"
"
"
c
c
c
???3cccèèè
^
^
^￥￥￥  
sss3
?3.2?%
íqDD
^?5.,,,,,,,,,,,,,,,,,,,,,,3
?3.2.15.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,3
?3.2.25.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,4
?3.3?RCè
^￥Y?.,,,,,,,,,,,,,,,,,,,,,,,,,,5
?3.3.15.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,5
?3.3.35.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,5
?3.3.45.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,6
?3.4?B¨L?è
^
s￥ ?1
íE.,,,,,,,,,,,,,,7
?3.4.15.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,7
?3.4.25.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,8
?3.4.35.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,10
?3.4.45.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,11
?3.4.55.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,12
?3.6?RLè
^￥Y?.,,,,,,,,,,,,,,,,,,,,,,,,,,13
?3.6.15.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,13
?3.6.25.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,14
?3.6.45.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,16
?3.6.55.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,17
1
List of Figures
153.2.1m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,3
253.2.2m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,4
353.3.1m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,5
453.3.3m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,6
553.3.4m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,6
653.4.1m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,7
753.4.2m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,8
853.4.2m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,9
953.4.3m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,10
1053.4.4m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,11
1153.4.5m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,12
1253.4.5m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,13
1353.6.1m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,14
1453.6.2m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,15
1553.6.4m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,16
1653.6.5m.,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,17
2
3èèè
^
^
^￥￥￥  
sss
3.2%%%
íííqqqDDDDDD
^
^
^???555
3.2.1
m1
Uòè
^D
^ -?)?×
,
k pD
^a ?è
@i￥
S′i(0+)
×
′i(1).
[333]
m1:53.2.1m
(1)m1(a)
Uè
^
iL(0+) = iL(0?) = 62A = 3A
i(0+) = 22+2 ￡3A = 1:5A
i(1) = 62￡2
2+2
￡ 12A = 3A
(2)m1(b)
Uè
^
uc(0+) = uc(0?) = 6V
i(0+) = 6?62 A = 0
i(1) = 62+2A = 1:5A
3
(3)m1(c)
Uè
^
iL1(0+) = iL1(0?) = 6A
iL2(0+) = iL2(0?) = 0
i(0+) = iL1(0+)?iL2(0+) = (6?0)A = 6A
i(1) = 0
(4)m1(d)
Uè
^
uc(0+) = uc(0?) = 62+2 ￡2V = 3V
i(0+) = 6?32+2A = 0:75A
i(1) = 62+2+2A = 1A
3.2.2
m2
Uè
^D
^ -)?×

k pD
^aiLuciS￥
S′×
′b
[333]
m2:53.2.2m
iL(0+) = iL(0?) = 15
10+10+ 15￡3015+30
￡ 3030+15A = 12 ￡ 3030+15A = 13A
uc(0+) = uc(0?) = (15?10￡0:5)V = 10V
iS(0+) = i1(0+)?iL(0+) = uc(0+)10?iL(0+) = (1010? 13)A = 23A
30?èE$
¤ ?è
@￥
S′1
,b
iL(1) = 0
uC(1) = 10￡ 1510+10V = 7:5V
iS(1) = 1510+10A = 34A
4
3.3 RCèèè
^
^
^￥￥￥YYY???
3.3.1
m3?I = 10mA,R1 = 3k?,R2 = 3k?,R3 = 6k?,C = 2?Fb 71S>
-è
^X)?×
b pt? 0
HuCi1iT
ì
HW￥M wLb
[333]
m3:53.3.1m
uc(0+) = uc(0?) = R3I = 6￡103 ￡10￡10?3V = 60V = U0
Dè ?íq1 ó￥?rèE
R = R1 + R2R3R
2 +R3
= (3+ 3￡63+6)k? = 5k?
HWè
= RC = 5￡103 ￡2￡10?6s = 0:01s
'5 p￥
^
,
{ ?Y?
è
@÷X$
¤#¤
uc = U0e?
t
= 60e?
t
0:01 = 60e?100tV
i1 =?CduCdt = U0R e?
t
= 605￡103e?100t = 12e?100tmA
3.3.3
è
^ ?m4
U 71S> -è
^X)?×
 p 71>a￥èaucb
[333]
uc(0+) = uc(0?) = 6￡103 ￡9￡10?3V = 54V
= 6￡36+3 ￡103 ￡2￡10?6s = 4￡10?3s
5
m4:53.3.3m
'5
^ p ?Y?uc5
79mA ?Xè
@÷ 7 p
,
{ ?M?u0c 7a
7uc(0+) =
0 p
,
Y?u00c Ka¤uc = u0c +u00cb
u0c = U0e?
t
= 54e?
t
4￡10?3V = 54e?250tV
u00c = U(1?e?
t
) = 18(1?e?250t)V
T?U = uc(1) = 3￡63+6 ￡103 ￡9￡10?3V = 18V
uc = (18+36e?250t)V
3.3.4
μBL?í÷=
?
N[m5(a)] ?%
íq?%μ
?
{ ?è
@i[ o? ?m5(b)
U]a 

èau￥o? ?m5(c)
Ub(1)u￥·
T (2)??
￥è
^i ??íq￥?
′b
[333]
m5:53.3.4m
6
(1)?m5(c) V¤
t = 0
H
u = 2(1?e?
t
)V
u(?) = 2(1?0:368)V = 2￡0:632V = 1:264V
t =1
H
u = 1:264e?
(t?1)
V
(2)??
￥è
^ ?m5(d)
Uby
u(1) = Ri = 2V
R￡1 = 2 R = 2?
?
= RC 1 = 2C C = 0:5F
3.4BBB¨¨¨LLL???èèè
^
^
^  
sss￥￥￥ ? ? ?111
í
í
íEEE
3.4.1
m6(a)
U￥è
^?u1B¨èa ?m6(b)
U
k pi3ucb
!uc(0?) = 1V
[333]
m6:53.4.1m
?¨ ?1
íE9
b
(1) puc
uc(0+) = uc(0?) = 1V
uc(1) = R3 uR
1 +R3
= 2￡ 42+2V = 2V
=
R2 + R1R3R
1 +R3
C =
1+ 2￡22+2
￡103 ￡1￡10?6s
= 2￡10?3s
7
?N¤
uc = uc(1)+[uc(0+)?uc(1)]e?
t
= [2+(1?2)e?
t
2￡10?3]V = (2?e?500t)V
(2) pi3
i3(0+) =
u
2 +
uc(0+)
1
1
2 +
1
1 +
1
2
1
R3
=
4
2 +
1
1
1
2 +
1
1 +
1
2
￡ 12mA = 34mA
i3(1) = uR
1 +R3
= 42+2mA = 1mA
?N¤
i3 = i3(1)+[i3(0+)?i3(1)]e?
t
= 1+(34?1)e?500tmA = (1?0:25e?500t)mA
3.4.2
è
^ ?m7
U pt? 0
H(1)è ?èauc(2)B?èêvB(3)A?èêvA￥
M
pbD
^ -è
^)?×
b
[333]
m7:53.4.2m
(1) pt? 0
H￥è ?èauc
t = 0?t = 0+￥è
^ ?m8(a)a(b)
U?N¤
8
m8:53.4.2m
uc(0+) = uc(0?) = 0?(?6)(5+25)￡103 ￡5￡103V = 1V
uc(1) = 6?(?6)(10+5+25)￡103 ￡5￡103V = 1:5V
= [(R1 +R3)==R2]C = 0:44￡10?6s
#
uc = [1:5+(1?1:5)e?
t
0:44￡10?6]V
= (1:5?0:5e?2:3￡106t)V
(2) pt? 0
H￥B?èêvB
VB(0+) =
6? 6?(?6)?1(10+25)￡103 ￡10￡103
V
= (6?3:14)V = 2:86V
VB(1) =
6? 6?(?6)(10+5+25)￡103 ￡10￡103
V
= (6?3)V = 3V
#
vB = [3+(2:86?3)e?2:3￡106t]V
= (3?0:14e?2:3￡106t)V
?i(1)VB(0?) = 07VB(0+) = 2:86V 6= VB(0?) (2)t = 0+￥è
^
?èE10k?25k??YV]Bè
@
?1 ó7èE5k??YV
6
Bè
@yN
D10k?25k??
^1 ó￥t = 1￥è
^? ???
M1 ó (3)t = 0+￥è
^?9
èE10k?25k??è
@￥
T0
^
6?(?6)?1
(10+25)￡103A
9
(3) pt? 0
HA?èêvA
VA(0+) =
6?(?6)?1
(10+25)￡103 ￡25￡10
3 +(?6)
V
= (7:86?6)V = 1:86V
VA(1) =
6?(?6)
(10+5+25)￡103 ￡25￡10
3 +(?6)
V
= (7:5?6)V = 1:5V
#
vA = [1:5+(1:86?1:5)e?2:3￡106t]V
= (1:5+0:36e?2:3￡106t)V
3.4.3
è
^ ?m9
UD
^ -X)?×

k pD
^a(t? 0)￥ucb
[333]
m9:53.4.3m
'5?¨ ?1
íE9
b
(1) ??
S′
uc(0+) = uc(0?) = (20￡103 ￡1￡10?3?10)V = 10V
(2) ??×
′
uc(1) =
10
10+10+20 ￡1￡10
3 ￡20￡103?10
V =?5V
(3) ??
HWè
| ?Xè
@÷ 7
^ ?Xèa÷
^bVè ?íq

Aé ?￥?rè
E1
R0 = 20￡(10+10)20+(10+10)k? = 10k?
10
#
= R0C = 10￡103 ￡10￡10?6s = 0:1s
?
^¤
uc = uc(1)+[uc(0+)?uc(1)]e?
t
=?5+[10?(?5)]e?
t
0:1
= (?5+15e?10t)V
3.4.4
μBRCè
^[m10(a)] 
{ ?èa ?m10(b)
Ub
!

zT = RCb
k
pμ

￥?U???v?
t = 2T
H
Puc = 0b
!uc(0?) = 0b
[333]
m10:53.4.4m
?t = 0?t = T
ùW
uc = 10(1?e?
t
)V
uc(T) = 10(1?e?1) = 6:32V
?t = T?t = 2T
ùW
u0c = U? +[uc(T)?U?]e?
t?T
T
t = 2T
Hu0c = 0'
U? +[uc(T)?U?]e?
2T?T
T = 0
U? +(6:32?U?)￡0:368 = 0
U? =?3:68V
11
3.4.5
m11? 71S5ê?1è
^)?×
bt = 0
H| 71Vê
?1?ê?2
k pt =?
Huc-′bt =?
H?| 71?ê?1
k
pt = 2￡10?2s
Huc-′bN
H| 71?2T￥ucM wLb èè
^
bèè
^￥
HWè
^?M?$
[333]
m11:53.4.5m
(1) t = 0
H| 71V1?2
uc(0?) = uc(0+) = 10V
uc = 10e
t
1
1 = (20+10)￡103 ￡ 13 ￡10?6s = 10?2s = 0:01s
uc(?1) = 10e?1V = 10￡0:368V = 3:68V
12
(2) t =?
H?| 71?1
uc(?1) = 3:68V
uc(1) = 10V
2 = 10￡103 ￡ 13 ￡10?6s = 13 ￡10?2s = 0:0033s
uc =
2
6410+(3:68?10)e?
(t?0:01)
2
3
75V
= (10?6:32e
(t?0:01)
2 )V
uc(0:02s) =
2
410?6:32e?
(0:02?0:01)
0:0033
3
5V
= (10?6:32e?3)V
= (10?6:32￡0:05)V
= 9:68V
(3) t = 0:02s
H| 71?2
uc = 9:68e
(t?0:02)
1 V
uc￥M wL ?m12
Ub
m12:53.4.5m
3.6 RLèèè
^
^
^￥￥￥YYY???
3.6.1
m13?R1 = 2?R2 = 1?L1 = 0:01HL2 = 0:02HU = 6Vb(1)
k
pS1>aè
^?è
@i1i2￥M
p (2)?>S1aè
^?r×?
H
>S2
k pi1i2￥M
pb
[333]
13
m13:53.6.1m
(1)? 71S1> -i1(0?) = i2(0?) = 0#[
,
Y?9
'
i1 = i2 = UR
1 +R2
(1?e
t
1)
T?
1 = L1 +L2R
1 +R2
= 0:01+0:021+2 s = 0:01s
#
i1 = i2 = 61+2(1?e?
t
0:01)A = 2(1?e?100t)A
è
^?r×
H
i1(1) = i2(1) = 2A
(2)?r×
H>S2ai1(0+) = i2(0+) = 2Ab>S2a?r×
H
i1(1) = UR
1
= 62A = 3Ai2(1) = 0
HWè
sY1
01 = L1R
1
= 0:012 s = 0:005s
2 = L2R
2
= 0:021 s = 0:02s
?
^¤
i1 = [3+(2?3)e?
t
0:005]A = (3?e?200t)A
i2 = [0+(2?0)e?
t
0:02]A = 2e?50tA
3.6.2
è
^ ?m14
UD
^ -X)?×
b?| 71V1￥ê?í?2￥ê?
a
k piiLb
[333]
14
m14:53.6.2m
(1) ??
S′
i(0?) =?3
1+ 2￡12+1
A =?95A
iL(0+) = iL(0?) = 22+1 ￡(?95)A =?65A
N?ii(0+) 6= i(0?)b
i(0+)?: ?èa?
p9
'
3 = ￡i(0+)+2[i(0+)?iL(0+)]
3 = i(0+)+2[i(0+)+ 65]
3 = 3i(0+)+ 125
i(0+) = 15A
(2) ??×
′
i(1) = 3
1+ 2￡12+1
A = 95A
iL(1) = 22+1 ￡ 95A = 65A
(3) ??
HWè
= LR
0
= 3
1+ 2￡12+1
s = 95s
15
?
^¤
i = i(1)+[i(0+)?i(1)]e?
t
=
2
495 +(15? 95)e?
5
9t
3
5A =
0
@95? 85e?
5
9t
1
AA
= (1:8?1:6e?
t
1:8)A
iL =
2
465 +(?65? 65)e?
5
9t
3
5A = (1:2?2:4e?
t
1:8)A
3.6.4
è
^ ?m15
U
k¨ ?1
íE pt? 0
H￥i1i2#iLbD
^ -è
^X)?
×
b
[333]
m15:53.6.4m
(1) ??
S′
iL(0+) = iL(0?) = 126 A = 2A
?ii1i2￥
S′??t = 0+￥è
^9
?
^?t = 0?￥è
^9
b?t = 0+￥è
^?¨: ??
p

i1(0+)+i2(0+) = iL(0+) = 2
6i1(0+)?3i2(0+) = 12?9 = 3
3-¤
i1(0+) = i2(0+) = 1A
16
(2) ??×
′
×
Hè?íq V
j1
^#
i1(1) = 126 A = 2A
i2(1) = 93A = 3A
iL(1) = i1(1)+i2(1) = (2+3)A = 5A
(3) ??
HWè
= LR
0
= 16￡3
6+3
s = 0:5s
?
^¤
i1 = [2+(1?2)e?
t
0:5]A = (2?e?2t)A
i2 = [3+(1?3)e?
t
0:5]A = (3?2e?2t)A
iL = [5+(2?5)e?
t
0:5]A = (5?3e?2t)A
3.6.5
? μèER = 1?#è?L = 0:2H￥èH?è L ?
m16?￥è
@i = 30A
H?è '?T7|è÷ Mb
!μèEL
^èEsY
1RL = 20?Rl = 1?°
@è÷èaU = 220V
kù?μ$
^a31
üV

HW?è ?
|è÷ M$
[333]
m16:53.6.5m
17
i(0+) = i(0?) = 2001+20+1A = 10A
i(1) = 2201+1A = 110A
= 0:21+1s = 0:1s
?
^¤
i = [110+(10?110)]e?
t
0:1A = (110?100e?10t)A
?i = 30A
H
30 = 110?100e?10t
e?10t = 110?30100 = 0:8s
t = 110 ln 10:8s = 0:02s
üV0:02s?è ?T7|è÷ Mb
18
19