\(R=p\dfrac{l}{S}\Rightarrow l=\dfrac{R.S}{p}=\dfrac{50.0,5.10^{-6}}{0,4.10^{-6}}=62,5\left(m\right)\)

\(432000\left(J\right)=120\left(Wh\right)\)

\(A=P.t\Rightarrow P=\dfrac{A}{t}=\dfrac{120}{2}=60\left(W\right)\)

\(R=p\dfrac{l}{S}\Rightarrow l=\dfrac{R.S}{p}=\dfrac{20.0,5.10^{-6}}{1,1.10^{-6}}=\dfrac{100}{11}\left(m\right)\)

\(\Rightarrow N=\dfrac{l}{\pi.d}=\dfrac{\dfrac{100}{11}}{\pi.2}\simeq1,5\left(vong\right)\)

\(R=p\dfrac{l}{S}=1,10.10^{-6}\dfrac{15}{0,3.10^{-6}}=55\Omega\)

\(\Rightarrow I=\dfrac{U}{R}=\dfrac{220}{55}=4A\)

\(S=2mm^2=2\cdot10^{-6}\left(m^2\right)\)

Điện trở dây:

  \(R=\rho\cdot\dfrac{l}{S}=1,7\cdot10^{-8}\cdot\dfrac{400}{2\cdot10^{-6}}=3,4\left(m\right)\)

\(R=p\dfrac{l}{S}=1,7.10^{-8}\dfrac{400}{2.10^{-6}}=3,4\Omega\)

\(R_Đ=\dfrac{U^2_Đ}{P_Đ}=\dfrac{6^2}{3}=12\Omega\)

Đèn sáng bình thường: \(I_m=I_Đ=\dfrac{6}{12}=0,5A\)

\(\Rightarrow R_m=\dfrac{U_m}{I_m}=\dfrac{9}{0,5}=18\Omega\)

Mà R₁ nt R_MN \(\Rightarrow R_1+R_{MN}=18\Rightarrow R_{MN}=18-12=6\Omega\)

Chọn B.

cảm ơn nhiều ah:>

25.

\(R=\dfrac{\left(R1+R2\right)R3}{R1+R2+R3}=\dfrac{\left(2+4\right)6}{2+4+6}=3\Omega\)

\(\Rightarrow I=\dfrac{U}{R}=\dfrac{6}{3}=2A\)

Chọn B

24 thì sao ah,cảm ơn nhìu nha<3

\(A=P.t=100.4.30=12000\left(Wh\right)=12\left(kWh\right)\)

\(\Rightarrow T=12.3000=36000\left(d\right)\)

cảm ơn nhiều ah:>

\(R=\dfrac{U^2}{P}=\dfrac{220^2}{100}=484\Omega\)