Bird J. Electrical Circuit Theory and Technology
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898 Electrical Circuit Theory and Technology
Determine the magnitude and phase of the current at the receiving
end if the sending-end voltage is 5
6
0
°
V rms. [0.340
6
61.67 mA]
6 The voltages at the input and at the output of a transmission line
properly terminated in its characteristic impedance are 10 V and 4 V
rms respectively. Determine the output voltage if the length of the
line is trebled. [0.64 V]
Characteristic impedance and propagation constant
7 At a frequency of 800 Hz, the open-circuit impedance of a length of
transmission line is measured as 500
6
35
°
# and the short-circuit
impedance as 300
6
15
°
#. Determine the characteristic impedance
of the line at this frequency. [387.3
6
25
°
#]
8 A transmission line has the following primary constants per loop
kilometre run: R D 12 #, L D 3mH,G D 4
µSandC D 0.02 µF.
Determine the characteristic impedance of the line when the
frequency is 750 Hz. [443.3
6
18.95
°
#]
9 A transmission line having negligible losses has primary constants:
inductance L D 1.0 mH/loop km and capacitance C D 0.20
µF/km.
Determine, at an operating frequency of 50 kHz, (a) the characteristic
impedance, (b) the propagation coefficient, (c) the attenuation and
phase-shift coefficients, (d) the wavelength on the line, and (e) the
velocity of propagation of signal in metres per second.
[(a) 70.71 # (b) j4.443 (c) 0; 4.443 rad/km
(d) 1.414 km (e) 70.71 ð 10
6
m/s]
10 At a frequency of 5 kHz the primary constants of a transmission line
are: resistance R D 12 #/loop km, inductance L D 0.50 mH/loop
km, capacitance C D0.01
µF/km and G D 60 µS/km. Determine
for the line (a) the characteristic impedance, (b) the propagation
coefficient, (c) the attenuation coefficient, and (d) the phase-shift
coefficient.
[(a) 248.6
6
13.29
°
# (b) 0.0795
6
65.91
°
(c) 0.0324 Np/km (d) 0.0726 rad/km]
11 A transmission line is 50 km in length and is terminated in its
characteristic impedance. At the sending end a signal emanates
from a generator which has an open-circuit e.m.f. of 20.0 V, an
internal impedance of 250 C j0# at a frequency of 1592 Hz. If the
line primary constants are R D 30 #/loop km, L D 4.0 mH/loop km,
G D 5.0
µS/km, and C D 0.01 µF/km, determine (a) the value of
the characteristic impedance, (b) the propagation coefficient, (c) the
attenuation and phase-shift coefficients, (d) the sending-end current,
(e) the receiving-end current, (f) the wavelength on the line, and
(g) the speed of transmission of a signal, in metres per second.
[(a) 706.6
6
17
°
# (b) 0.0708
6
70.14
°
(c) 0.024 Np/km; 0.067 rad/km
(d) 21.1
6
12.58
°
mA (e) 6.35
6
178.21
°
mA
(f) 94.34 km (g) 150.2 ð 10
6
m/s]
Transmission lines 899
Distortion on transmission lines
12 A cable has the following primary constants: resistance R D
90 #/loop km, inductance L D 2.0 mH/loop km, capacitance C D
0.05
µF/km and conductance G D 3.0 µS/km. Determine the value
to which the inductance should be increased to satisfy the condition
for minimum distortion. [1.5 H]
13 A condition of minimum distortion is required for a cable. Its
primary constants are: R D 40 #/loop km, L D 2.0 mH/loop km,
G D 2.0
µS/km and C D 0.08 µF/km. At a frequency of 100 Hz
determine (a) the increase in inductance required, (b) the propagation
coefficient, (c) the speed of signal transmission and (d) the
wavelength on the line.
[(a) l.598 H (b) 8.944 Cj22510
3
(c) 2.795 ð 10
6
m/s (d) 27.93 km]
Reflection coefficient
14 A coaxial line has a characteristic impedance of 100 # and is termi-
nated in a 400 # resistive load. The voltage measured across the
termination is 15 V. The cable is assumed to have negligible losses.
Calculate for the line the values of (a) the reflection coefficient,
(b) the incident current, (c) the incident voltage, (d) the reflected
current, and (e) the reflected voltage.
[(a) 0.60 (b) 93.75 mA (c) 9.375 V
(d) 56.25 mA (e) 5.625 V]
15 A long transmission line has a characteristic impedance of
400j50# and is terminated in an impedance of (i) 400 C j50#,
(ii) 500 Cj60# and (iii) 400
6
0
°
#. Determine the magnitude of
the reflection coefficient in each case.
[(i) 0.125 (ii) 0.165 (iii) 0.062]
16 A transmission line which is loss-free has a characteristic
impedance of 600
6
0
°
# and is connected to a load of impedance
400 C j300#. Determine (a) the magnitude of the reflection
coefficient and (b) the magnitude of the sending-end voltage if the
reflected voltage is 14.60 V [(a) 0.345 (b) 42.32 V]
Standing-wave ratio
17 A transmission line has a characteristic impedance of 500
6
0
°
# and
negligible loss. If the terminating impedance of the line is
320 C j200# determine (a) the reflection coefficient and (b) the
standing-wave ratio. [(a) 0.319
6
61.72
°
(b) 1.937]
18 A low-loss transmission line has a mismatched load such that the
reflection coefficient at the termination is 0.5
6
135
°
. The character-
istic impedance of the line is 60 #. Calculate (a) the standing-wave
900 Electrical Circuit Theory and Technology
ratio, (b) the load impedance, and (c) the incident current flowing if
the reflected current is 25 mA.
[(a) 3 (b) 113.93
6
43.32
°
# (c) 50 mA]
19 The standing-wave ratio on a mismatched line is calculated as 2.20. If
the incident power arriving at the termination is 100 mW, determine
the value of the reflected power.
[14.06 mW]
20 The termination of a coaxial cable may be represented as a 150 #
resistance in series with a 0.20
µH inductance. If the characteristic
impedance of the line is 100
6
0
°
# and the operating frequency is
80 MHz, determine (a) the reflection coefficient and (b) the standing-
wave ratio. [(a) 0.417
6
138.35
°
(b) 2.43]
21 A cable has a characteristic impedance of 70
6
0
°
#. The cable is
terminated by an impedance of 60
6
30
°
#. Determine the ratio of the
maximum to minimum current along the line. [1.77]
45 Transients and Laplace
transforms
At the end of this chapter you should be able to:
ž determine the transient response of currents and voltages in
R–L, R–C and L –R–C series circuits using differential
equations
ž define the Laplace transform of a function
ž use a table of Laplace transforms of functions commonly met
in electrical engineering for transient analysis of simple
networks
ž use partial fractions to deduce inverse Laplace transforms
ž deduce expressions for component and circuit impedances in
the s-plane given initial conditions
ž use Laplace transform analysis directly from circuit diagrams
in the s-plane
ž deduce Kirchhoff law equations in the s-plane for determining
the response of R–L, R –C and L –R –C networks, given
initial conditions
ž explain the conditions for which an L –R –C circuit response
is over, critical, under or zero-damped and calculate circuit
responses
ž predict the circuit response of an L –R –C network, given
non-zero initial conditions
45.1 Introduction
A transient state will exist in a circuit containing one or more energy
storage elements (i.e., capacitors and inductors) whenever the energy
conditions in the circuit change, until the new steady state condition is
reached. Transients are caused by changing the applied voltage or current,
or by changing any of the circuit elements; such changes occur due to
opening and closing switches. Transients were introduced in Chapter 17
where growth and decay curves were constructed and their equations
stated for step inputs only. In this chapter, such equations are developed
analytically by using both differential equations and Laplace trans-
forms for different waveform supply voltages.
45.2 Response of R–C
series circuit to a step
input
Charging a capacitor
A series R–C circuit is shown in Figure 45.1 (a).
A step voltage of magnitude V is shown in Figure 45.1(b). The capacitor
in Figure 45.1(a) is assumed to be initially uncharged.
902 Electrical Circuit Theory and Technology
vv
V
CR
CR
i
Switch
0
V
t
(b)
(a)
Figure 45.1
From Kirchhoff’s voltage law, supply voltage,
V D
v
C
C v
R
45.1
Voltage
v
R
D iR and current i D C
d
v
C
dt
, hence
v
R
D CR
d
v
C
dt
Therefore, from equation (45.1)
V D
v
C
C CR
d
v
C
dt
45.2
This is a linear, constant coefficient, first order differential equation. Such
a differential equation may be solved, i.e., find an expression for voltage
v
C
, by separating the variables.(See Engineering Mathematics or Higher
Engineering Mathematics)
Rearranging equation (45.2) gives:
V
v
C
D CR
d
v
C
dt
and
d
v
C
dt
D
V
v
C
CR
from which,
d
v
C
V v
C
D
dt
CR
and integrating both sides gives:
dv
C
V v
C
D
dt
CR
Hence lnV
v
C
D
t
CR
C k45.3
where k is the arbitrary constant of integration
(To integrate
dv
C
V v
C
make an algebraic substitution,
u D V
v
C
—see Engineering Mathematics or Higher Engineering
Mathematics)
When time t D 0,
v
C
D 0, hence lnV D k
Thus, from equation (45.3), lnV
v
C
D
t
CR
lnV
Rearranging gives:
lnV lnV
v
C
D
t
CR
ln
V
V v
C
D
t
CR
by the laws of logarithms
i.e.,
V
V v
C
D e
t
CR
and
V
v
C
V
D
1
e
t/CR
D e
t/CR
V v
C
D Ve
t/CR
V Ve
t/CR
D v
C
Transients and Laplace transforms 903
v
C
V
t
0
v
C
=
V
(1−
e
−
t/CR
)
Figure 45.2
i.e., capacitor p.d.,
v
c
= V
1 − e
−t=CR
45.4
This is an exponential growth curve, as shown in Figure 45.2.
From equation (45.1),
v
R
D V v
C
D V
V
1 e
t/CR
from equation (45.4)
D V V C Ve
t/CR
i.e., resistor p.d.,
v
R
= Ve
−t=CR
45.5
This is an exponential decay curve, as shown in Figure 45.3.
0
t
V
v
R
v
R
=
Ve
−
t
/
CR
Figure 45.3
In the circuit of Figure 45.1 (a), current i D C
d
v
C
dt
Hence i D C
d
dt
V
1 e
t/CR
from equation (45.4)
i.e., i D C
d
dt
V Ve
t/CR
D C
0 V
1
CR
e
t/CR
D C
V
CR
e
t/CR
i.e., current,
i =
V
R
e
−t=CR
45.6
where
V
R
is the steady state current, I.
This is an exponential decay curve as shown in Figure 45.4.
t
V
R
i
0
i
=
e
−
t
/
CR
R
V
Figure 45.4
After a period of time it can be determined from equations (45.4) to (45.6)
that the voltage across the capacitor,
v
C
, attains the value V, the supply
voltage, whilst the resistor voltage,
v
R
, and current i both decay to zero.
Problem 1. A 500 nF capacitor is connected in series with a
100 k resistor and the circuit is connected to a 50 V, d.c. supply.
Calculate (a) the initial value of current flowing, (b) the value of
current 150 ms after connection, (c) the value of capacitor voltage
80 ms after connection, and (d) the time after connection when the
resistor voltage is 35 V.
904 Electrical Circuit Theory and Technology
(a) From equation (45.6), current, i D
V
R
e
t/CR
Initial current, i.e., when t D 0, i
0
D
V
R
e
0
D
V
R
D
50
100 ð 10
3
D 0.5mA
(b) Current, i D
V
R
e
t/CR
hence, when time t D 150 ms or 0.15 s,
i D
50
100 ð 10
3
e
0.5/500ð10
9
100ð10
3
D 0.5 ð 10
3
e
3
D 0.5 ð 10
3
0.049787 D 0.0249 mA or 24.9 mA
(c) From equation (45.4), capacitor voltage,
v
C
D V1 e
t/CR
When time t D 80 ms,
v
C
D 501 e
80ð10
3
/500ð10
3
ð100ð10
3
D 501 e
1.6
D 500.7981
D 39.91 V
(d) From equation (45.5), resistor voltage,
v
R
D Ve
t/CR
When v
R
D 35 V,
then 35 D 50e
t/500ð10
9
ð100ð10
3
i.e.,
35
50
D e
t/0.05
and ln
35
50
D
t
0.05
from which, time, t D0.05ln0.7
D 0.0178 s or 17.8ms
Discharging a capacitor
If after a period of time the step input voltage V applied to the circuit of
Figure 45.1 is suddenly removed, by opening the switch, then
from equation (45.1),
v
R
C v
C
D 0
or, from equation (45.2), CR
d
v
C
dt
C
v
C
D 0
Rearranging gives:
d
v
C
dt
D
1
CR
v
C
and separating the variables gives:
d
v
C
v
C
D
dt
CR
and integrating both sides gives:
dv
C
v
C
D
dt
CR
Transients and Laplace transforms 905
from which, ln v
C
D
t
CR
C k45.7
where k is a constant.
At time t D 0 (i.e., at the instant of opening the switch),
v
C
D V
Substituting t D 0and
v
C
D V in equation (45.7) gives:
lnV D 0 C k
Substituting k D lnV into equation (45.7) gives:
ln
v
C
D
t
CR
C lnV
and ln
v
C
lnV D
t
CR
ln
v
C
V
D
t
CR
and
v
C
V
D e
t/CR
from which,
v
C
= Ve
−t=CR
45.8
i.e., the capacitor voltage,
v
C
, decays to zero after a period of time, the
rate of decay depending on CR, which is the time constant, t (see
Section 17.3, page 260). Since
v
R
C v
C
D 0 then the magnitude of the
resistor voltage,
v
R
, is given by:
v
R
= Ve
−t=CR
45.9
and since i D C
d
v
C
dt
D C
d
dt
Ve
t/CR
D CV
1
CR
e
t/CR
i.e., the magnitude of the current,
i =
V
R
e
−t=CR
45.10
Problem 2. A d.c. voltage supply of 200 V is connected across
a5
µF capacitor as shown in Figure 45.5. When the supply is
suddenly cut by opening switch S, the capacitor is left isolated
except for a parallel resistor of 2 M. Calculate the p.d. across the
capacitor after 20 s.
From equation (45.8), v
C
D Ve
t/CR
After 20 s, v
C
D 200e
20/5ð10
6
ð2ð10
6
D 200 e
2
D 2000.13534
D 27.07 V
S
200 V
2 MΩ
+
−
5 µF
Figure 45.5
906 Electrical Circuit Theory and Technology
45.3 Response of R–L
series circuit to a step
input
Current growth
A series R –L circuit is shown in Figure 45.6. When the switch is closed
and a step voltage V is applied, it is assumed that L carries no current.
From Kirchhoff’s voltage law, V D
v
L
C v
R
Voltage v
L
D L
di
dt
and voltage
v
R
D iR
Hence V D L
di
dt
C iR 45.11
This is a linear, constant coefficient, first order differential equation.
Again, such a differential equation may be solved by separating the
variables.
V
Switch
LR
v
L
v
R
i
Figure 45.6
Rearranging equation (45.11) gives:
di
dt
D
V iR
L
from which,
di
V iR
D
dt
L
and
di
V iR
D
dt
L
Hence
1
R
lnV iR D
t
L
C k45.12
where k is a constant
Use the algebraic substitution u D V iR to integrate
di
V iR
At time t D 0, i D 0, thus
1
R
lnV D 0 C k
Substituting k D
1
R
lnV in equation (45.12) gives:
1
R
lnV iR D
t
L
1
R
lnV
Rearranging gives:
1
R
[lnV lnV iR] D
t
L
and ln
V
V iR
D
Rt
L
Hence
V
V iR
D e
Rt
L
and
V iR
V
D
1
e
Rt/L
D e
Rt/L
V iR D Ve
Rt/L
V Ve
Rt/L
D iR
Transients and Laplace transforms 907
i
=
V
R
1−
e
−
Rt/L
V
R
i
0
t
Figure 45.7
and current,
i =
V
R
1 − e
−Rt=L
45.13
This is an exponential growth curve as shown in Figure 45.7.
The p.d. across the resistor in Figure 45.6,
v
R
D iR
Hence
v
R
D R
V
R
1 e
Rt/L
from equation 45.13
i.e.,
v
R
= V .1 − e
−Rt=L
/
45.14
which again represents an exponential growth curve.
The voltage across the inductor in Figure 45.6,
v
L
D L
di
dt
i.e.,
v
L
D L
d
dt
V
R
1 e
Rt/L
D
LV
R
d
dt
[1 e
Rt/L
]
D
LV
R
0
R
L
e
Rt/L
D
LV
R
R
L
e
Rt/L
i.e.,
v
L
= Ve
−Rt=L
45.15
Problem 3. A coil of inductance 50 mH and resistance 5 is
connected to a 110 V, d.c. supply. Determine (a) the final value
of current, (b) the value of current after 4 ms, (c) the value of the
voltage across the resistor after 6 ms, (d) the value of the voltage
across the inductance after 6 ms, and (e) the time when the current
reaches 15 A.
(a) From equation (45.13), when t is large, the final, or steady state
current i is given by:
i D
V
R
D
110
5
D 22 A
(b) From equation (45.13), current, i D
V
R
1 e
Rt/L
When t D 4ms, i D
110
5
1 e
54ð10
3
/50ð10
3
D 221 e
0.40
D 220.32968 D 7.25 V
(c) From equation (45.14), the voltage across the resistor,
v
R
D V1 e
Rt/L