B4:TT : R =176Ω ; U = 200V; t=20p=1200s

=> Q=?

Giai:

NL do day dan toa ra la:

Q= I².R.t = \(\left(\dfrac{U}{R}\right)^2.R.t=\dfrac{U^2}{R}.t=\dfrac{200^2}{176}.1200\approx272727,3\left(J\right)\)

B3: TT:\(U=U_{đm}=220V\Rightarrow P=P_{đm}=1000W\)

V=3l = 3.10^-3m³; \(t^o_1=30^oC\);\(t^o_2=100^oC\)

c=4200J/kg.K ; D=1000kg/m³

=> t =? h

GIAI:

khoi luong nc mang di dun:

m= D.V = 3.10^-3.1000=3(kg)

NL làm nong nc den 100 :

Q₁= m.c.\(\Delta t^o\) = 3.4200(100-20)= 1008000(J)

NL toan phan de dun nc:

Q= I².R.t = I².U/I . t = U.I.t = P.t = 1000t(J)

vì NL lam nong am va NL toa ra moi truong khong dang ke nen: Q=Q₁

=> 1000t = 1008000

=> t= 1008 (s) = \(\dfrac{7}{25} \left(h\right)\)

|x+5| = 2

+ x+5 = 2 => x =-3

+x+5 = -2 => x =-7

Vậy x thuộc { -3 ; -7}

TT: \(U=220V\) ; \(P=1000W\)

t = 1,5h = 5400s

=> Q=? J, Cal

GIAI:

nhiet luong toa ra cua bep:

Q = I².R.t = U.I.t = P.t = 1000.5400= 5400000(J) = 1296000(Cal)

đề câu 4 bị thiếu công suất bn ơi

C1: TT: R1 = 7,5\(\Omega\) ; R2= 5 ; R3= 15;U= 12V

=> a, R_tđ ; b, I ; c, P_1,2,3

GIAI:

a, dien tro tuong duong la:

\(R_{tđ}=\left(\dfrac{1}{R_3}+\dfrac{1}{R_2}\right)+R_1=\left(\dfrac{1}{15}+\dfrac{1}{5}\right)+7,5\approx7,77\left(\Omega\right)\)

b, cuong do dong dien:

\(I=\dfrac{U}{R_{tđ}}=\dfrac{12}{7,77}\approx1,54\left(A\right)\)

vì (R2//R3) nt R1 => I= I1 = I' = 1,54A

hieu dien the dien tro 1: \(U1=I1.R1=1,54.7,5=11,55\left(V\right)\)

Vì R2//R3 nên hieu dien the la:

\(U'=U2=U3=U-U1=12-11,55=0,45\left(V\right)\)

cuong do qua dien tro 2 la:

\(I2=\dfrac{U2}{R2}=\dfrac{0,45}{5}=0,09\left(A\right)\)

=> I3 = I' - I2 = 1,54-0,09= 1,45 A

c, cong suat cua cac dien tro:

P1= U1 . I1 = 11,55 . 1,54\(\approx17,79\left(W\right)\)

P2= U2.I2= 0,09.0,45= 0,0405(W)

P3 = U3.I3 = 1,45.0,45= 0,6525(W)

B15: TT: U=28V ; I= 2A; l = 11,2m ;

S= 0,4mm²= 4.10^-7m²

=> a, R ; b, p?

GIAI:

a, dien tro cua day: \(R=\dfrac{U}{I}=\dfrac{28}{2}=14\left(\Omega\right)\)

b, dien tro suat cua chat lam day:

\(R=\dfrac{l.p}{S}\Rightarrow p=\dfrac{R.S}{l}=\dfrac{14.4.10^{-7}}{11,2}=5.10^{-7}\left(\Omega m\right)\)

B16: TT: \(U_{đm}=220V\) ; \(P_{đm}=900W\)

t = 20p = 1200s ; U =220 ;

=> a,\(I_{đm}\) ; b, R ; c, A

GIAI:

a, cuong do dong dien dinh muc:

\(I_{đm}=\dfrac{P_{đm}}{U_{đm}}=\dfrac{900}{220}=\dfrac{45}{11}\left(A\right)\)

b, vì nó hoat dong binh thuong nen:

\(P_{đm}=U_{đm}.I_{đm}=I^2_{đm}.R\Rightarrow R=\dfrac{P_{đm}}{I^2_{đm}}=900:\left(\dfrac{45}{11}\right)^2\approx53,8\left(\Omega\right)\)

c, vì U = Uđm₌ = 220V

=> I = Iđm₌ 45/11 (A)

dien nang tieu thu la: A= I².R.t = \(\left(\dfrac{45}{11}\right)^2.53,8.1200=1080446,3\left(J\right)\)

B17:TT:t=5.30=150h;700đ/Kwh;P =150W

=> a, P'=; b,A? ; c,T

GIAI:

a, cong suat dien trung binh cua ca khu dan cu moi ngay:

P'= P. 45 = 45.150= 6750(W)=6,75(kW)

b, dien nang ca khu dan cu tieu thu la:

A = P'. t = 6,75.150= 1012,5(kWh)

c, so tien phai tra la:

T = A. 700 = 700.1012,5 = 708750 (đồng)

B13: TT: R₁=6Ω ; R₂=12Ω; R₃= 16Ω

U= 24V ; t = 30s

=> a, R_tđ ; b, I,I₁,I₂,I₃ ; c, A

GIAI:

a, vì mắc // nên \(\dfrac{1}{R_{tđ}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{16}=\dfrac{5}{16}\)

\(\Rightarrow R_{tđ}=\dfrac{16}{5}=3,2\left(\Omega\right)\)

vì mắc // nên U = U₁ = U₂ = U₃ = 24V

b,cuong do dong dien la:

\(I=\dfrac{U}{R_{tđ}}=\dfrac{24}{3,2}=7,5\left(A\right)\)

\(I_1=\dfrac{U}{R_1}=\dfrac{24}{6}=4\left(A\right)\)

\(I_2=\dfrac{U}{R_2}=\dfrac{24}{12}=2\left(A\right)\)

\(I_3=\dfrac{U}{R_3}=\dfrac{24}{16}=1,5\left(A\right)\)

c,cong suat la: P = U.I = 24.7,5=180(W)

dien nang tieu thu: \(A=\dfrac{P}{t}=\dfrac{180}{30}=6\left(J\right)\)

B14: TT: l =15m ; S=1,5mm²=1,5.10^-6m²

U = 28V ; p= 1,1.10^-4 Ωm

=> I =?

GIAI:

dien tro cua day dan: \(R=\dfrac{l.p}{S}=\dfrac{15.1,1.10^{-4}}{1,5.10^{-6}}=1100\left(\Omega\right)\)

cuong do dong dien: \(I=\dfrac{U}{R}=\dfrac{28}{1100}\approx0,03\left(A\right)\)

ko có thoi gian sử dụng sao tính đc dien nang tieu thụ bn ơi

TT: \(U=U_{đm}=220V\) ; \(P_{đm}=1000W\)

V = 3l = 3.10^-3m^{3 ;}\(t^o_1=20^oC\);\(t^o_2=100^oC\)

c= 4200J/kg.K ; D= 1000kg/m³ => t=? h

GIAI: vì HĐT định mức = HĐT nguon dien nên công suất định mức = công suất làm việc

khoi luong nc mang di nau:

m = D.V = 3kg

nhiet luong de nc soi:

\(Q_{ci}=m.c.\Delta t^o=3.4200\left(100-20\right)=1008000\left(J\right)\)

vì bỏ qua NL làm nóng am và toa ra moi truong ngoai => \(Q_{toa}=Q_{ci}\)

<=> P.t = 1008000

=> \(t=\dfrac{1008000}{P}=\dfrac{1008000}{1000}=1008\left(s\right)=\dfrac{7}{25}\left(h\right)\)