: , , .
.
4
, R=8 , L=19,1 =265 , , (169 sinωt +45,4 sin 3 ωt). £=50 .
, S .
U= f(1) U1=169 sinωt () U3=45,4 sin 3 ωt (). . .
(£=50 )
XL1=2π£1L=2*3,14*50* 0,0191=6
1 1
XC1 2π£1C 2*3,14*50*256*10-6 12
: U1=169 sinωt ()
R=8 , XL1=6 , XC1=12 .
Z1= 2+(XL1- XC1)2 = 2+(6-12)2=10 .
U1m 169
U1 , U1m=169, U1 ≈ 120 B
U1 120
I1 Z1 10 12 A
: U3=45,4 sin 3 ωt, f3=3f1
R 8 .
(XL=2π£L). , 3 3 , . .
XL3 = 3* XL1 = 3*6=18
(= 2π£) 3 3 . .
3 12
3 3 3 4
Z3= 3+(XL3-Xc3)2= 2+(18-4)2=16,1
U3m 45,4
U3 32,2 B
U3 32,2
I3 Z3 16,1 2 A
I= 12+I23= 2+22=12,2 A
U= 12+U32= 2+32,22=124 B.
S=U*I=124*12,2=1512,8 Ba.
P=I2R=12,22*8=1190 .
41-50 .
5
=0,4 , R=2,5 , U= 50 .
t , t0=0, t1=t, t2=2t, t3=3t, t4=4t, t5=5t.
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|
U=f(t)
1)
t=R*C=2,5*106*0,4*10-6=1c,
( R , ).
2)
1
U1=U(1-e r)
=2,718
t0=0
0
U=U(1-e r )=50(1-e0)=50(1-1)=0 B
t1 = t
r
U1=U (1- er )=50(1-e-1)=50(1-0.367)=31.65 B
t2=2t
2r
U2=U (1- er )=50(1-e-2)=50(1-0.135)=42.25 B
t3=3t
3r
U3=U (1- er )=50(1-e-3)=50(1-0.0498)=47.51 B
t4=4t
4r
U4=U (1- er )=50(1-e-4)=50(1-0.0183)=49.09 B
t5=5t
5r
U5=U (1- er )=50(1-e-5)=50(1-0.0067)=49.07 B
.
U()
50
40
U=f(t)
30
20
10
1 2 3 4 6 t(c)