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Файл:Theoretical fundamentals of electrical engineering. P.II. Transients in linear electrical circuits with lamped parameters. A collection of advanced pr
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11
opp opp
2opp 2 opp 2
di di
u (t) r i (t) L M
dt dt
= + − =
50t 150t 150t
80 80 e (0,3 0,1) (200 e ) 80 40 e V
− − −
= − + − = +
.
Problem 3.8.
For the circuit shown in Fig. 3.8, it is known:
E 80V=
;
14
r r 20= =
;
23
r r 80= =
;
L 0,02H=
.
Calculate
1
i (t)
and
5
i (t)
.
Fig. 3.8
Answer:
1600t
1
i (t) 2 1,2 e A
−
= −
;
1600t
5
i (t) 1,5 1,5 e A
−
= −
.
Problem 3.9.
For the circuit shown in Fig. 3.9, it is known:
E 30V=
;
1
r 150 =
;
1
L 0,2H=
;
M 0,4H=
.
Calculate
1
i (t)
and
2
u (t)
.
Fig. 3.9
Answer:
750t
1
i (t) 0,2 0,2 e A
−
= −
;
750t
2
u (t) 60 e V
−
=
.
Solution
Current in the transformer primary winding:
12
1
1
r
t
L
750t
1 1ss 1f
11
EE
i (t) i i e 0,2 0,2 e A
rr
−
−
= + = − = −
.
Voltage of the secondary open winding of the transformer (taking into account the location of
the like clips of windings):
750t 750t
1
2
di (t)
u (t) M 0,4 750 0,2 e 60 e V
dt
−−
= = =
.
Problem 3.10.
For the circuit shown in Fig. 3.10, it is known:
E 15V=
;
1
r 10=
;
2
r5=
;
L 10H=
.
At the moment of time
t0=
the key closes, and then it opens in 1s. Calculate
L
i (t)
after
the first and the second switchings and the current
L
i (t)
value after 2s after the second switching.
Fig. 3.10
Answer:
0,5t
L
i (t) 3 3 e A; t [0;1];
−
= −
( )
1,5 t 1
L
i (t) 1,18 e A; t 1;
−−
=
L
i (3) 0,059A
.
Solution
Coil current after the first switching:
2
r
t
0,5t
L
L
22
EE
i (t) e 3 3 e A
rr
−
−
= − = −
;
0,51
L
i (1) 3 3 e 1,18A
−
= −
.
Coil current after the second switching:
12
rr
(t 1)
1,5(t 1)
L
LL
i (t) 0 i (1) e 1,18 e A
+
−−
−−
= + =
.
In 2s after the second switching
1,5 2 3
L
i (3) 1,18 e 1,18 e 0,059A
− −
= =
.
Problem 3.11.
For the circuit shown in Fig. 3.11, it is known:
J 4A=; r 10=
;
L 0,1H=
.
13
Calculate
L
i (t)
and current
L
i (t)
value in 0,005s after switching.
Fig. 3.11
Answer:
L
i (0,005) 1,264A=
Solution
Let’s use the classical method of transients calculation.
pt 200t
L Lss Lf
J
i (t) i i A e 2 A e A
2
−
= + = + = +
, where
1
2r
p 200 s
L
−
= − = −
.
L
i (0 ) 2 A 0 A 2
+
= + = = −
.
200t
L
i (t) 2 2 e A
−
= −
.
200 0,005 1
L
i (0,005) 2 2 e 2 2 e 1,264A
− −
= − = − =
.
Problem 3.12.
For the circuit shown in Fig. 3.12, it is known:
E 300V=; r 50=
;
12
r r 100= =
;
L 0,15H=; C 20 F=
.
Calculate:
1)
C
u (t)
;
2) Heat energy released in resistors r1 and r2 after switching
12
r r heat
Q
.
Fig. 3.12
Answer:
1000t
C
u (t) 300 150 e V
−
= −
;
12
r r heat
Q (t 0) 0,169J
.
Solution
1. The voltage on the capacitor we find by the classical method.
14
pt
C Css Cf
u (t) u u E A e= + = +
;
1
C
1
p 1000s ; A u (0 ) E
rC
−
+
= − = − = −
;
CC
12
12
E 300
u (0 ) u (0 ) E r i(0 ) E r 300 50 150V
rr
50 50
r
rr
+ − −
= = − = − = − =
+
+
+
;
1000t
C
u (t) 300 150 e V
−
= −
.
2. The heat energy released in resistors r1 and r2 after switching is the magnetic field energy
of inductance L accumulated by the coil by the time of switching, dissipated on these resistors in
the form of heat after switching, that is:
12
22
2
L
r r heat L
i(0 ) 3
i (0 )
22
Q (t 0) W (0 ) L 0,15 0,15 0,169J
2 2 2
−
−
−
= = = =
.
Problem 3.13.
For the circuit shown in Fig. 3.13, it is known:
1
j (t) 10 cos(100t)A=
;
2
j (t) 1A=
;
12
r r 3= =
;
3
r 1,5=
;
3
r 1,5=
;
L 0,03H=
;
5
C 10 F
−
=
.
Calculate
1
i (t)
after switching.
Fig. 3.13
Answer:
5
1
0t
Ai (t) 2,5 e
−
=
Solution
After switching at
t0
:
L
ab
1
11
d i (t)
u (t)
L
i (t)
r r dt
= − = −
.
( )
12
1
2
rr
t
L r r
pLt 50t
2L ss Lf
j A e 1 A e 1 A e Ai (t) i i
−
+
−
=+ −= − + = + = − +
.
LL
A i (0 ) 1 i (0 ) 1
+−
= + = +
.
We will calculate the current
L
i (0 )
−
using the superposition method:
L L L
i (0 ) i (0 ) i (0 )
− − −
=+
.
1. We will calculate a partial current
L
i (0 )
−
from the current source j1 by a symbolic
method.
15
At
t0
:
( )
j90 j45
2
Lm 1m
2
r
3
I J 10 e 5 2 e A
r j L 3 j3
= = =
+ +
;
L
i (t) 5 2 sin(100t 45 )A
= +
;
L
i (0 ) 5 2 sin(45 ) 5A
−
= =
.
2. Partial current
L
i (0 )
−
from current source j2:
L2
i (0 ) j 1A
−
= − = −
.
So,
L
i (0 ) 5 1 4A
−
= − =
.
Finally we get:
LL
A i (0 ) 1 i (0 ) 1 5
+−
= + = + =
;
5
L
0t
Ai 1e( 5t)
−
−= +
.
( )
( )
50t
50t 5
1
0t
d
0,03
i A1
15
(t) 0,0 250 2 5
e
ee,
3 dt
−
−−
= − = − − =
−+
.
Problem 3.14.
For the circuit shown in Fig. 3.14, it is known:
M 0,5 L=
.
Calculate the mutual-inductance e.m.f. right after switching
M
e (0 )
+
.
Fig. 3.14
Answer:
M
E
e (0 ) V
9
+
=−
.
Solution
LL
M
di di
e (t) M 0,5 L
dt dt
= − = −
.
At
t0
:
L Lss Lf
i (t) i i=+
.
The scheme of the postcommutation circuit is shown in Fig. 3.14a.
16
Fig. 3.14a
Analysis of the schemes in Fig. 3.14 and Fig. 3.14a allows us to get:
Lss
i
E
r
=
;
t
Lf
p
Ai e=
;
1
r p L r
r 2 0 p s
r p L 3 L
−
+ = = −
+
;
L Lss L Lss
1 E E 2
A i (0 ) i i (0 ) i E
rr
2 r 3 r
r
rr
+−
= − = − = − = −
+
+
;
r
L
t
3L
A
2
i
r
E
e
r
(t) E
3
−
=−
.
Then
L
r
t
3L
M
2
dE
3r
di
e (t) 0,5 L 0,5 L
d
E
e
r
t dt
−
= −
−
= − =
rr
tt
3 L 3 L
r
e e V
21
0,5 L E E
3r L 93
−−
= − − − = −
;
M
E
e (0 ) V
9
+
=−
.
17
3.2. Electrical circuits with capacitors
Problem 3.15.
Capacitor with capacitance
C 0,02 F=
has imperfect insulation. Measurements showed
that the voltage of the charged capacitor decreases by 2 times during
t 140s=
. The voltage was
measured by a voltmeter V of an electrostatic system with almost ideal insulation. Calculate the
capacitor insulation resistance r and the time constant of the capacitor τ (Fig. 3.15).
Fig. 3.15
Answer:
4
r 10 M
;
200s=
.
Solution
Discharge of the capacitor through the resistance is a transient process that is described by the
function
t
rC
cc
u (t) u (0) e
−
=
. It follows from the task conditions that
cc
1
u (t t) u (t)
2
+ =
, or
t t t
r C r C
cc
1
u (0) e u (0) e
2
+
−−
=
. Hence it follows:
4
8
t t 140
ln2 r 10 M
r C C ln2 2 10 ln2
−
= = =
.
Capacitor time constant
10 8
r C 10 2 10 200s
−
= = =
.
Problem 3.16.
For the circuit shown in Fig. 3.16, it is known:
E 150V=
;
1
r 75=
;
2
r 25=
;
C 0,04F=
.
At the moment of time,
t0=
the key closes, and in one second it opens.
Calculate
C
u (t)
and the value
C
u
in two seconds after the second switching.
18
Fig. 3.16
Answer:
t
C
u (t) 150 150 еV
−
= −
;
C
u (3) 57,51V
.
Solution
After closing the key at a time
t0=
:
2
t
rC
C
u (t) E E е
−
= − =
t
150 150 еV
−
−
.
After opening the key at a time
t 1s=
:
( )
12
t1
t1
r r C
4
CC
u (t) u (1) е 94,818 е V
−
−
−
−
+
=
.
Two seconds after opening the key
2
4
C
u (3) 94,818 е 57,51V
−
=
.
Problem 3.17.
For the circuit shown in Fig. 3.17, it is known:
1
E 150V=
;
2
E 50V=
;
1
r5=
;
2
r 10=
;
r 100=
;
C 100 F=
.
At the moment of time
t0=
the key is moved from position 1 to position 2.
Calculate
C
u (t)
;
i(t)
.
Fig. 3.17
Answer:
100t
C
u (t) 50 150e V
−
=−
;
100t
i(t) 1,5 e A
−
= −
.
Solution
Before switching the key, the steady-state regime circuits are considered. Then we have after
switching the key:
19
( )
t
rC
C 2 C 2
u (t) E u (0 ) E е
−
+
= + − =
t
100t
1
rC
2 2 2
12
E
E r E е 50 150e V
rr
−
−
= + − − = −
+
.
( )
6 3 100t 100t
C
d
i(t) C u (t) 100 10 15 10 e 1,5 e A
dt
− − −
= − = − = −
.
Problem 3.18.
For the circuit shown in Fig. 3.18, it is known:
J 20A=
;
1
r4=
;
23
r r 5= =
;
4
r 15=
;
C 100 F=
.
Calculate transient currents
1
i (t)
;
2
i (t)
;
3
i (t)
;
4
i (t)
;
C
i (t)
.
Fig. 3.18
Answer:
1
i (t) J 20A==
;
1667t
2
i (t) 16 4 e A
−
= +
;
1667t
3
i (t) 4 4 e A
−
= −
;
1667t
4
i (t) 4 2,67 e A
−
+
;
1667t
C
i (t) 6,67 e A
−
−
.
Solution
1.
1
i (t) J 20A==
.
2.
C
C
du (t)
i (t) C
dt
=
.
( )
2 3 4
2 3 4
1
t
r r r
C
r r r
pt
2
C Css Cf 4ss 4 4
234
r
u (t) u u i r A e J r A e
r r r
−
+
++
= + = + = +
++
1667t
60 A e V
−
+
;
where
C Css 2
A u (0) u J r 60 40 V= − = − =
;
so
1667t
C
u (t) 60 40 e V
−
= +
;
then
20
( )
6 1667t 1667t
C
C
du (t)
i (t) C 100 10 40 1667 e 6,67 e A
dt
− − −
= = − −
.
3.
1667t
C
4
4
u (t)
i (t) 4 2,67 e A
r
−
= +
.
4.
1667t 1667 t 1667t
3 C 4
i (t) i (t) i (t) 6,67 e 4 2,67 e 4 4 e A
− − −
= + = − + + = −
.
5.
1667t
23
i (t) J i (t) 16 4 e A
−
= − = +
.
Problem 3.19.
For the circuit shown in Fig. 3.19, it is known:
( )
u 100 sin 1000t 90 V= +
;
12
r r 5= =
;
12
C C 100 F= =
.
Calculate
C1
u
in 0,5ms after switching.
Fig. 3.19
Answer:
C1
t 0,5ms
u (t) 84,2V
=
.
Solution
Let’s use the classical method:
C1 C1ss C1f
u (t) u u=+
.
1. The root of the characteristic polynomial.
( )
1 2 1 2 1 2
12
1 2 1 2
C C r r p C C
11
Z(p) r r
p C p C p C C
+ + +
= + + + =
;
( )
1
12
1 2 1 2
CC
p 2000s
r r C C
−
+
= − = −
+
;
pt
C1 C1ss
u (t) u A e= +
.
2.
C1 C1ss
A u (0 ) u
+
=−
.
3. Independent initial conditions.
C1 C1
u (0 ) u (0 ) 100V
+−
==
.
The initial condition for the second capacitor we define from the pre-commutation
steady-state regime of AC circuit with symbolic method.
mC2 mC2
C2 C2
t0
u (0 ) u (0 ) U sin 1000t arg U
••
+−
=
= = + =
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