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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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51
Fig. 3.48.3
In this scheme
ab
50
U (p) V s
p 500
=
+
, and the remaining energy sources take zero values.
Now we can get operator images
1
I (p)
;
2
I (p)
of the currents sought and their originals
1
i
;
2
i
:
( ) ( )
o
500t 1000t
ab 1 1
11
o
11
U (p) L i (0)
500
I (p) A s 1 e 1 e A i (t)
r pL p 500 p 1000
−−
+
= = = − =
+ + +
;
( ) ( )
C
ab
o
500t 1000t
22
o
1
u (0)
U (p)
0,5p
p
I (p) A s 0,5 e 1 e A i (t)
1
p 500 p 1000
r
pC
−−
−
= = = − + =
+ +
+
.
Problem 3.49.
For the circuit shown in Fig. 3.49, it is known:
J 1A=; r 100=
;
L 0,1H=; C 10μF=
.
Calculate
u(t)
;
L
u (t)
after switching.
Fig. 3.49
Answer:
u(t) 100V=
;
1000t
L
u (t) 100 e V
−
=
.
Solution
The circuit has zero independent initial conditions. Parameters of parallel section of circuit
r C/ /r L−−
satisfy condition of "indifferent" resonance of currents
L
r 100
C
= = =
. So
we can do an equivalent replacement of the parallel section for its input resistance (Fig. 3.49.1)
52
Fig. 3.49.1
The steady-state regime of direct current comes right after switching in the branch with a
source of energy and the voltage on the parallel section immediately assumes constant value
u(t) J 100V= =
.
Return to the initial diagram (Fig. 3.49) in post-switching regime. In the r-L branch of
post-switching circuit (Fig. 3.49), a transient occurs when switched on to a constant voltage u (t)
and the current in the branch is equal
r
t
1000t
L
L
U
i (t) 1 e 1 1 e A
r
−
−
= − = −
.
Then coil voltage is:
1000t 1000t
L
L
di (t)
u (t) L 0,1 1000 e 100 e V
dt
−−
= = =
.
Problem 3.50.
For the circuit shown in Fig. 3.50, it is known:
U 200V=
;
r 10=
;
L 0,01H=
;
C 100μF=
.
Calculate in the simplest way
i(t)
;
cd
u (t)
after switching.
Fig. 3.50
Answer:
i(t) 10A=
;
cd
u (t) 0 V=
.
Solution
The circuit has zero independent initial conditions. Parameters of parallel section of circuit
r C/ /r L−−
satisfy condition of "indifferent" resonance of currents
L
r 10
C
= = =
. So
we can do an equivalent replacement of the parallel section for its input resistance (Fig. 3.50.1).
53
Fig. 3.50.1
1. In this circuit (Fig. 3.50a) the current sought immediately takes a steady-state value
U
i(t) 10A
r
==
+
after switching. The voltage in section ab will also immediately take a
steady-state constant value
ab
u (t) i(t) 100V= =
.
2. Then, returning to the initial circuit in the post-switching regime (Fig. 3.50), we calculate
the transient currents in the parallel section when switching on for DC voltage
ab
U 100V=
by
any method:
( )
r
t
1000t
ab
L
L
U
i (t) 1 e 10 1 e A
r
−
−
= − = −
;
1
t
1000t
ab
rC
C
U
i (t) e 10 e A
r
−
−
= =
.
The voltage sought
cd
u (t)
is easily calculated from the potential diagram a-c-d-b (ideally),
assuming the point b potential to be 0V:
cd c d
u (t) = − =
1000t 1000t
LC
100 i (t) r 0 i (t) r 100 100 100 e 100 e 0V
−−
= − − + = − + − =
.
54
3.4. Ill-posed (incorrect) problems
In this section, we consider methods for calculating transients for such switchings, in which
the 1st or 2nd switching laws are “as if” violated, if we assume that the switching itself is ideal, i.e.
instantaneous.
In such problems, the entire transient process can be divided into two stages [5]: "fast
transition" (FT) and "slow transition" (ST). The switching process is as if "stretched" from time t =
0- (this is the time immediately before the start of the FT) to time t = 0
+
(this is the time of the end
of the RT and the start of the ST). During this period of time (FT) in the electric circuit there is a
very fast, but not instantaneous, redistribution of currents in branches with inductance coils or
redistribution of electric charges between capacitors in accordance with the generalized 1st (for
magnetic flux linkage) or 2nd (for electric charges) switching laws, respectively. Generalized
switching laws can be formulated as follows:
1st generalized switching law
In any mesh of the post-switching electrical circuit the algebraic sum Ψ of magnetic flux
linkages of coils does not change during switching: Ψ(0
+
) = Ψ(0-).
2nd generalized switching law
For any node of the post-switching electrical circuit, the total electric charge of capacitors q
in the branches connected by the node does not change during switching, i.e. q(0
+
) = q(0-).
The 1st and 2nd generalized switching laws are used to calculate independent initial
conditions for time t = 0
+
:
iL (0 +) and uC (0 +).
After the end of the FT, the "normal" transient process (ST) begins, the calculation of which is
easy to carry out by classic method.
The operator method (Laplace transform method of circuit analysis) allows us to
calculate the transient process for incorrect switchings, as if "not noticing" the FT stage, and to
avoid using generalized switching laws. When constructing an equivalent operator circuit, we must
use independent initial conditions for the moment of time t=0
-
:
iL(0-) and uC(0-).
Problem 3.51.
For the circuit shown in Fig. 3.51, it is known:
J 6A=
;
1
r2=
;
1
L 0,4H=
;
2
r4=
;
2
L 0,2H=
.
Calculate
1
i (t)
;
2
i (t)
.
Fig. 3.51
Answer:
10t
1
i (t) 4 2 e A
−
= −
;
10t
2
i (t) 2 2 e A
−
= +
.
55
Solution
Classic method.
To calculate independent initial conditions
1
i (0 )
+
,
2
i (0 )
+
at
t0
+
=
, we apply the 1st
generalized switching law
(0 ) (0 )
+−
=
and the 1st Kirchhoff’s law:
1 1 2 2
12
L i (0 ) L i (0 ) 0;
i (0 ) i (0 ) J 6.
++
++
− =
+ = =
From here we get:
12
i (0 ) 2A,i (0 ) 4A
++
==
. Further, the calculation of the ST stage by the
classical method is not difficult.
12
12
rr
t
LL
10t
2
1 1ss 1f 1 1
12
r
i (t) i i J A e 4 A e A
rr
+
−
+
−
= + = + = +
+
.
1 1 1
i (0 ) 2 4 A A 2
+
= = + = −
.
10t
1
i (t) 4 2 e A
−
= −
.
( )
10t 10t
21
i (t) J i (t) 6 4 2 e 2 2 e A
−−
= − = − − = +
.
Operator method
We apply the operator method as usual. The equivalent operator circuit is shown in Fig.
3.51.1 (that scheme is built taking into account the fact that when
t0
−
=
(see the preamble to
section 3.4) independent initial conditions take zero values:
12
i (0 ) i (0 ) 0A
−−
==
.
Fig. 3.51.1
Let’s write down an operator image of the current in the 1st branch and find its original:
( )
22
1
1 1 2 2
J r pL 6 4 0,2p 40 2p
I (p)
p r pL r pL p 2 0,4p 4 0,2p p p 10
+ + +
= = = =
+ + + + + + +
( )
( )
o
10t 10t 10t
1
o
10 1
4 2 A s 4 1 e 2 e 4 2 e A i (t)
p p 10 p 10
− − −
= + = − + = − =
+ +
.
Then
21
i (t) J i (t)= − =
10t
2 2 e A
−
+
.
We check the solution fidelity according to the 2nd law of Kirchhoff, because this law was
not used in the above calculations:
( )
10t 10t
12
1 1 1 2 2 2
di di
r i L r i L 8 4 e 8 4 e 0
dt dt
−−
+ − + = + − + =
.
REMARK:
This case of transient refers to incorrect switching when de-energized parallel branches
kk
r – L
(
k 1,n=
) are connecting to an ideal direct current source. Analysis of the obtained solution
shows that at the first moment after switching, i.e. at t = 0
+
, the current of the current source is
56
distributed over the branches
kk
r – L
inversely in proportion to the inductances of the branches
L
k
. In the steady state mode, of course, the current of the current source is distributed along the
kk
r – L
k
of branches.
Problem 3.52.
For the circuit shown in Fig. 3.52, it is known:
E 100V=
;
1
r 10=
;
2
r2=
;
12
L L 0,2H==
;
M 0,1H=
.
Calculate the transformer secondary winding current.
Fig. 3.52
Answer:
10t
2
i (t) 5 e A
−
= −
.
Solution
In this problem, incorrectness is related to disconnecting from the power source of the
inductive branches with non-zero currents.
Classic method.
To calculate independent initial condition
2
i (0 )
+
at
t0
+
=
, we apply the 1st generalized
switching law
(0 ) (0 )
+−
=
for the second winding mesh:
2 2 1 2 2 1
L i (0 ) M i (0 ) L i (0 ) M i (0 )
+ + − −
− = −
.
It’s clear that before switching at
t0
−
=
the following is true:
1
1
E
i (0 ) 10A
r
−
==
;
2
i (0 ) 0A
−
=
.
Certainly, after FT before ST at
t0
+
=
1
i (0 ) 0A
+
=
.
Then we have
22
1
E
L i (0 ) M
r
+
= −
;
2
21
ME
i (0 ) 5A
Lr
+
= − = −
.
When compiling the characteristic polynomial
z(p)
, it should be borne in mind that during
the time of ST
t0
+
the current in the primary winding is zero
1
i (t) 0A=
and the mutual
inductance with the primary winding does not affect the processes in the secondary winding:
57
22
z(p) r pL 0= + =
;
1
2
2
r
p 10s
L
−
= − = −
.
Then we get:
10t 10 t
2 2ss 2f
i (t) i (t) i (t) 0 5 e 5 e A
−−
= + = − = −
.
Problem 3.53.
For the circuit shown in Fig. 3.53, it is known:
U 60V=
;
12
r r 3= =
;
1
L 0,1H=
;
2
L 0,2H=
;
M 0,1H=
.
Calculate the current
1 2 12
i (t) i (t) i (t)==
after switching.
Fig. 3.53
Answer:
12t
12
i (t) 10 0,4 e A
−
= −
.
Solution
In this problem, incorrectness is associated with different values of currents in the coils at the
moment
t0
−
=
right before they are connected in serial.
Classic method.
To calculate independent initial condition
12
i (0 )
+
at
t0
+
=
, we apply the 1st generalized
switching law
(0 ) (0 )
+−
=
for the mesh
1 1 2 2
r L r L U− − − −
:
1 2 12
+ + +
==
;
1
i (0 ) 12A
−
=
;
2
i (0 ) 8A
−
=
.
( )
1 2 12 1 1 2 2 2 1
L L 2M i (0 ) L i (0 ) M i (0 ) L i (0 ) M i (0 )
+ − − − −
+ + = + + +
.
( )
1 1 2 2 1 2
12
12
L i (0 ) L i (0 ) M i (0 ) i (0 )
i (0 ) 9,6A
L L 2M
− − − −
+
+ + +
==
++
.
Further, the calculation of the ST stage by the classical method is not difficult.
12
12
rr
t
L L 2M
12t
12 12ss 12f 1
12
U
i (t) i i A e 10 0,4 e A
rr
+
−
++
−
= + = + = −
+
.
Operator method
We apply the operator method as usual, but taking into account the recommendations from the
preamble to section 3.4. The equivalent operator circuit after switching is shown in Fig. 3.53.1.
58
Fig. 3.53.1
( )
1 1 2 2 2 1
12
1 2 1 2
U 60
L i (0 ) L i (0 ) M i (0 ) i (0 ) 4,8
pp
I (p)
r r pL pL 2pM 0,5 p 12
− − − −
+ + + + +
= = =
+ + + + +
( ) ( )
( )
o
12t 12t 12t
12
o
120 9,6
A s 10 1 e 9,6 e 10 0,4 e A i (t)
p p 12 p 12
− − −
= + = − + = − =
+ +
.
Problem 3.54.
At what ratio of the element parameters in the circuit shown in Fig. 3.54, the forced regime is
set immediately after switching?
Fig. 3.54
Answer:
1 1 2 2
r C r C =
.
Solution
In this problem, the incorrectness is related to the different voltage values of the capacitors at
the time
t0
−
=
right before they are connected in, as shown in Fig. 3.54.1 (by default, we consider
the second capacitor to be discharged before switching).
Fig. 3.54.1
59
The condition of the problem will be fulfilled if there would be no ST stage after switching,
i.e.
C12 C12ss
u (0 ) u+=
.
The forced (steady-state) component of capacitors voltage is
C12ss 2
12
E
ur
rr
=
+
.
To calculate independent initial condition
C12
u (0 )
+
at
t0
+
=
, we apply the 1st generalized
switching law
q(0 ) q(0 )
+−
=
for the lower node of the circuit, taking into account that
at
t0
+
=
:
C1 C2 C12
u (0 ) u (0 ) u (0 )
+ + +
==
;
at
t0
+
=
:
C1
u (0 ) E
−
=
;
C2
u (0 ) 0V
−
=
.
Let’s write down the 1st generalized switching law
( )
1 2 C12 1 C1 2 C2
C C u (0 ) C u (0 ) C u (0 )
+ − −
+ = +
.
Then
1 C1
1
C12
1 2 1 2
C u (0 )
C
u (0 ) E
C C C C
−
+
= =
++
.
1
2
1 2 1 2
CE
Er
C C r r
=
++
, from here follows
1 1 2 2
r C r C =
.
Problem 3.55.
For the circuit shown in Fig. 3.55, it is known:
E 60V=
;
3
r 10=
;
1
C 100μF=
;
2
C 200μF=
.
Calculate voltage on capacitors after switching.
Fig. 3.55
Answer:
3,33t
C12
u (t) 60 40 e V
−
= −
.
Solution
In this problem, incorrectness is associated with different values of the voltages of the
capacitors at the moment
t0
−
=
right before they are connected in parallel (by default, we
consider the second capacitor to be discharged before switching).
Classical method
It’s clear that:
60
at
t0
+
=
:
C1 C2 C12
u (0 ) u (0 ) u (0 )
+ + +
==
;
and at
t0
−
=
:
C1
u (0 ) E 60V
−
==
;
C2
u (0 ) 0V
−
=
.
To calculate an independent initial condition
C12
u (0 )
+
at
t0
+
=
, we apply the 2nd
generalized switching law
q(0 ) q(0 )
+−
=
for the lower node of the circuit.
( )
1 2 C12 1 C1 2 C2
C C u (0 ) C u (0 ) C u (0 )
+ − −
+ = +
.
Then
6
1 C1
1
C12
6
1 2 1 2
C u (0 )
C 100 10
u (0 ) E 60 20V
C C C C 300 10
−
−
+
−
= = = =
+ +
.
Further, the calculation of the ST stage by the classical method is not difficult:
12
r
t
CC
3,33t
C12 C12ss C12f
u (t) u (t) u (t) E A e 60 40 e V
−
+
−
= + = + = −
.
Operator method
We apply the operator method as usual, but taking into account the recommendations from the
preamble to section 3.4. The equivalent operator circuit after switching is shown in Fig. 3.53.1.
Fig. 3.55.1
C1
1
o
C12 ab
o
12
u (0 )
1
E(p) pC
60 3,33 20
rp
U (p) U (p) ... V s
1
p (p 3,33) p 3,33
pC pC
r
−
+
= = = = + =
+ +
++
( )
o
3,33t 3,33t 3,33t
C12
o
60 1 e 20 e 60 40 e V u (t)
− − −
= − + = − =
.
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