\(R=p\dfrac{l}{S}=5,5.10^{-8}\dfrac{0,2}{0,01.10^{-6}}=1,1\Omega\)

\(P=UI=220\left(\dfrac{220}{1,1}\right)=44000\left(W\right)\)

\(A=UIt=220\left(\dfrac{220}{1,1}\right).0,5=22000\left(Wh\right)=22\left(kWh\right)=79200000\left(J\right)\)

Điện trở dây: 

  \(R=\rho\cdot\dfrac{l}{S}=5,5\cdot10^{-8}\cdot\dfrac{0,2}{0,01\cdot10^{-6}}=1,1\Omega\)

Công suất bóng đèn:

 \(P=\dfrac{U^2}{R}=\dfrac{220^2}{1,1}=44000W\)

 Công của đèn thực hiện:

 \(A=Pt=44000\cdot30\cdot60=79200000J=22kWh\)

Tóm tắt:

U = 220V

P = 1000W

V = 2,5l = 2,5kg

t₀ = 20⁰

\(\Delta t\) = 14p35'

Giải 

a. \(Q_{nc}=m.c.\Lambda t=2,5.4200\left(100-20\right)=840000\left(J\right)\)

b. \(H=\dfrac{Q_{nc}}{Q_{tp}}.100\%=\dfrac{840000}{1000.875}.100\%=96\%\)

\(\Rightarrow Q'=\dfrac{2Q}{H}=\dfrac{2.840000}{0,96}=1750000\left(J\right)\)

\(\Rightarrow Q_{tp}=Q'.30=1750000.30=52500000\left(J\right)\simeq14,6\left(kWh\right)\)

\(\Rightarrow T=Q_{tp}.1500\simeq14,6.1500\simeq21900\left(dong\right)\)

Ta có: \(\dfrac{R1}{R2}=\dfrac{l1}{l2}\Rightarrow R2=\dfrac{l2.R1}{l1}=\dfrac{8.3}{2}=12\Omega\)

\(R_{tđ}=\dfrac{60}{9}=\dfrac{20}{3}\Omega\)

\(\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}\)\(\Rightarrow\dfrac{1}{3R_3}+\dfrac{1}{\dfrac{3}{2}R_3}+\dfrac{1}{R_3}=\dfrac{1}{\dfrac{20}{3}}\Rightarrow R_3=13,3\Omega\)

\(\Rightarrow\left\{{}\begin{matrix}R_1=13,3\cdot3=39,9\Omega\\R_2=13,3\cdot2=26,6\Omega\end{matrix}\right.\)

Do \(R_1//R_2//R_3\)\(\Rightarrow U_1=U_2=U_3=U_m=60V\)

\(\Rightarrow I_1=\dfrac{60}{39,9}=1,504A\)

    \(I_2=\dfrac{60}{26,6}=2,256A\)

    \(I_3=\dfrac{60}{13,3}=4,5A\)

a. \(P=UI\Rightarrow I=\dfrac{P}{U}=\dfrac{100}{110}=\dfrac{10}{11}A\)

b. \(A=Pt=100.\dfrac{1}{3}=33,\left(3\right)\left(W\right)\)

a) Cường độ dòng điện bóng đèn khi hđ bình thường:

    \(I=\dfrac{P}{U}=\dfrac{100}{110}=\dfrac{10}{11}A\)

b) Điện năng tiêu thụ:

   \(A=UIt=110\cdot\dfrac{10}{11}\cdot20\cdot60=120000J\)

Phải mắc nối tiếp. Vì: \(U_{mach}=U_1+U_2=110+110=220\left(V\right)\)

a. \(P=P1+P2=100+75=175\left(W\right)\)

\(I=I1+I2=\left(\dfrac{P1}{U1}\right)+\left(\dfrac{P2}{U2}\right)=\left(\dfrac{100}{220}\right)+\left(\dfrac{75}{220}\right)=\dfrac{35}{44}\left(A\right)\)(R1//R2)

b. \(I=I1=I2=\dfrac{U}{R}=\dfrac{220}{\left(\dfrac{220^2}{100}\right)+\left(\dfrac{220^2}{75}\right)}=\dfrac{15}{77}A\left(R1ntR2\right)\)

 \(\left\{{}\begin{matrix}U1=I1.R1=\dfrac{15}{77}.\left(\dfrac{U1^2}{P1}\right)=\dfrac{15}{77}.\left(\dfrac{220^2}{100}\right)=\dfrac{660}{7}V\\U2=I2.R2=\dfrac{15}{77}.\left(\dfrac{U2^2}{P2}\right)=\dfrac{15}{77}.\left(\dfrac{220^2}{75}\right)=\dfrac{880}{7}V\end{matrix}\right.\)

\(P_{nt}=U_{nt}.I_{nt}=220.\dfrac{15}{77}=\dfrac{300}{7}\left(W\right)\)