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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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31
33
C
t0
t0
i (0 ) i (0 )
21
CF
du (t) du(t)
12 36 12
dt
dt
+
+
++
=
=
= = = =
−+
.
We calculate resistance r using the roots of the characteristic polynomial of the circuit:
1 r pL
Z(p) 0
pC r pL
= + =
+
or
2
11
p p 0
r C L C
+ + =
;
1
1
p 2s
−
=−
;
1
2
p 6s
−
=−
.
By Vieta’s formulas
12
1
pp
rC
+ = −
.
Then
1
1
8s
rC
−
=
;
1
r 1,5
8C
= =
.
Calculation of the currents in the branches of the circuit
( ) ( )
2t 6t 2t 6t
1
1
i (t) 6 e 6 e 4 e e A
r
− − − −
= − = −
;
6t 2t
3
du(t)
i (t) C 3 e e A
dt
−−
= = −
;
2t 6t
2 1 3
i (t) J i (t) i (t)(A) 2 3e e (A)
−−
= − − = − +
.
Problem 3.29.
For the circuit shown in Fig. 3.29, it is known:
J 2A=; r 1,5=
;
L 1,0H=
;
1
CF
12
=
.
Calculate transient currents in the circuit branches and transient voltage
u(t)
.
Fig. 3.29
Answer:
( )
2t 6t
1
i (t) 4 e e A
−−
= −
;
2t 6t
2
i (t) 2 3 e e A
−−
= − +
;
2t 6t
3
i (t) e 3 e A
−−
= − +
;
( )
2t 6t
u(t) 6 e e V
−−
= −
.
Solution
1
u(t)
i (t) A
r
=
;
32
C
3
du (t)
i (t) C A
dt
=
;
2 1 3
i (t) J i (t) i (t)A= − −
;
C
u(t) u (t)=
.
We calculate the transient voltage at the capacitor by the classical method:
C Css Cf Cf Cf
u (t) u (t) u (t) 0 u (t) u (t)= + = + =
.
1. The roots of the characteristic polynomial.
1 r pL
Z(p) 0
pC r pL
= + =
+
,
2
p 8p 12 0+ + =
;
1
1,2
p 4 2s
−
= −
;
1
1
p 2s
−
=−
;
1
2
p 6s
−
=−
.
2t 6t
C 1 2
u (t) u(t) A e A e V
−−
= = +
.
2. Initial conditions.
CC
u (0 ) u (0 ) 0V
+−
==
;
C
2
C3
12
t0
u (0 )
J i (0 )
du (t) i (0 )
J i (0 ) i (0 ) J V
r
24
dt C C C C s
+
+
+
+
++
=
−−
−−
= = = = =
.
3. Undefined coefficients.
c 1 2
u (0 ) A A 0V
+
= + =
;
C
12
t0
du (t)
V
2 A 6 A 24
dt s
+
=
= − − =
;
1
A 6A=
;
2
A 6A=−
.
Finally we get
( )
2t 6t
C
u(t) u (t) 6 e e V
−−
= = −
;
( )
2t 6t
1
u(t)
i (t) 4 e e A
r
−−
= = −
;
2t 6t
C
3
du (t)
i (t) C u(t) e 3 e A
dt
−−
= = = − +
;
2t 6t
2 1 3 1
i (t) J i (t) i (t) i (t) 2 3 e e A
−−
= − − = = − +
.
Problem 3.30.
For the circuit shown in Fig. 3.30a, it is known:
E 20V=
;
r 0,5k=
;
L 1mH=; C 0,02 F=
;
the time dependence of the current source
J(t)
is shown in Fig. 3.30b.
Calculate the transient voltage on the capacitor
C
u (t)
.
33
Fig. 3.30a Fig. 3.30b
Answer:
55
1,3810 t 3,62 10 t
C1
u (t) 70 58,74 e 8,74 e V
− −
= − +
.
Solution
A graph of the current intensity of the current source against time allows us to consider the
problem as a calculation of the transient process when the DC source is switched on to the circuit at
a time
t0=
.
We find the transient voltage on the capacitor by the classical method:
C Css Cf
u (t) u (t) u (t)=+
.
1. The roots of the characteristic polynomial.
1
Z(p) r pL 0
pC
= + + =
;
2 5 10
p 5 10 p 5 10 0+ + =
;
5 10 10 5 5 1
1,2
p 2,5 10 6,25 10 5 10 2,5 10 1,12 10 s
−
= − − = −
;
51
1
p 1,38 10 s
−
= −
;
51
2
p 3,62 10 s
−
= −
.
55
1,3810 t 3,62 10 t
Cсв 1 2
u (t) A e A e V
− −
= +
.
2. Initial conditions.
CC
u (0 ) u (0 ) E 20V
+−
= = =
;
LL
i (0 ) i (0 ) 0A
+−
==
;
6
CC
L
6
t0
du (t) i (0 )
J(0 ) i (0 ) J(0 ) 0,1 V
5 10
dt C C C 0,02 10 s
+
+
+ + +
−
=
−
= = = = =
.
3. Forced component.
Css Lss
t
u u r J(t) E
→
+ − =
;
Css Lss
t
u E u r J(t) 20 0 500 0,1 70 V
→
= − + = − + =
.
4. Undefined coefficients
c 1 2
5 5 6
C
12
t0
u (0 ) 70 A A 20V;
du (t)
V
1,38 10 A 3,62 10 A 5 10 ;
dt s
+
+
=
= + + =
= − − =
1
A 58,74V=−
;
2
A 8,74V=
;
34
55
1,3810 t 3,6210 t
C
u (t) 70 58,74 e 8,74 e V
− −
= − +
.
Problem 3.31.
For the circuit shown in Fig. 3.31, it is known:
r 20=
;
L 0,01H=
;
1
C 125 F=
;
2
C 500 F=
;
C1
u (0 ) 200V
−
=
;
C2
u (0 ) 50V
−
=
.
Calculate capacitor voltages, current and inductance voltage in transient mode. Check
solution.
Fig. 3.31
Answer:
1000t
C1
u (t) 80 120 (1 1000t) e V
−
= + +
;
1000t
C2
u (t) 80 30 (1 1000t) e V
−
= − +
;
( )
1000t
L
di
u (t) L 150 1 1000t e V
dt
−
= = −
.
Solution
We find transient voltages on capacitors using the classical method:
C1 C1ss C1f
u (t) u (t) u (t)=+
;
C2 C2ss C2f
u (t) u (t) u (t)=+
.
1. The roots of the characteristic polynomial.
12
11
Z(p) r pL 0
pC pC
= + + + =
or
2
12
12
r C C
p p 0
L L C C
+
+ + =
;
2 3 6
p 2 10 p 10 0+ + =
;
1
12
p p p 1000s
−
= = = −
- root of multiplicity 2.
( )
1000t
C1f 1 2
u (t) A A t e V
−
= +
;
( )
1000t
C2f 3 4
u (t) A A t e V
−
= +
.
2. Initial conditions.
By problem conditions
C1 C1
u (0 ) u (0 ) 200V
+−
==
;
C2 C2
u (0 ) u (0 ) 50V
+−
==
;
i(0 ) i(0 ) 0A
+−
==
;
35
C1
t0
11
du (t)
i(0 ) 0 V
0
dt C C s
+
+
=
−
= = =
;
C2
t0
22
du (t)
i(0 ) 0 V
0
dt C C s
+
+
=
= = =
.
3. Forced components.
The post-commutation circuit is sequential. Therefore, the same amount of electricity q will
flow through the capacitors for a period of time from the moment of switching till 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 immediately 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 have
C1ss C2ss
u u 0.− + =
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:
3
66
200 50
q 15 10 C
11
125 10 500 10
−
−−
−
= =
+
.
Then the forced voltage components on the capacitors:
63
C1ss C1 1
C1ss
6
11
Q u (0 ) C q
200 125 10 15 10
u 80V
С C 125 10
−−
−
−
−
−
= = = =
;
63
C2ss C2 2
C2ss
6
22
Q u (0 ) C q
50 500 10 15 10
u 80V
С C 500 10
−−
−
−
+
+
= = = =
.
4. Undefined coefficients.
c1 1
u (0 ) 80 A 200V
+
= + =
;
C1
12
t0
du (t)
V
1000 A A 0
dt s
+
=
= − + =
.
1
A 120V=
;
3
2
V
A 120 10
s
=
.
In the same way we calculate
3
A 30V=−
;
3
4
V
A 30 10
s
= −
.
Then:
36
1000t
C1
u (t) 80 120 (1 1000t) e V
−
= + +
;
1000t
C2
u (t) 80 30 (1 1000t) e V
−
= − +
;
C2
2
du
i(t) C
dt
= =
( )
( )
6 6 1000t 1000t
500 10 30 1000 1000 10 t e 15000t e A
− − −
= − − − =
.
Or
C1
1
du
i(t) C
dt
= − =
( )
( )
6 6 1000t 1000t
125 10 120 1000 1000 10 t e 15000t e A
− − −
= − − − =
;
( )
1000t
L
di
u (t) L 150 1 1000t e V
dt
−
= = −
.
The solution can be checked according to Kirchhoff's 2nd law for the circuit at
post-commutation regime, since this law was not used in the above calculations:
L C2 С1
r i u u u 0 + + − =
;
( )
1000t 1000t
20 15000t e 150 1 1000t e
−−
+ − +
1000t 1000t
80 30 (1 1000t) e 80 120 (1 1000t) e 0
−−
+ − + − − + =
.
Problem 3.32.
For the circuit shown in Fig. 3.32, it is known:
U 120V=
;
r 20=
;
L 10mH=
;
C 100 F=
; key K2 has closed τ=1ms later than key
K
1
.
Calculate the current in the key K2.
Fig. 3.32
Answer:
500 (t τ)
2
i (t) 4,415 e 4,415A
− −
= −
.
Solution
1. Consider the circuit after closing the key K1 at a time interval
0 t 1ms
. The
characteristic polynomial of the circuit has multiples roots:
1
Z(p) r pL 0
pC
= + + =
;
2
r1
p p 0
L L C
+ + =
;
37
2 3 6
p 2 10 p 10 0+ + =
;
1
12
p p p 1000s
−
= = = −
- root multiple of 2.
This is a critical transient regime (Fig. 3.32a), where
1ms=
.
Fig. 3.32a
Hence, in respect to zero independent initial conditions (we believe that the capacitor was
discharged by default), currents in the circuit and the coil are described by the function
pt
L ss f ss 1 2
i (t) i(t) i i i (A A t) e
= = + + + = =
p t 1000t 1000t
L
3
u (0 ) 120
0 0 t e t e 12000t e A
L 10 10
− −
+
−
+ + = =
=
.
2. Consider the circuit after closing key K2 at
t 1ms =
.
By the time
t
−
=
the key K2 has closed current in the circuit and the coil reaches the
maximum value
31
Lmax max
i i 12000 10 e 4,415A
−−
= =
. At the moment of time
t+=
the circuit breaks down into two separate circuits (Fig. 3.32b). In the r-C circuit, the capacitor
will continue being charged by transient current
1
(t )
500 (t )
rC
max
i(t) i e 4,415 e A
− −
− −
= =
up to the source voltage
U 120V=
, and in the L-circuit, due to the lack of resistance, a
steady-state DC mode will immediately occur wherein
Lmax max
i i 4,415A==
.
Fig. 3.32b
3. The current
2
i
in the key K2 after its closure can be found according to the 1st Kirchhoff
law from the scheme in Fig. 3.32:
500 (t )
2L
i (t) i(t) i (t) 4,415 e 4,415A
− −
= − = −
.
Note that switching the key K2 will not lead to a jump in the current i(t) in the r-C circuit,
since the coil current reaches a local maximum by the switching moment, which means that the
38
voltage on the coil
L
L
di
uL
dt
=
at the moment t = t- is zero. Therefore, closing the key K2 will
not change the voltage distribution on the elements of the mesh r-C at the moment t =t
+
.
Problem 3.33.
For the circuit shown in Fig. 3.33, it is known:
E 200V=
;
12
r r 10= =
;
L 0,6H=
; the roots of the characteristic polynomial
1
1
p 50s
−
=−
;
1
2
200
ps
3
−
=−
.
Calculate currents in the circuit branches.
Fig. 3.33
Answer:
200
t
50t
3
2
i (t) 10 80 e 90 e A
−
−
= − +
;
200
t
50t
3
C
i (t) 40 e 60 e A
−
−
= −
;
200
t
50t
3
1
i (t) 10 40 e 30 e A
−
−
= − +
.
Solution
Here it is advisable to find all the sought currents through the capacitor voltage
c
u (t)
:
C
C
du
iC
dt
=
;
C
2
u
ir=
and
1 C 2
i i i=+
.
Problem 3.34.
For the circuit shown in Fig. 3.34, it is known:
E 2V=
;
( )
2t 6t
i(t) 2 e e A
−−
= −
;
C
u (0 ) 0V
−
=
.
Calculate the circuit parameters r; L; C and
C
u (t)
.
39
Fig. 3.34
Answer:
r2=
;
L 0,25H=
;
1
CF
3
=
;
2t 6t
C
u (t) 2 3 e 1 e V
−−
= − +
.
Solution
It is recommended to apply the Vieta’s formulas (see problems 3.26 - 3.28).
Problem 3.35.
For the circuit shown in Fig. 3.35, it is known:
1
CF
7
=
;
2t 14t
i(t) e e A
−−
=−
;
C
u (0 ) 0V
−
=
.
Calculate circuit parameters r; L; E.
Fig. 3.35
Answer:
r4=
;
L 0,25H=
;
E 3V=
.
Solution
It is recommended to apply the Vieta’s formulas (see problems 3.26 - 3.28).
Problem 3.36.
For the circuit shown in Fig. 3.36, it is known:
U 120V=
;
L 1,0H=
;
1
CF
12
=
.
Calculate:
1.
3
i (t)
if
r 1,5=
;
40
2.
2
i (t)
and
3
i (t)
if
r3=
;
3.
3
i (t)
if
r 2 3=
.
Fig. 3.36
Answer:
1.
6t 2t
3
i (t) 120 e 40 e A
−−
= −
;
2.
( ) ( )
2 3 t 2 3 t
2
i (t) 40 3 40 3 e 240t e A
− −
= − −
;
( ) ( )
2 3 t 2 3 t
3
i (t) 40 3 e 240t e A
− −
= −
;
3.
( )
( )
( )
( )
3 t 3 t
3
i (t) 20 3 e cos 3t 20 e sin 3t
−−
= + =
( )
( )
3t
40 e sin 3t 60 A
−
= +
.
Solution
Here it is advisable to find all the sought currents through the capacitor voltage
c
u (t)
:
C
3
du
iC
dt
=
;
C
1
Uu
i
r
−
=
and
2 1 3
i i i=−
.
Problem 3.37.
For the circuit shown in Fig. 3.37, it is known:
U 120V=
;
r 1,5=
;
L 1,0H=
;
1
CF
12
=
.
Calculate the transient currents of the branches and the voltage on the coil.
Fig. 3.37
Answer:
2t 6t
1
i (t) 80 90 e 10 e A
−−
= − +
;
2t 6t
2
i (t) 30 e 30 e A
−−
= −
;
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