Question

Answer 1

a. \(R_b=p\dfrac{l}{S}=0,5\cdot10^{-6}\dfrac{6}{0,2\cdot10^{-6}}=15\Omega\)

b. \(MCD:\left(R_d//R\right)ntR_b\)

Ta có: \(R_d=\dfrac{U_d^2}{I_d}=\dfrac{20}{0,8}=25\Omega\)

\(\rightarrow R_{AM}=\dfrac{R_dR}{R_d+R}=\dfrac{25\cdot12}{25+12}=\dfrac{300}{37}\Omega\)

\(\rightarrow R_{td}=R_{AM}+R_b=\dfrac{300}{37}+15=\dfrac{855}{37}\Omega\)

Để đèn sáng bình thường: \(\left\{{}\begin{matrix}I_d=0,8A\\I_R=\dfrac{U_d}{R}=\dfrac{20}{12}=\dfrac{5}{3}A\end{matrix}\right.\)\(\Rightarrow I_b=I_d+I_R=0,8+\dfrac{5}{3}=\dfrac{37}{15}A\)

\(\Rightarrow R_b'=\dfrac{U-U_b}{I_b}=\dfrac{24-20}{\dfrac{37}{15}}=\dfrac{60}{37}\Omega\)