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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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21
( )
m
mC2
C2
t0
I jx sin 1000t arg U
••
=
= − + =
( )
m
mC2
C2
1 2 C2
t0
U
jx sin 1000t arg U
r r jx
•
•
=
= − + =
+−
( )
j90
mC2
t0
100 e
j10 sin 1000t arg U
10 j10
•
=
= − + =
−
( )
j45
t0
50 2 e sin 1000t 45 50V
=
= + =
.
4. Forced component.
The post-commutation circuit is sequential. Therefore, the same amount of electricity (i.e. the
charge) q will flow through the capacitors for a period of time from the moment of switching to the
steady state mode (in the absence of drains and sources of electric charges), i.e. the charge of the
capacitor C1 during this period of time will change by the final value -q and the charge of the
capacitor C2 will change by the final value + q (the signs are selected depending on the positive
directions of the current and voltage on the capacitors):
C1ss 1 C1ss C1
Q C u Q (0 ) q
+
= = −
;
C2ss 2 C2ss C2
Q C u Q (0 ) q
+
= = +
.
Capacitor charges right after switching can be found from the pre-commutation regime:
C1 C1 1 C1
Q (0 ) Q (0 ) C u (0 )
+−
= = −
;
C2 C2 2 C2
Q (0 ) Q (0 ) C u (0 )
+−
= = −
.
According to the 2nd law of Kirchhoff for a forced regime we can compose equation
C1ss C2ss
u u 0− + =
;
or
C1ss C2ss C1 C2
1 2 1 2
Q Q Q (0) q Q (0) q
0
C C C C
−+
− + = − + =
.
From here we calculate the value of q:
4
C1 C2
12
u (0 ) u (0 )
q 25 10 C
11
CC
−
−−
−
= =
+
.
Then the forced voltage components on the capacitors will be:
44
C1ss C1 1
C1ss
4
11
Q u (0 ) C q
100 10 25 10
u 75V
С C 10
−−
−
−
−
−
= = = =
;
44
C2ss C2 2
C2ss
4
22
Q u (0 ) C q
50 10 25 10
u 75V
С C 10
−−
−
−
+
+
= = = =
.
5.
C1 C1ss
A u (0 ) u 100 75 25V
+
= − = − =
;
2000t 1
C1
t 0,5ms
t 0,5ms
u (t) 75 25 e 75 25 e 84,2 V
−−
=
=
= + = +
.
Problem 3.20.
For the circuit shown in Fig. 3.20, it is known:
C1
u (0 ) 120V
−
=
;
C2
u (0 ) 0V
−
=
;
r 1k=
;
1
C 200 F=
;
2
C 100 F=
.
Calculate transient voltages
C1
u (t)
;
C2
u (t)
and current
i(t)
.
22
Fig. 3.20
Answer:
15t
C1
u (t) 80 40 e V
−
= +
;
15t
C2
u (t) 80 80 e V
−
= −
;
15t
i(t) 0,12 e A
−
=
.
Solution
Let’s use the classical method:
C1 C1ss C1f
u (t) u u=+
;
C2 C2ss C2f
u (t) u u=+
.
1. Root of characteristic polynomial
1 2 1 2
1 2 1 2
1 1 C C r p C C
Z(p) r 0
p C p C p C C
+ +
= + + = =
;
1
12
12
CC
p 15s
r C C
−
+
= − = −
;
pt
C1 C1ss 1
u (t) u A e= +
;
pt
C2 C2ss 2
u (t) u A e= +
.
2.
1 C1 C1ss
A u (0 ) u
+
=−
;
2 C2 C2ss
A u (0 ) u
+
=−
.
3. Independent initial conditions
C1 C1
u (0 ) u (0 ) 120V
+−
==
;
C2 C2
u (0 ) u (0 ) 0V
+−
==
.
4. Forced component
Following the explanations given in the previous problem, we define the amount of charge q
that has moved during the transient process from capacitor C1 to capacitor C
2
:
4
C1 C2
12
u (0 ) u (0 )
q 80 10 C
11
CC
−
−−
−
= =
+
.
Forced voltages on capacitors
64
1ss C1 1
C1ss
4
11
Q u (0 ) C q
120 200 10 80 10
u 80V
С C 200 10
−−
−
−
−
−
= = = =
.
4
2ss
C2ss
4
22
Q
q 80 10
u 80V
С C 100 10
−
−
= = = =
.
Thus
C1ss C2ss
u u 80V==
.
1 C1 C1ss
A u (0 ) u 120 80 40V
+
= − = − =
;
15t
C1
u (t) 80 40 e V
−
= +
.
23
2 C1 C1ss
A u (0 ) u 0 80 80V
+
= − = − = −
;
15t
C2
u (t) 80 80 e V
−
= −
.
5.
6 15t 15t
C2
2
du (t)
i(t) C 100 10 1200 e 0,12 e A
dt
− − −
= = =
.
Problem 3.21.
For the circuit shown in Fig. 3.21, it is known:
E 300V=
;
r 50=
;
12
r r 100= =
;
L 0,15H=; C 20 F=
.
Calculate:
1.
L
i (t);
2. Heat energy released in resistor r2 after switching
2
r heat
Q
.
Fig. 3.21
Answer:
1000t
L
i (t) 2 0,5 e A
−
= −
;
2
r heat
Q 0,225J=
.
Solution
Let’s use the classical method of transients calculation:
pt
L Lss Lf Lss
i (t) i i i A e= + = +
.
1. The root of the characteristic polynomial.
1
Z(p) pL r r 0= + + =
;
1
2
rr
p 1000s
L
−
+
= − = −
.
2.
L Lss
A i (0 ) i
+
=−
.
3. Independent initial conditions.
2
LL
12
12
12
E r 300 100
i (0 ) i (0 ) 1,5A
r r 100 100
r r 100 100
r 50
r r 100 100
+−
= = = =
++
++
++
;
C C r2 2 L 2
u (0 ) u (0 ) i (0 ) r i (0 ) r 1,5 100 150V
+ − − −
= = = = =
.
4. Forced component.
Lss
2
E 300
i2А
r r 50 100
= = =
++
;
L Lss
A i (0 ) i 1,5 2 0,5
+
= − = − = −
.
5.
1000t
L
i (t) 2 0,5 e A
−
= −
.
24
After switching, the capacitor is completely discharged through resistor r2. Therefore, after
switching the energy
2
r heat
Q
will dissipate in the resistor r2 in form of heat. That energy is equal
to the electric field energy WC accumulated in the capacitor by the time of switching. Thus,
2
2
62
C
r heat C
C u (0 )
20 10 150
Q W 0,225J
22
−
−
= = = =
.
25
3.3. Electrical circuits with two energy storage devices
3.3.1 Use of the Law of Conservation of Energy
Problem 3.22.
For the circuit shown in Fig. 3.22, it is known:
e 100 2 sin(1000t 45 )V= +
;
r 10=
;
L 0,01H=; C 100 F=
.
Calculate the highest values of current and voltage at the inductance after switching.
Fig. 3.22
Answer:
Lhigh
u 100 2 V=
;
Lhigh
i 10 2 A=
.
Solution
It can be seen that the circuit is in an ideal current resonance state before switching, therefore
there is no current in the e.m.f. source branch, and
LC
uue==
.
After switching at t=0
t0
, nothing will change in the circuit, in the LC-section of the
circuit there will be a continuous oscillatory process of exchange of electric energy
2
C
C
Cu
W
2
=
and magnetic energy
2
L
L
Li
W
2
=
between the energy storing devices. At the same time,
CL
u (t) u (t) e 100 2 sin(1000t 45 )V= = = +
;
L
11
i (t) e dt 100 2 sin(1000t 45 ) dt 10 2 cos(1000t 45 )A
L 0,01
= = + = +
.
Accordingly
Lнаиб
u 100 2 V=
;
Lнаиб
i 10 2 A=
.
The highest magnetic and electrical energies of the coil and capacitor will be
( )
2
2
Lhigh
М high L high
0.01 10 2
Li
W W 1,0J
22
= = = =
;
( )
2
6
2
Chigh
E high C high
100 10 100 2
Cu
W W 1,0J
22
−
= = = =
.
Electromagnetic energy of the circuit is constant after switching:
EM E M Emax E max
W W W W W 1,0J= + = = =
.
26
Problem 3.23.
For the circuit shown in Fig. 3.23, it is known:
e 100 2 sin(314t 45 )V= +
;
1
r L 10
C
= = =
.
Calculate the current i highest value after switching.
Fig. 3.23
Answer:
10 2 A
.
Solution
See task 3.22.
Problem 3.24.
For the circuit shown in Fig. 3.24, it is known:
e 100 2 sin( t 45 )V= +
;
1
r L 10
C
= = =
.
Calculate the capacitor current highest value after switching.
Fig. 3.24
Answer:
10 2 A
.
Solution
See task 3.22.
27
Problem 3.25.
For the circuit shown in Fig. 3.25, it is known:
e 380 2 sin( t 60 )V= +
;
L 120 =
;
1
600
C
=
.
Calculate the highest coil current value after switching.
Fig. 3.25
Answer:
Lhigh
i 2,83A
.
Solution
Due to the law of energy conservation, the electromagnetic energy
EМ
W
have stored by the
time of switching in the in LC-mesh without loss doesn’t change infinitely long after switching.
2
2
C
L
EМ EМ C L
C u (0 )
L i (0 )
W (t 0) W (0 ) W (0 ) W (0 )
22
−
−
− − −
= = + = +
.
After switching, the inductance coil and capacitor exchange magnetic and electrical energy
infinitely, respectively. Maximum values of electric
Cmax
W
and magnetic
Lmax
W
energies
coincide with circuit energy
EM Cmax Lmax
W W W==
.
The maximum value (it is also the largest value) of exchange magnetic energy is determined
by the highest value of coil current after switching:
2
Lhigh
Lmax
Li
W
2
=
. Then
Lmax
EМ
Lhigh
2W
2W
i
LL
==
;
where
2
2
C
L
EМ
C u (0 )
L i (0 )
W
22
−
−
=+
.
Calculation
Lhigh
i
C
3
u (0 ) e(0) 380 V
2
−
= =
.
L
L)
t 0 t 0
1 du 1 de
i (0 ) dt dt
L dt L dt
−−
−
==
= = =
( )
t0
1 1 19
380 2 cos t 60 A
L
62
−
=
= − + = −
.
28
2
2
C
L
22
CL
Lhigh
2 2 2 2
C L C L
C u (0 )
L i (0 )
C u (0 ) L i (0 )
22
i2
LL
CC
u (0 ) i (0 ) u (0 ) i (0 )
LL
−
−
−−
− − − −
+
+
= = =
= + = + =
2
2
1 3 19 1
380 2,83A
120 600 2 6 2
= + −
.
3.3.2 Calculation of circuit parameters and transient processes
Problem 3.26.
In the circuit shown in Fig. 3.26, it is known:
the transient current
( )
2t 6t
i(t) 2 e e A
−−
= −
;
the e.m.f. source
E 2V=
;
С
u (0 ) 0V
−
=
.
Calculate circuit parametersr; L; C.
Fig. 3.26
Answer:
r2=
;
L 0,25H=
;
1
CF
3
=
.
Solution
To calculate the inductance of the coil we use the initial conditions. Let's write down the 2nd
law of Kirchhoff at the moment of time
t0
+
=
:
LC
i(0 ) r u (0 ) u (0 ) E
+ + +
+ + =
;
here
i(0 ) 0A
+
=
;
( )
L
t0
di
u (0 ) L L 4 12 8 L
dt
+
+
=
= = − + =
;
C
u (0 ) 0V
+
=
.
Then
E
8 L E L 0,25(H)
8
= = =
.
We calculate the remaining parameters of the circuit using the roots of the characteristic
polynomial of the circuit.
29
1
Z(p) r pL 0
pC
= + + =
or
2
r1
p p 0
L L C
+ + =
;
1
1
p 2s
−
=−
;
1
2
p 6s
−
=−
.
By Vieta’s formulas
12
r
pp
L
+ = −
;
12
1
pp
LC
=
.
Then
( )
12
rV
p p 8 r 8 L 2
Ls
= − + = = =
;
( ) ( )
2
12
2
1 V 1 1
p p 2 6 12 C F
LC s 12 L 3
= = − − = = =
.
Problem 3.27.
In the circuit shown in Fig. 3.27, it is known:
the transient current
6t 2 t
i(t) 90 e 10 e A
−−
= −
;
the applied voltage
U 120V=
;
С
u (0 ) 0V
−
=
.
Calculate circuit parameters r; L; C.
Fig. 3.27
Answer:
r 1,5=
;
1
CF
12
=
;
L 1,0H=
.
Solution
We calculate resistance r from the initial conditions.
L
i (0 ) 0A
+
=
;
C
u (0 ) 0V
+
=
.
Let's write down the 1st and 2nd laws of Kirchhoff at the moment of time
t0
+
=
:
rL
i (0 ) i(0 ) i (0 ) i(0 ) 80A
+ + + +
= − = =
;
Cr
u (0 ) i (0 ) r U
++
+ =
;
r
U 120
r 1,5
i (0 ) 80
+
= = =
.
We calculate the remaining parameters of the circuit using the roots of the characteristic
polynomial of the circuit.
1
1
p 2s
−
=−
;
1
2
p 6s
−
=−
.
1 r pL
Z(p) 0
pC r pL
= + =
+
or
2
11
p p 0
r C L C
+ + =
.
30
By Vieta’s formulas
12
1
pp
rC
+ = −
;
12
1
pp
LC
=
.
Then
1V
8
r C s
=
;
11
CF
8 r 12
==
;
2
2
1V
12
L C s
=
;
1
L 1,0H
12 C
==
.
Problem 3.28.
For the circuit shown in Fig. 3.28, it is known:
the transient voltage
( )
2t 6t
u(t) 6 e e V
−−
= −
;
current of the current source
J 2A=
.
Calculate transient currents in the circuit branches.
Fig. 3.28
Answer:
( )
2t 6t
1
i (t) 4 e e A
−−
= −
;
2t 6t
2
i (t) 2 3 e e A
−−
= − +
;
6t 2t
3
i (t) 3 e e A
−−
= −
.
Solution
( )
2t 6t
1
u(t) 1
i (t) 6 e 6 e A
rr
−−
= = −
;
( )
2t 6t
C
3
du (t)
du(t)
i (t) C C C 12 e 36 e A
dt dt
−−
= = = − +
;
2 1 3
i (t) J i (t) i (t)A= − −
.
Calculation the circuit parameters r; C
We calculate the capacitance of the capacitor using initial conditions.
22
i (0 ) i (0 ) 0A
+−
==
;
1
u(0 ) 0
i (0 ) 0V
rr
+
+
= = =
.
Let's write down the 1st law of Kirchhoff at the moment of time
t0
+
=
:
3 1 2
i (0 ) J i (0 ) i (0 ) 2 0 0 2A
+ + +
= − − = − − =
.
Then
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