Question

Zadanie 1 z załącznika, odnośnie prądu

Answer 1

R1=R2=4; R3=6; E=3V
dla rw=0; E=U
1/R1,2=(1/4+1/4)=1/2
R1,2=2 Rz=R1,2+R3=2+6=8
I=U/Rz=3/8
I=I1+I2
I1R1=I2R2
U1=U2 U1=U-U3=U-IR3= 3-3*6/8=0,75
I1=U1/R1=0,75/4= 3/16 A
I2=U2/R2=3/16 A
U1=I1R1= (3/16)*4=12/16=3/4 V
dla R1≠R2 I1/I2=R2/R1 dla R1>R2; I1/I2<1; I1<I2
dla R1>R2; I1>I2 2.
E=6 V; r=2 Ω; I1=0,5
R=(E-I1r)/I1= (6-0,5*2)/0,5=10 Ω
dla 3R I2=E/(r+3R)= 6/(2+3*10)=0,1875 A

3. m=20 t=2*10^4 kg; v=50 km/h= 50/3,6=13,88 m/s; f=0,01; η=0,75; U=1000 V
Moc P=Fv/η=fmgv/η
P=UI I=fmgv/ηU= 0,01*2*10^4*9,81*13,88/0,75*1000=36,3101
I=36 A
sprawdzenie
P= UI=1000*36=36 kW
P= 0,01*2*10^4*9,81*13,88/0,75=36310,08 W
P=36 kW