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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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MINISTRY OF TRANSPORT OF THE RUSSIAN FEDERATION
FEDERAL STATE AUTONOMOUS
EDUCATIONAL INSTITUTION OF HIGHER EDUCATION
"RUSSIAN UNIVERSITY OF TRANSPORT"
Department of power supply on electric railways
THEORETICAL FUNDAMENTALS OF ELECTRICAL ENGINEERING
PART II
TRANSIENTS IN LINEAR ELECTRICAL CIRCUITS WITH LAMPED PARAMETERS
PROBLEM BOOK
Advanced problems
For students of the specialty 23.05.05
"Train support systems"
programs
"Railway Power supply"
"Telecommunications and railway transport networks"
"Automation and telemechanics in railway transport"
Moscow - 2023
UDK 621.3
V 82
Theoretical fundamentals of electrical engineering. Part II. Transients in linear electrical
circuits with lamped parameters: A collection of advanced problems for students of electrical
specialties and specializations of the university : Vlasov S.P., Volyntsev V.V., Kruchinin E.V. M.: RUT (MIIT), 2023– 75 pages.
This edition presents the second part of the collection of advanced problems on the
theoretical fundamentals of electrical engineering (TFEE). The first part [8] contains advanced
problems on linear DC and AC electrical circuits. The second part of this collection contains
tasks on transients in linear electrical circuits with lamped parameters. Most of the tasks of the
collection are characterized by increased complexity, unusual posing of the question,
extraordinary, non-trivial methods of solving, surprise of the result.
The collection of tasks is intended for independent in-depth work of students on the TFEE
course; it can also be useful to teachers in organizing the preparation of the most successful
students for participation in the TFEE Olympiads.
All comments and wishes on the content of the manual should be sent to the authors at the
Department of Electric Power and Transport of the RUT (MIIT).
© RUT (MIIT), 2023
3
PREFACE
This collection of problems of increased complexity on TFEE is the next (2nd) part of the
collection. The 1st part is devoted to the problems for DC and AC electrical circuits in steady
regimes [8]. This collection contains the problems on calculation of transients in linear electrical
circuits with lamped parameters.
The basis of the collection is the of the TFEE Olympiads at MIIT, compiled by Associate
Professor V.F. Klimov. Some problems are borrowed from common TFEE problem books edited by
K. M. Polivanov [1], P. A. Ionkin [2], M. R. Shebes [3], from the problem book of programmed
tasks according to the TFEE of Lviv University [4], as well as from problem books of MIIT on the
TFEE [5, 7] and from the Olympic problem book MPEI [6].
A feature of the task is the presence of problems equipped with detailed solutions, however, it
also has problems designed for independent solution. Such tasks are provided with answers.
Various methods of solving are used: classical, operator, Duhamel integral method, so-called
"ill-posed" problems are also considered.
The authors believe that the proposed collection of tasks will be useful for students for
independent work in TFEE, as well as for teachers when organizing students' classes in preparation
for Olympiads on the theoretical fundamentals of electrical engineering.
Authors.
4
Further:
1. In problems, unless otherwise specified, we consider circuits at steady-state regime
before switching, with uncharged capacitors if there is any.
2. international designations of physical quantities will be used in the text and figures.
Arbitrary SI units having special designation names (GOST 8_417-2002) are given in
the table below.
Table.
Unit
Expression via
main and
derivative SI
units
Name
Dimension
Name
Designation
Inter-pe
ople
the
Russian
Flat angle
1
radian
rad
рад
m·m
–1
=1
Frequency
T
-1
hertz
Hz
Гц
s–1
Energy, work, quantity of
heat
L2MT
-2
joule
J
Дж
m
2
·kg·s
–2
Power
L2MT
-3
watt
W
Вт
m
2
·kg·s
–3
Electric charge, quantity of
electricity
TI
coulomb
C
Кл
s·A
Electric current strength
I
ampere
A А
Electrical voltage, electrical
potential, electrical potential
difference, electromotive
force (e.m.f.)
L2MT
-3I -1
volt
V В m
2
·kg·s–3·A
–1
Electric capacity
L–2M
-1T 4I2
farad
F Ф m
–2
·kg–1·s4·A
2
Electrical resistance
L2MT
-3I -2
ohm
Ω
Ом
m
2
·kg·s–3·A
–2
Electrical conductivity
L–2M
-1T 3I2
siemens
S
См
m
–2
·kg–1·s3·A
2
Magnetic induction flux,
magnetic flux
L2MT
-2I -1
weber
Wb
Вб
m
2
·kg·s–2·A
–1
Magnetic flux density,
magnetic induction
MT
-2I -1
tesla
T
Тл
kg·s–2·A
–1
Inductance, mutual
inductance
L2MT
-2I -2
henry
H
Гн
m
2
·kg·s–2·A
–2
5
3. TRANSIENTS IN LINEAR ELECTRICAL CIRCUITS WITH LAMPED
PARAMETERS
3.1. Electrical circuits with inductive coils
Problem 3.1.
The coil winding resistance of powerful electric machine can be determined at direct current
by the ammeter/voltmeter method according to the scheme shown in Fig. 3.1.
During measurements, the readings
I 10 A=
;
mV
U 210mV=
were recorded.
Millivoltmeter resistance is
mV
r 1 k=
and
mV 0
rr
, so we can neglect the current through a
millivoltmeter in calculations. The coil resistance of the winding turned out to be
mV
0
U
0,21
r 0,021
I 10
= = =
.
In what sequence and why in that sequence should the experimental circuit be disassembled
after performing the measurements?
Fig. 3.1
Answer:
1. Millivoltmeter disconnecting.
2. Key K2 switching from position 1 to position 2.
3. Key K1 opening.
Solution
We should note that:
1. . Before opening key K1, we shall disconnect the millivoltmeter. If you open the key K1 in
the presence of a millivoltmeter in the circuit (we assume t = 0 when the key K1 opens), the current
in the winding
( ) ( )
i 0 I i 0
+−
==
will close through the millivoltmeter and the voltage on it will
reach a value
3
mV mV
U r i(0 ) 10 10 10 kV
+
= − = =
that can damage the millivoltmeter.
2. Opening the key K1 without millivoltmeter can lead to a breakdown in the insulation of the
winding of the tested electrical machine: an electric arc (incorrect switching) will burn for some
(short) time between the contacts of the key K1, the winding current will rapidly decrease, which
will lead to a significant value of self-induction e.m.f.
L
di
eL
dt
= −
, which can lead to a
breakdown in the insulation of the winding wires.
6
3. In order to avoid the negative consequences noted above in pp. 1 and 2, the circuit shall
include selector key K2 and discharge resistor rd. The value of the discharge resistor shall be such
that the voltage on the winding of the electric machine after turning off the millivoltmeter and
switching over the key K2 from position 1 to position 2 does not exceed the allowable value for the
winding.
d al
r i(0 ) U
+
−
.Then the key K1 may be opened.
So, the disassembling order for the measuring circuit must be as follows:
1. Millivoltmeter disconnecting.
2. Key K2 switching from position 1 to position 2.
3. Key K1 opening.
Problem 3.2.
To accelerate the process of reducing the current in the exciting winding of an electric
machine with parameters L; r the winding is connected without breaking the circuit to the resistor r1
(Fig. 3.2). It is known:
E 50V=; L 0,5H=; r1=
;
1
r 10=
.
Calculate the winding current i and voltage u.
Calculate the required time t1 for current i reducing by 50 times compared to the initial value.
What will be the time t1 with a smaller resistance value
1
r 0,1=
?
Fig. 3.2
Answer:
22t
i(t) 50 e A
−
=
;
22t
u(t) 500 e V
−
= −
;
1
t 0,178s
;
1
t 1,78s
.
Solution
Current in inductance before switching:
E 50
i(0 ) 50A i(0 )
r1
−+
= = = =
.
Differential equation after switching:
1
di
(r r ) i L 0
dt
+ + =
.
Characteristic equation:
1
pL r r 0+ + =
,
from where
1
1
r r 11
p 22s
L 0,5
−
+
= − = −=-
.
General solution of differential equation:
22t
ss f
i(t) i i 0 A e
−
= + = +
;
here
ss
i
- steady-state transient component of the current (forced component);
7
f
i
- free transient component of the current.
The initial condition
i(0 ) 50A
+
=
;
steady-state current component (forced component)
ss
i 0A=
, therefore
22t
i(t) 50 e A
−
=
;
22t
di
u(t) i(t) r L 500 e V
dt
−
= + = −
, or simpler
22t
1
u(t) i(t) r 500 e V
−
= − = −
.
At the moment
1
t
we have
1
22t
50
50e A
50
−
=
;
1
ln0,02
t 0,178s
22
=
−
.
Let
1
r 0,1=
. Then
1
1
rr
p 2,2s
L
−
+
= − = −
and
2,2t
i 50 e A
−
=
. Then at a time
1
t
we
get
1
2,2t
50
50 e A
50
−
=
1
ln0,02
t 1,78s
2,2
=
−
.
The greater is the resistance r1, the faster is the current reduction. However, the initial voltage
on the exciting winding may exceed the amount allowed for insulation of the winding.
For example, when
1
r 10=
:
1
u(0 ) r i(0 ) 10 50 500V
++
= − = − = −
; and when
1
r 100=
:
1
u(0 ) r i(0 ) 100 50 5000V
++
= − = − = −
; and there is the risk of breaking down in
the insulation of the winding at this voltage level.
Problem 3.3.
Determine how many times the voltage at the voltmeter terminals UV will increase right after
switching off the excitation winding of the DC generator. The voltmeter resistance
V
r 1k=
;
0
r 0,5=
(Fig. 3.3).
Fig. 3.3
Answer: 2000 times.
Solution
Before the key is opened, voltage on the voltmeter
V
UU=
. Immediately after opening the key, the voltage on the voltmeter
V
V V V V
00
r
U
U i(0 ) r i(0 ) r r U 2000V
rr
+−
= = = = =
.
8
Problem 3.4.
In the circuit shown in Fig. 3.4 it is known:
U = 120 V; r1 = 10 Ω; r2 = 30 Ω; L = 0,1 (H).
Calculate i(t) and u
L
(t).
Fig. 3.4
Answer:
100t
i(t) 12 9 e A
−
= −
;
100t
L
u (t) 90 e V
−
=
.
Problem 3.5.
In the circuit shown in Fig. 3.5, it is known:
U = 120 V; r1 = 10 Ω; r2 = 30 Ω; L = 0,1 H.
Calculate i(t) and u
L
(t).
Fig. 3.5
Answer:
400t
i(t) 3 9 e A
−
= +
;
400t
L
u (t) 360 e V
−
= −
.
Problem 3.6.
In the circuit shown in Fig. 3.6, the key K2 is opened in ∆t after closing the key K1. What
should be the ratio of resistances r1 and r2 so that when the key K2 is opened, the transient process
stops instantly?
9
Fig. 3.6
Answer:
1
r
t
1
L
2
r
e1
r
=−
.
Solution
1. . When key K
1
is closed
1
r
t
L
1
U
i(t) (1 e );
r
−
= −
by the moment of key K2 opening, current i
will reach the value
1
r
t
L
1
U
i( t) (1 e )
r
−
= −
.
2. In order for the transient process to stop instantaneously after opening the key K2, the
current i by the moment of key K2 opening must reach a steady-state value (forced current)
ss
i
in
the postcommutation circuit:
ss
12
U
i( t ) i( t ) i
rr
−+
= = =
+
. Then, taking into account p.1, we get:
1
r
t
L
1 1 2
UU
(1 e )
r r r
−
− =
+
; from where
1
r
t
1
L
2
r
e1
r
=−
.
Problem 3.7.
For the circuit shown in Fig. 3.7, it is known:
1
r 10=
;
1
L 0,1H=
;
2
r 20=
;
2
L 0,3H=
;
M 0,1H=
;
U 120V=
.
Calculate the current and voltage at each coil at their aiding and opposite connection.
Fig. 3.7
10
Answer:
50t
aid
i (t) 4 4 e A
−
= −
;
150t
opp
i (t) 4 4 e A
−
= −
;
1aid
u (t) 40V=
;
2aid
u (t) 80V=
;
150t
1opp
u (t) 40 40 e V
−
= −
;
150t
2opp
u (t) 80 40 e V
−
= +
.
Solution
Let’s use the classical method of transients calculation.
Let us write down the differential equation of the state of the postcommutation circuit taking
into account the mutual-inductance voltages of coils:
1 1 2 2
di di di di
r i L M r i L M U
dt dt dt dt
+ + + =
.
Here, the upper sign "+" corresponds to the aiding connection of the coils, the lower signs "-"
– to the opposite connection. Then we calculate the roots of characteristic polynomials in both
cases.
1 2 1 2
r r p (L L 2 M) 0+ + + =
;
1
12
1
12
r r 30
p 50s
L L 2 M 0,6
−
+
= − = − = −
+ +
- at aiding connection of coils;
1
12
2
12
r r 30
p 150s
L L 2 M 0,2
−
+
= − = − = −
+ −
- at opposite connection of coils.
Transient current at aiding connection
1
pt
50t
aid ss f 1 1
12
U
i (t) i i A e 4 A e A
rr
−
= + = + = +
+
.
At
t0
+
=
:
1
i(0 ) 4 A i(0 ) 0A
+−
= + = =
;
1
A4=−
.
So
50t
aid
i (t) 4 4 e A
−
= −
.
Transient current at opposite connection
150t
opp
i (t) 4 4 e A
−
= −
.
Voltages on coils at aiding connection:
50t 50t
aid aid
1aid 1 aid 1
di di
u (t) r i (t) L M 40 40 e (0,1 0,1) (200 e ) 40V
dt dt
−−
= + + = − + + =
;
50t 50t
aid aid
2aid 2 aid 2
di di
u (t) r i (t) L M 80 80 e (0,3 0,1) (200 e ) 80V
dt dt
−−
= + + = − + + =
.
Thus, in the circuit in question, at aiding connection of the coils their voltages reach steady-state
values at the time of switching immediately.
Voltages on coils at opposite connection:
opp opp
1opp 1 opp 1
di di
u (t) r i (t) L M
dt dt
= + − =
50t 150t 150t
40 40 e (0,1 0,1) (200 e ) 40 40 e V
− − −
= − + − = −
;
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