{59}} = \dfrac{{96}}{{59}}A\\
{P_d} = I_R^2{R_d} = {\left( {\dfrac{{36}}{{59}}} \right)^2}.24 = 8,94W\\
H = \dfrac{{{R_N}}}{{{R_N} + {r_b}}}.100\%  = \dfrac{{\dfrac{{96}}{{11}}}}{{\dfrac{{96}}{{11}} + 2}}.100\%  = 81,4\% \\
{I_{dm}} = \dfrac{{{P_{dm}}}}{{{U_{dm}}}} = \dfrac{6}{{12}} = 0,5A
\end{array}\)

Do \({I_d} > {I_{dm}} \Rightarrow \) đèn sáng hơn bình thường

b) Ta có: 

\(\begin{array}{l}
{R_{dR}} = {R_d} + R = 24 + R\\
{R_N} = \dfrac{{{R_1}{R_{dR}}}}{{{R_1} + {R_{dR}}}} = \dfrac{{12.\left( {24 + R} \right)}}{{12 + 24 + R}} = \dfrac{{12\left( {24 + R} \right)}}{{R + 36}}\\
I = \dfrac{{{E_b}}}{{{R_N} + {r_b}}} = \dfrac{{24}}{{\dfrac{{12\left( {24 + R} \right)}}{{R + 36}} + 2}} = \dfrac{{24\left( {R + 36} \right)}}{{14R + 360}}\\
{I_d} = \dfrac{{{R_1}}}{{{R_1} + {R_{dR}}}}.I = \dfrac{{12}}{{R + 36}}.\dfrac{{24\left( {R + 36} \right)}}{{14R + 360}} = \dfrac{{288}}{{14R + 360}}\\
 \Rightarrow \dfrac{{288}}{{14R + 360}} = 0,5 \Rightarrow R = \dfrac{{108}}{7}\Omega 
\end{array}\)