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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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41
2t 6t
3
i (t) 80 120 e 40 e A
−−
= − +
;
2t 6t
L
u (t) 180 e 60 e V
−−
= −
.
Solution
See problem 3.33.
Problem 3.38.
For the circuit shown in Fig. 3.38, it is known:
E 400V=
;
12
r r 10= =
;
L 0,6H=
;
3
C 10 F
−
=
;
C
u (0 ) 100V
−
=
.
Calculate the transient currents of the branches and the voltage on the coil.
Fig. 3.38
Answer:
200
t
50t
3
1
i (t) 20 10 e 10 e A
−
−
= + −
;
200
t
50t
3
2
i (t) 20 20 e 30 e A
−
−
= + −
;
200
t
50t
3
3
i (t) 10 e 20 e A
−
−
= − +
;
200
t
50t
3
L
u (t) 300 e 400 e V
−
−
= − +
.
Solution
See problem 3.33.
Problem 3.39.
For the circuit shown in Fig. 3.39, it is known:
U 120V=
;
r 2 3=
;
L 1,0H=
;
1
CF
12
=
;
C
u (0 ) 0V
−
=
.
Calculate branch transient currents and capacitor voltage.
42
Fig. 3.39
Answer:
( )
( )
3t
1
i (t) 40 e sin 3t 60 A
−
== +
;
( )
( )
3t
2
i (t) 40 e sin 3t A
−
=
;
( )
( )
3t
3
i (t) 40 e sin 3t 60 A
−
= − −
;
( )
( )
3t
С
u (t) 120 80 3 e sin 3t 60 V
−
= + −
.
Solution
See problem 3.33.
Problem 3.40.
For the circuit shown in Fig. 3.40, it is known:
J 3A=; r 1,5=
;
L 1,0H=
;
1
CF
12
=
.
Calculate the transient currents of the branches and the node voltage.
Fig. 3.40
Answer:
2t 6t
u(t) 9 e 9 e V
−−
= −
;
2t 6t
1
i (t) 6 e 6 e A
−−
= −
;
2t 6t
2
i (t) 3 4,5 e 1,5 e A
−−
= − +
;
2t 6t
3
i (t) 1,5 e 4,5 e A
−−
= − +
.
Fig. 3.40.1 shows the time diagrams of all transients.
Fig. 3.40.1
43
Solution
See problem 3.33.
Problem 3.41.
For the circuit shown in Fig. 3.41, it is known:
U 120V=
;
2t 6t
1
i (t) 80 90 e 10 e A
−−
= − +
.
Calculate circuit parameters r; L; C and transient currents
2
i (t)
;
3
i (t)
.
Fig. 3.41
Answer:
r 1,5=
;
L 1,0H=
;
1
CF
12
=
;
2t 6t
2
i (t) 80 120 e 40 e A
−−
= − +
;
2t 6t
3
i (t) 30 e 30 e A
−−
= −
.
Solution
It is recommended to apply the Vieta’s formulas (see problems 3.26 - 3.28) and also see
problem 3.33.
Problem 3.42.
For the circuit shown in Fig. 3.42, it is known:
U 60V=
;
2t 6t
1
i (t) 40 80 e 80 e A
−−
= − +
.
Calculate circuit parameters r; L; C and transient currents
2
i (t)
;
3
i (t)
.
44
Fig. 3.42
Answer:
r 1,5=
;
L 1,0H=
;
1
CF
12
=
;
2t 6t
2
i (t) 40 60 e 20 e A
−−
= − +
;
2t 6t
3
i (t) 20 e 60 e A
−−
= − +
.
Solution
It is recommended to apply the Vieta’s formulas (see problems 3.26 - 3.28) and also see
problem 3.33.
3.3.3 Circuits satisfying the condition of indifferent resonance of currents
A feature of the following tasks is the presence of a parallel section
r C/ /r L−−
in the
circuit, the parameters of which satisfy the condition of "indifferent" resonance of currents
L
r
C
= =
.
From the theory of AC circuits, it is known that the input resistance of such section is purely
resistive and equal , which allows equivalently replacing the parallel section for this resistance
. Since this is true for a circuit regime with sinusoidal current any frequency, such a replacement
will be equivalent for a circuit with current of any form, including in transient circuit regime, if the
initial conditions for C and L are zero.
Problem 3.43.
For the circuit shown in Fig. 3.43, it is known:
U 100V=; r 100=
;
L 0,1H=; C 10μF=
.
Calculate the current in the switch after switching.
Fig. 3.43
Answer:
i 1(A)=
.
Solution
There are zero independent initial conditions in the circuit.
The parameters of the elements of the circuit section with parallel branches r-L//r-C satisfy
the condition of indifferent resonance:
45
6
L 0,1
100 r
C 10 10
−
= = = =
.
The input resistance of the circuit segment with parallel branches r-L//r-C is purely resistive
and is equal
100 =
.
In a branch with a source of energy, steady-state regime of direct current comes right after
switching and the current in the key is
U
i 1A==
.
The desired current i can also be found by calculating the transient process.
In r-L and r-C branches, transients occur. Transient currents of branches:
( )
r
t
100t
L
1
U
i 1 e 1 1 e A
r
−
−
= − = −
;
t
100t
rC
2
U
i e 1 e A
r
−
−
= =
.
Then
12
i i i 1A= + =
.
Problem 3.44.
For the circuit shown in Fig. 3.44, it is known:
r 100=
;
L 0,1H=; C 10μF=
; coil voltage is constant:
L
u (t) 100V=
.
Calculate
C
i (t)
.
Fig. 3.44
Answer:
C
i (t) 1A=
.
Solution
The problem is solved using known relations from the theory of linear electrical circuits.
Solution plan:
1. Knowing u
L
(t) and the connection of the self-induction voltage with the current in the coil,
we find
LL
1
i (t) u (t) dt 1000t A
L
= =
taking into account that
L
i (0 ) 0A
+
=
.
2. This allows, according to the Ohm’s law and the 2nd Kirchhoff law, to find the voltage at
the input of the circuit
5
LL
u(t) r i (t) u (t) 100 10 t V= + = +
.
3. We shall take in account that the circuit parameters satisfy the condition of "indifferent"
resonance of currents
L
r 100
C
= = =
, and that in the circuit there are zero
46
independent initial conditions. So we can equivalently replace the circuit with its input
resistance and find the current at the input of the circuit according to the Ohm’s law
3
u(t)
i(t) 1 10 t А= = +
.
4. Finally, we find the current in the capacitor according to the 1st Kirchhoff law
CL
i (t) i(t) i (t) 1 А= − =
.
Problem 3.45.
For the circuit shown in Fig. 3.45, it is known:
12
r r r 100= = =
;
L 0,2H=
;
2
L 0,01H=
;
E 200V=
;
6
C 10 F
−
=
.
Calculate currents in the circuit branches.
Fig. 3.45
Answer:
3
r
t
10 t
L
E
i(t) 1 e 1 e A
r
+
−
−
= − = −
+
;
34
10 t 10 t
1
11
i (t) e e A
99
−−
= −
;
34
10 t 10 t
2
10 1
i (t) 1 e e A
99
−−
= − +
.
Solution
The circuit has zero independent initial conditions. Parameters of parallel section of circuit
1 2 2
r C/ /r L−−
satisfy condition of "indifferent" resonance of currents
2
12
L
r r 100
C
= = = =
. So we can do an equivalent replacement of the parallel section for
its input resistance (Fig. 3.45.1).
47
Fig.3.45.1
In this circuit, it is easy to find a transient current in an energy source branch:
3
r
t
10 t
L
E
i(t) 1 e 1 e A
r
+
−
−
= − = −
+
.
After constructing an operator image of the post-commutation circuit in Fig. 3.45, we can find
the operator image and then the original of the transient current in one of the branches of the
parallel section, for example
1
rC−
:
( ) ( )
2 2 2
1
34
12
1 2 2
Z (p) r pL
1 1 1000
I (p) I(p) A s
1
Z (p) Z (p) p p 1000
p 10 p 10
r r pL
pC
+
= = − =
++
+ +
+ + +
.
The original current in the first branch we find by the Laplace transformation pairs table:
( )
3 4 3 4
10 t 10 t 10 t 10 t
1
1000 1 1
i (t) e e e e A
9000 9 9
− − − −
= − = −
.
We find current in the second branch according to the 1st law of Kirchhoff:
34
10 t 10 t
21
10 1
i (t) i(t) i (t) 1 e e A
99
−−
= − = − +
.
Problem 3.46.
For the circuit shown in Fig. 3.46, it is known:
E 100V=
;
r 10=
;
L 10mH=
;
C 100μF=
.
Calculate
i(t)
.
Fig. 3.46
Answer:
( )
500t
i(t) 10 1 e A
−
= −
.
48
Solution
The circuit has zero independent initial conditions. Parameters of parallel section of circuit
1 2 2
r C/ /r L−−
satisfy condition of "indifferent" resonance of currents
2
L
r 10
C
= = =
.
So we can do an equivalent replacement of the parallel section for its input resistance (Fig.
3.46.1).
Fig. 3.46.1
Further calculation of the sought current is not difficult:
( )
( )
r
t
rL
500t
1
E
i(t) i (t) 1 e 10 1 e A
r
rr
r
−
+
−
= = − = −
+ +
+
.
Problem 3.47.
For the circuit shown in Fig. 3.47, it is known:
r 100=
;
L 0,1H=
;
1
L 0,01H=
;
6
C 10 F
−
=
;
E 200V=
.
Calculate:
1)
L
i (t)
;
2)
ab
u (t)
in 1ms after switching.
Fig. 3.47
Answer:
1000t
L
i (t) 2 e A
−
=
;
ab
u (0,001) 273,6 V
.
Solution
The independent initial conditions for C and L1 are zero, and the parameters of the parallel
section of the circuit
1
r C/ /r L−−
satisfy the condition of "indifferent" resonance of currents
49
1
L
r 100
C
= = =
. So we can do an equivalent replacement of the parallel section for its
input resistance (Fig. 3.47.1).
Fig. 3.47.1
Further calculation of the sought current and voltage is not difficult:
1.
t
1000t
L
L Lss Lf
E
i (t) i i e 2 e A
r
−
−
= + = =
.
2.
1000t
L
ab bd L
di (t)
u (t) E u (t) E u (t) E L 200 (1 e )V
dt
−
= − = − = − = +
.
1
ab
u (0,001) 200 (1 e ) 273,6 V
−
= +
.
Problem 3.48.
For the circuit shown in Fig. 3.48, it is known:
U 100V=
;
1 2 3
r r r 100= = =
;
14
L L 0,1H==
;
C 10μF=
.
Calculate transient currents of the circuit branches.
Fig. 3.48
Answer:
500t 1000t
1
i (t) 1 e 1 e А
−−
= −
;
500t 1000t
2
i (t) 0,5 e 1 e А
−−
=− +
;
500t
3
i (t) 1 0,5 e A
−
= −
;
500t
4
i (t) 1 1 e A
−
= −
.
Solution
50
The circuit has zero independent initial conditions. Parameters of parallel section of circuit
2 1 1
r C/ /r L−−
satisfy condition of "indifferent" resonance of currents
1
12
L
r r 100
C
= = = =
. So, we can do an equivalent replacement of the parallel section for
its input resistance (Fig. 3.48.1).
Fig. 3.48.1
1. Further calculation of the sought currents
3
i
;
4
i
is not difficult:
3
34
r
t
(r p) L
500t
4 4ss 4f
33
UU
i (t) i i 1 1 e A
rr
−
+
−
= + = − = −
;
4
500t
4
ab L 4
di (t)
u (t) u (t) L 50 e V
dt
−
= = =
;
500t
ab
12
u (t)
i (t) 0,5 e A
−
= =
;
500t
3 12 4
i (t) i (t) i (t) 1 0,5 e A
−
= + = −
.
2. Now let's go back to the original scheme and do a technique that can be called the
"separation" of the circuit, namely: we select and leave branches 1, 2, 4 and get the scheme
shown in Fig. 3.48.2.
Fig. 3.48.2
To calculate the transient process in this circuit, it is advisable to use the operator method
(Laplace transform method of circuit analysis). The equivalent operator circuit is shown in Fig.
3.48.3.
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