ZESTAW C Zad. R=100 Ω, U=400Ω Iźr=? U=Um/ √2 Um=U*√2 =565 Im=Um/ R=565/100=5,65 Iźr=2/ T ∫ 0π/2 Im sinωt dt =2Im/T ∫ 0π/2 sinωt dt= =2Im/ T (-1/ ω cosωT/2)=2Im/ T( -1/2πf *cos 2πf*1/ f/ 2)= =2Im/ T*T/2 π=Im/ π(cosπ - cos0)=2Im/ π=3,6 Zad. u(t)=282,8 sin(314t - π/3), P=400W, I=10A, R=?, C=? P=U*I cos ϕ U=282,8/ √2=200V cosϕ=400/200*10=0,2 ϕ=78,4 ϕu=60 ϕ =ϕu -ϕi ϕi=60-78,4=18,4 U=200 ej60=100+j173,2 I=9,4-j3,1 Z=4,1+j19,7 19,7=1/ 314C C=1/314*1/ 19,7=161 µF Zad. Z1=R1+j XL1 IŹR=E1/ Z1 Y1=1/ Z1 Y2= -1/ j Xc Y2=Y1+Y2 Z2=1/ Y2 E2=IŹR*Z2 Iż=E1+E2/ Z1+Z2 ZESTAW B Zad. R=20 Ω, t=2000T, f=50Hz U=Um/ √2 Um=U*√2=400V Im=Um/R=20A I= √1/T∫0T/2 Im2*sin2ωtdt=√ Im2 / T*1/2(∫0T/2 dt - ∫0T/2 cos2ωtdt= = √Im / 4=Im / 2 Im=10A T=1/f Q=R*I2*T=20*100*2000*1/50=80kJ Zad. E=126V, I2=6,26A, Rw=6 Ω, R1 i R2=? I=E/ 2Rw=126/ 2*6=10,5A I=I1+I2 I1=I – I2=10,5-6,26=4,24A U=I*Rw=10,5*6=63V UAB=E1–U=126-63=63 I1=UAB/R1 →R1=UAB/I1=63/4,24=15Ω I2= UAB/R2 →R2= UAB/I2=63/6,26=10Ω Pmax gdy Rw=R Pźr=E*I=126*10,5=1323W PRW=Rw*I2=661,5W PR1=R1*I12=264,6W PR2=R2*I22=391,8W Zad. XL3= ωL3 E2=I2*jXL3 Z11=R1+j( ωL1+XM+ωL2+XM) Z22=R2+j(XM+ ωL2- XM- 1/ ωC+XL3) Z12=j(ωL2+XM) Z11*Il – Z12*Ill=E1 - Z12*Il+Z22*Ill=E2 ZESTAW E Zad. I=20A, f=50Hz A1,A2=? I= √1/T∫0T i2(t)= √1/T∫0T Im2*sin2ωtdt= = √Im2/T∫0T1/2(1-cos2ωt)dt=√Im2/2T ∫0Tdt-Im2/2T ∫0T cos2ωtdt= = √Im2/2T T=Im/√2 Im=I √2=20*1,41=28,5A I= √1/T∫0T/2 Im2*sin2ωtdt=√ Im2 / T*1/2(∫0T/2 dt - ∫0T/2 cos2ωtdt= = √Im / 4=Im / 2 I1=I2=28,2/2=14,1A Zad. R=Xc=10 Ω, Uc=50V u(t), i(t), p(t)=? Z=R–jXc U=I*Z U=I*(R-jXc) U=50+j50 ϕUc=0 I=Uc/ -jXc=50/ -j10=j5 i(t)=5 √2 sin(ωt+π/2) UR=I*R=j5*10=j50V ϕUR=π/2 u(t)=70√2 sin(ωt+π/2) ϕ=45-90= -45 p(t)=U*I cosϕ - U*I cos(2ωt+ϕu+ϕi) Zad. U=220V, P=5kW, , f=50Hz, cos ϕ=0,7 ϕ=46 ϕu=0 ϕi= - 46 P=U*I cos ϕ I=5000/ 220*0,7=32A QL=Qc QL=U*I sin ϕ=5064 Ic=Uc/ Xc 5064=U*U/Xc sin ϕ 5064=U2/ Xc sinϕ=7Ω ω=2πf=314 Xc=1/ωC C=1/Xcω=1/7*314=4,5*10-4
(…)
…=8Rx+8R
2Rx=8R Rx=4R Rx=40Ω
Iźr1=E1/ Rx =30/40=0,75A Gwx=1/ Rx G=1/ R
Iźro=Iźr1+Iźr2=2+0,75=2,75A Rz=40*10/50=8Ω
UR=Iźro*Rz=22V IR=UR/R=2,2 P=UR*IR=48,4W
Zad.
Dane: ω=100 R=20Ω L=200mH IL=2A
Obliczyć: u(t), i(t), p(t)
XL=ωL=100*200*10-3=20Ω
Z=R*jXL / R*jXL=20-j20 / 20+j20* 20-j20 / 20-j20
=58000+8000 / 400+400=(10+j10)Ω
UL=XL*IL=2A*j20=j40 → woltomierz=0
IR=Uc/ R=j40/20=j2A
I=2+2jA → amperomierz =√22…
…
ϕu=60O ϕ=ϕu-ϕi ϕi=23O
Z=U/I=80
Z=80(cos37O+jsin23O)=80(0,8+j0,4)=64+j32
Y=1/Z=1/64+j32*64-j32/64-j32=64-j32/5120=0,0125-j0,00625
Y=1/R+!/jXL R=1/0,0125=80 XL=1/0,00625=160
I(t)=1,25*√2 sin(314t+23O)
p(t)=100-125cos(628t+83O)
Zad. W szeregowym połączeniu RLC woltomierze V1,V2,V3 wskazują po
100 V. wyznaczyć wskazania V,V12,V23 dla kombinacji połączeń podanej
w tablicy.
Lp.
Z1
Z2
Z3
1
C
L
R
2
L
L
C
3…
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