Bird J. Electrical Circuit Theory and Technology
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938 Electrical Circuit Theory and Technology
s
s C
R
L
needs to be divided out:
1
s C
R
L
s
s C
R
L
R
L
Thus
s
s C
R
L
1
R/L
s C
R
L
Hence v
L
s D
VL
R
VL
R
1
R/L
s C
R
L
D
VL
R
VL
R
C
VL
R
R/L
s C
R
L
D
VL
R
R
L
1
s C
R
L
D V
1
s C
R
L
Thus inductor current v
L
D L
1
V
1
s C
R
L
D VL
1
1
s C
R
L
i.e., v
L
= Ve
−Rt=L
from 3 of Table 45.1, as previously obtained in equation (45.15),
page 907.
Since V D
v
L
C v
R
in Figure 45.19,
resistor voltage,
v
R
D V v
L
D V Ve
Rt/L
D V .1 − e
−Rt=L
/ as previ-
ously obtained in equation (45.14), page 907.
Problem 33. For the circuit of Figure 45.19 of Problem 32, a
ramp of V volts/s is applied to the input terminals, instead of a
step voltage. Determine expressions for current i, inductor
v
L
and
resistor voltage
v
R
(i) The circuit for the s-domain is shown in Figure 45.21.
SL
v
L
(
s
)
Rv
R
(
s
)
V
s
2
Figure 45.21
Transients and Laplace transforms 939
(ii) From Kirchhoff’s voltage law:
V
s
2
D IsR C sL
(iii) Current Is D
V
s
2
R C sL
D
V
s
2
L
s C
R
L
D
V/L
s
2
s C
R
L
Using partial fractions:
V/L
s
2
s C
R
L
D
A
s
C
B
s
2
C
C
s C
R
L
D
As
s C
R
L
C B
s C
R
L
C Cs
2
s
2
s C
R
L
from which,
V
L
D As
s C
R
L
C B
s C
R
L
C Cs
2
when s D 0,
V
L
D 0 C B
R
L
C 0 from which, B D
V
R
when s D
R
L
,
V
L
D 0 C 0 C C
R
L
2
from which,
C D
V
L
L
2
R
2
D
VL
R
2
Equating s
2
coefficients: 0 D A C C from which, A DC D
VL
R
2
Thus Is D
V/L
s
2
s C
R
L
D
A
s
C
B
s
2
C
C
s C
R
L
D
VL/R
2
s
C
V/R
s
2
C
VL/R
2
s C
R
L
45.39
(iv) Current, i D
L
1
fIsgD
VL
R
2
L
1
1
s
C
V
R
L
1
1
s
2
C
VL
R
2
L
1
1
s C
R
L
D
VL
R
2
1 C
V
R
t C
VL
R
2
e
Rt/L
from 2, 7 and 3 of Table 45.1
940 Electrical Circuit Theory and Technology
i.e., i=
V
R
t
−
VL
R
2
.1 − e
.−Rt=L/
/
Inductor voltage,
v
L
s D IssL D sL
VL/R
2
s
C
V/R
s
2
C
VL/R
2
s C
R
L
from equation (45.39) above
D
VL
2
R
2
C
VL/R
s
C
VL
2
/R
2
s
s C
R
L
D
VL
2
R
2
C
VL
sR
C
VL
2
R
2
s
s C
R
L
The division
s
s C
R
L
was shown on page 938,
and is equivalent to 1
R/L
s C
R
L
Hence
v
L
s D
VL
2
R
2
C
VL
sR
C
VL
2
R
2
1
R/L
s C
R
L
D
VL
sR
VL
2
R
2
R
L
1
s C
R
L
D
VL
sR
VL
R
1
s C
R
L
Thus v
L
D L
1
fv
L
sgD
VL
R
L
1
1
s
VL
R
L
1
1
s C
R
L
i.e., inductor voltage, v
L
D
VL
R
VL
R
e
Rt/L
D
VL
R
1 − e
.−Rt=L/
Resistor voltage, v
R
s D IsR D R
VL/R
2
s
C
V/R
s
2
C
VL/R
2
s C
R
L
from equation (45.39)
Transients and Laplace transforms 941
D
VL
sR
C
V
s
2
C
VL
R
s C
R
L
hence v
R
D L
1
VL
sR
C
V
s
2
C
VL
R
s C
R
L
D
VL
R
C Vt C
VL
R
e
Rt/L
from2,7and3Table45.1
i.e.,
v
R
D Vt −
VL
R
.1
− e
.−Rt=L/
/
Problem 34. At time t D 0, a sinusoidal voltage 10 sin ωt is
applied to an L–R series circuit. Determine an expression for the
current flowing.
(i) The circuit is shown in Figure 45.22 and the s-domain circuit
is shown in Figure 45.23, the 10 sinωt input voltage becoming
10
ω
s
2
C ω
2
in the s-domain from 5 of Table 45.1
(ii) From Kirchhoff’s voltage law:
10ω
s
2
C ω
2
D IssL C IsR
0
−10
10
V
t
Li
R
Figure 45.22
(iii) Hence current, Is D
10ω
s
2
C ω
2
R C sL
D
10ω
s
2
C ω
2
L
s C
R
L
D
10ω/L
s
2
C ω
2
s C
R
L
Using partial fractions:
10ω/L
s
2
C ω
2
s C
R
L
D
As C B
s
2
C ω
2
C
C
s C
R
L
D
As C B
s C
R
L
C C
s
2
C ω
2
s
2
C ω
2
s C
R
L
10
w
s
2
+ w
2
sL
I(s)
R
Figure 45.23
942 Electrical Circuit Theory and Technology
hence
10ω
L
D As C B
s C
R
L
C Cs
2
C ω
2
When s D
R
L
10ω
L
D 0 CC
R
L
2
C ω
2
from which, C D
10ω
L
R
2
L
2
C ω
2
D
10ω
L
L
2
R
2
C L
2
ω
D
10ωL
R
2
C ω
2
L
2
Equating s
2
coefficients,
0 D A C C, from which, A DC D
10ωL
R
2
C ω
2
L
2
Equating constant terms,
10ω
L
D B
R
L
C Cω
2
10ω
L
Cω
2
D B
R
L
from which, B D
L
R
10ω
L
Cω
2
D
10ω
R
Lω
2
R
10ωL
R
2
C ω
2
L
2
D
10ωR
2
C ω
2
L
2
Lω
2
10ωL
RR
2
C ω
2
L
2
D
10ωR
2
C 10ω
3
L
2
10ω
3
L
2
RR
2
C ω
2
L
2
D
10ωR
R
2
C ω
2
L
2
Hence
Is D
10ω/L
s
2
C ω
2
s C
R
L
D
As C B
s
2
C ω
2
C
C
s C
R
L
D
10ωL
R
2
C ω
2
L
2
s C
10ωR
R
2
C ω
2
L
2
s
2
C ω
2
C
10ωL
R
2
C ω
2
L
2
s C
R
L
Transients and Laplace transforms 943
D
10ω
R
2
C ω
2
L
2
L
s C
R
L
sL
s
2
C ω
2
C
R
s
2
C ω
2
(iv) Current, i D L
1
fIsg
D
10!
R
2
Y !
2
L
2
Le
.−Rt=L/
− Lcos!t Y
R
!
sint
from3,6and5oftable45.1
Problem 35. In the series-parallel network shown in Figure 45.24,
a 5 V step voltage is applied at the input terminals. Determine an
expression to show how current i varies with time.
In the sdomain, Zs D 15 C
104 C 0.1s
10 C 4 C 0.1s
D 15 C
40 C s
14 C 0.1s
D
1514 C 0.1s C 40 C s
14 C 0.1s
D
210 C 1.5s C 40 C s
14 C 0.1s
D
250 C 2.5s
14 C 0.1s
i
5 V step
100 mH4 Ω
10 Ω
15 Ω
Figure 45.24
Since in the s-domain the input voltage is V/s then
Is D
Vs
ZS
D
5/s
250 C 2.5s
14 C 0.1s
D
514 C 0.1s
s250 C 2.5s
D
70 C 0.5s
s250 C 2.5s
D
70
s250 C 2.5s
C
0.5s
s250 C 2.5s
D
70
2.5ss C 100
C
0.5
2.5s C 100
i.e., Is D
28
ss C 100
C
0.2
s C 100
Using partial fractions:
28
ss C 100
D
A
s
C
B
s C 100
D
As C 100 C Bs
ss C 100
from which, 28 D As C 100 C Bs
944 Electrical Circuit Theory and Technology
When s D 028D 100 A and A D 0.28
When s D100 28 D 0 100 B and B D0.28
Hence Is D
0.28
s
0.28
s C 100
C
0.2
s C 100
D
0.28
s
0.08
s C 100
and current, i D
L
1
fIsgD0.28 − 0.08e
−100t
from 2 and 3 of
Table 45.1
45.8 L–R–C series
circuit using Laplace
transforms
An L –R–C series circuit is shown in Figure 45.25 with a step input
voltage V.Inthes-domain, the circuit components are as shown in
Figure 45.26 and if the step is applied at time t D 0, the s-domain supply
voltage is V/s.
Hence
V
s
D Is
R C sL C
1
sC
from which, current,
Is D
V/s
R C sL C 1/sC
D
V/s
1/s
sR C s
2
L C 1/C
D
V
sR C s
2
L C 1/C
D
V
L
s
2
C sR/L C 1/LC
D
V/L
s
2
C R/Ls C 1/LC
The denominator is made into a perfect square (as in Problem 18):
Hence Is D
V/L
s
2
C
R
L
s C
R
2L
2
C
1
LC
R
2L
2
i
L
R
C
V
Figure 45.25
sL
R
V
s
I(s)
1
sC
Figure 45.26
Transients and Laplace transforms 945
D
V/L
s C
R
2L
2
C
1
LC
R
2L
2
2
45.40
or Is D
V/L
1
LC
R
2L
2
1
LC
R
2L
2
s C
R
2L
2
C
1
LC
R
2L
2
2
and current, i D L
1
fIsg
From 13 of Table 45.1,
L
1
ω
s C a
2
C ω
2
D e
at
sinωt, hence
current, i
=
V =L
1
LC
−
R
2L
2
e
−.R=2L/ t
sin
1
LC
−
R
2L
2
t
45.41
Problem 36. For the circuit shown in Figure 45.27 produce an
equation which shows how the current varies with time. Assume
zero initial conditions when the switch is closed.
5 Ω 0.1 H 20 µF
i
2 V
Figure 45.27
In the s-domain, applying Kirchhoff’s voltage law gives:
2
s
D Is
5 C 0.1s C
1
20 ð 10
6
s
and current Is D
2
s
5 C 0.1s C
5 ð 10
4
s
D
2
5s C 0.1s
2
C 5 ð 10
4
946 Electrical Circuit Theory and Technology
D
2
0.1
s
2
C
5
0.1
s C
5 ð 10
4
0.1
D
20
s
2
C 50s C 5 ð10
5
D
20
/
s
2
C 50s C 25
2
0
C
/
5 ð 10
5
25
2
0
D
20
s C 25
2
C
499 375
2
D
20
p
499 375
p
499 375
s C 25
2
C
499375
2
D
20
706.7
706.7
s C 25
2
C 706.7
2
Hence current, i D L
1
fIsgD0.0283e
25t
sin706.7t,
from 13 of Table 45.1,
i.e., i D 28.3e
−25t
sin706.7t mA
Damping
The expression for current Is in equation (45.40) has four possible solu-
tions, each dependant on the values of R, L and C.
Solution 1. When R D 0, the circuit is undamped and, from equa-
tion (45.40),
Is D
V/L
s
2
C
1
LC
From Chapter 28, at resonance, ω
r
D
1
LC
hence
Is D
V/L
s
2
C ω
2
r
D
V
ω
r
L
ω
r
s
2
C ω
2
r
Hence current, i D
L
1
fIsgD
V
!
r
L
sin!
r
t from 5 of Table 45.1
which is a sine wave of amplitude
V
ω
r
L
and angular velocity ω
r
rad/s.
This is shown by curve A in Figure 45.28.
Transients and Laplace transforms 947
i
+
0
−
A
B
C
D
t
Figure 45.28
Solution 2. When
R
2L
2
<
1
LC
, the circuit is underdamped and the
current i is as in equation (45.41). The current is oscillatory which is
decaying exponentially. This is shown by curve B in Figure 45.28.
Solution 3. When
R
2L
2
D
1
LC
, the circuit is critically damped and
from equation (45.40),
Is D
V/L
s C
R
2L
2
and current, i D L
1
V/L
s C
R
2L
2
D
V
L
te
−.Rt=2L/
45.42
from 12 of Table 45.1
The current is non-oscillatory and is as shown in curve C in Figure 45.28.
Solution 4. When
R
2L
2
>
1
LC
, the circuit is overdamped and from
equation (45.40),
Is D
V/L
s C
R
2L
2
R
2L
2
1
LC
2