Question

[Lưu Xuân Thắng 11]

Thầy cô giúp em ạ!

Answer 1

[Bear]

a. 

\(ξ_b=nξ=8.1,5=12V\)

\(r_b=nr=8.0,125=1Ω\)

b.

\(R_d=\dfrac{U_{dm}^2}{P_{dm}}=\dfrac{6^2}{4}=9\Omega\)

\(R_{pd}=\dfrac{R_p.R_d}{R_p+R_d}==2,25\Omega\)

\(R_N=R_1+R_{pd}=4,75+2,25=7\Omega\)

\(I=\dfrac{\text{ }\xi_b}{R_N+r_b}=\dfrac{12}{7+1}=1,5A\)

\(U_p=U_d=U_{pd}=IR_{pd}=1,5.2,25=3,375V\)

\(I_p=\dfrac{U_p}{R_p}=\dfrac{3,375}{3}=1,125A\)

\(m_{Cu}=\dfrac{AI_pt}{Fn}=\dfrac{64.1,125.1930}{96500.2}=0,72g\)

c.

Đèn sáng yếu bởi vì \(U_d=3,375V< U_{dm}=6V\)

d.

\(P_1=\dfrac{\xi_b^2R_1}{\left(R_1+R_{pd}+r_b\right)^2}=\dfrac{144R}{\left(3,25+R_1\right)^2}\)

\(P_1\) max \(\Leftrightarrow R_1=3,25\Omega\)

\(P_1\) max \(\dfrac{144.3,25}{\left(3,25+3,25\right)^2}=11,077W\)

Answer 2

[zero]

tham khảo :3

 a. 

${\xi }_b=n\xi =8.1,5=12V$

$r_b=nr=8.0,125=1\Omega$

b.

$R_d=\frac{U_{dm}^2}{P_{dm}}=\frac{6^2}{4}=9\Omega$

$R_{pd}=\frac{R_p.R_d}{R_p+R_d}==2,25\Omega$

$R_N=R_1+R_{pd}=4,75+2,25=7\Omega$