Bird J. Electrical Circuit Theory and Technology
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738 Electrical Circuit Theory and Technology
(i.e., instantaneous current D capacitance ðrate of change of p.d.)
The instantaneous value of power to the capacitor,
p D
vi watts D v
C
d
v
dt
watts
The energy supplied to the capacitor during time dt
D power ð time D
vC
d
v
dt
dt
D C
v dv joules
Thus the total energy supplied to the capacitor when the p.d. is increased
from 0 to V volts is given by
W
f
D
V
0
Cv dv D C
v
2
2
V
0
i.e., energy stored in the electric field,
W
f
=
1
2
CV
2
joules
40.15
Consider a capacitor with dielectric of relative permittivity ε
r
, thickness
d metres and area A square metres. Capacitance C D Q/V, hence energy
stored D
1
2
Q/VV
2
D
1
2
QV joules.
The electric flux density, D D Q/A, from which Q D DA.
Hence the energy stored D
1
2
DAV joules.
The electric field strength, E D V/d, from which V D Ed.
Hence the energy stored D
1
2
DAEd joules. However Ad is the volume
of the field.
Hence energy stored per unit volume,
!
f
=
1
2
DE joules/cubic metre
40.16
Since D/E D ε
0
ε
r
, then D D ε
0
ε
r
E. Hence, from equation (40.16), the
energy stored per unit volume,
ω
f
D
1
2
ε
0
ε
r
EE
i.e.,
!
f
=
1
2
"
0
"
r
E
2
joules/cubic metre
40.17
Also, since D/E D ε
0
ε
r
, then E D D/ε
0
ε
r
. Hence from equation (40.16),
the energy stored per unit volume,
ω
f
D
1
2
D
D
ε
0
ε
r
Field theory 739
i.e.,
!
f
=
D
2
2"
0
"
r
joules/cubic metre
40.18
Summarizing,
energy stored in a capacitor D
1
2
CV
2
joules
and energy stored per unit volume of dielectric
D
1
2
DE =
1
2
"
0
"
r
E
2
D
D
2
2"
0
"
r
joules/cubic metre
Problem 12. Determine the energy stored in a 10 nF capacitor
when charged to 1 kV, and the average power developed if this
energy is dissipated in 10
µs.
From equation (40.15),
energy stored,W
f
D
1
2
CV
2
D
1
2
10 ð 10
9
10
3
2
D 5mJ
average power developed D
energy dissipated, W
time, t
D
5 ð 10
3
J
10 ð 10
6
s
D 500 W
Problem 13. A capacitor is charged with 5 mC. If the energy
stored is 625 mJ, determine (a) the voltage across the plates and
(b) the capacitance of the capacitor.
(a) From equation (40.15),
energy stored, W
f
D
1
2
CV
2
D
1
2
Q
V
V
2
D
1
2
QV
from which voltage across the plates,
V D
2 ð energy stored
Q
D
2 ð 0.625
5 ð 10
3
D 250 V
(b) Capacitance C D
Q
V
D
5 ð 10
3
250
F D 20 ð 10
6
F D 20 mF
740 Electrical Circuit Theory and Technology
Problem 14. A ceramic capacitor is to be constructed to have
a capacitance of 0.01
µF and to have a steady working poten-
tial of 2.5 kV maximum. Allowing a safe value of field stress
of 10 MV/m, determine (a) the required thickness of the ceramic
dielectric, (b) the area of plate required if the relative permittivity
of the ceramic is 10, and (c) the maximum energy stored by the
capacitor.
(a) Field stress E D V/d, from which thickness of ceramic dielectric,
d D
V
E
D
2.5 ð10
3
10 ð 10
6
D 2.5 ð 10
4
m D 0.25 mm
(b) Capacitance C D ε
0
ε
r
A/d for a two-plate parallel capacitor. Hence
cross-sectional area of plate,
A D
Cd
ε
0
ε
r
D
0.01 ð 10
6
0.25 ð 10
3
8.85 ð 10
12
10
D 0.0282 m
2
or 282 cm
2
(c) Maximum energy stored,
W
f
D
1
2
CV
2
D
1
2
0.01 ð 10
6
2.5 ð 10
3
2
D 0.0313 J or 31.3mJ
Problem 15. A 400 pF capacitor is charged to a p.d. of 100 V.
The dielectric has a cross-sectional area of 200 cm
2
and a relative
permittivity of 2.3. Calculate the energy stored per cubic metre of
the dielectric.
From equation (40.18), energy stored per unit volume of dielectric,
ω
f
D
D
2
2ε
0
ε
r
Electric flux density
D D
Q
A
D
CV
A
D
400 ð 10
12
100
200 ð 10
4
D 2 ð 10
6
C/m
2
Hence energy stored,
ω
f
D
D
2
2ε
0
ε
r
D
2 ð 10
6
2
28.85 ð 10
12
2.3
D 0.0983 J=m
3
or 98.3mJ=m
3
Field theory 741
Further problems on energy stored in electric fields may be found in
Section 40.9, problems 16 to 23, page 755.
40.5 Induced e.m.f. and
inductance
A current flowing in a coil of wire is accompanied by a magnetic flux
linking with the coil. If the current changes, the flux linkage (i.e., the
product of flux and the number of turns) changes and an e.m.f. is induced
in the coil. The magnitude of the induced e.m.f. e in a coil of N turns is
given by
e
= N
df
dt
volts
where d)/dt is the rate of change of flux.
Inductance is the name given to the property of a circuit whereby
there is an e.m.f. induced into the circuit by the change of flux linkages
produced by a current change. The unit of inductance is the henry, H.A
circuit has an inductance of 1 H when an e.m.f. of 1 V is induced in it
by a current changing uniformly at the rate of 1 A/s.
The magnitude of the e.m.f. induced in a coil of inductance L henry is
given by
e
= L
di
dt
volts
where di/dt is the rate of change of current.
If a current changing uniformly from zero to I amperes produces a
uniform flux change from zero to ) webers in t seconds then (from above)
average induced e.m.f., E
av
D N)/t D LI/t, from which
inductance of coil, L
=
Nf
I
henry
Flux linkage means the product of flux, in webers, and the number of
turns with which the flux is linked. Hence flux linkage D N). Thus since
L D N)/I, inductance D flux linkages per ampere.
40.6 Inductance of a
concentric cylinder (or
coaxial cable)
Skin effect
When a direct current flows in a uniform conductor the current will
tend to distribute itself uniformly over the cross-section of the conductor.
However, with alternating current, particularly if the frequency is high,
the current carried by the conductor is not uniformly distributed over
the available cross-section, but tends to be concentrated at the conductor
surface. This is called skin effect. When current is flowing through a
conductor, the magnetic flux that results is in the form of concentric
circles. Some of this flux exists within the conductor and links with the
current more strongly near the centre. The result is that the inductance
of the central part of the conductor is greater than the inductance of the
conductor near the surface. This is because of the greater number of flux
742 Electrical Circuit Theory and Technology
linkages existing in the central region. At high frequencies the reactance
X
L
D 2fL of the extra inductance is sufficiently large to seriously
affect the flow of current, most of which flows along the surface of the
conductor where the impedance is low rather than near the centre where
the impedance is high.
Inductance due to internal linkages at low frequency
When a conductor is used at high frequency the depth of penetration of
the current is small compared with the conductor cross-section. Thus the
internal linkages may be considered as negligible and the circuit induc-
tance is that due to the fields in the surrounding space. However, at very
low frequency the current distribution is considered uniform over the
conductor cross-section and the inductance due to flux linkages has its
maximum value.
Consider a conductor of radius R, as shown in Figure 40.17, carrying
a current I amperes uniformly distributed over the cross-section. At all
points on the conductor cross-section
current density, J D
current
area
D
I
R
2
amperes/metre
2
Figure 40.17
Consider a thin elemental ring at radius r and width υr contained within
the conductor, as shown in Figure 40.17. The current enclosed by the ring,
i D current density ð area enclosed by the ring
D
I
R
2
r
2
i.e. i D
Ir
2
R
2
amperes
Magnetic field strength, H D Ni/l amperes/metre.
At radius r, the mean length of the flux path, l D 2r (i.e., the circum-
ference of the elemental ring) and N D 1 turn.
Hence at radius r,
H
r
D
Ni
l
D
1Ir
2
/R
2
2r
D
Ir
2R
2
ampere/metre
and the flux density, B
r
D
0
r
H
r
D
0
r
Ir
2R
2
tesla
Flux ) D BA webers. For a 1 m length of the conductor, the cross-sec-
tional area A of the element is υr ð 1 m
2
(see Figure 40.17). Thus the
flux within the element of thickness υr,
) D
0
r
Ir
2R
2
υr webers
Field theory 743
The flux in the element links the portion r
2
/R
2
, i.e., r
2
/R
2
of the total
conductor. Hence linkages due to the flux within radius r
D
0
r
Ir
2R
2
υr
r
2
R
2
D
0
r
Ir
3
2R
4
υr weber turns
Total linkages per metre due to the flux in the conductor
D
R
0
0
r
Ir
3
2R
4
dr D
0
r
I
2R
4
R
0
r
3
dr
D
0
r
I
2R
4
r
4
4
R
0
D
0
r
I
2R
4
R
4
4
D
1
4
0
r
I
2
weber turns
Inductance per metre due to the internal flux D internal flux linkages per
ampere
D
1
4
m
0
m
r
2p
or
m
8p
henry=metre
It is seen that the inductance is independent of the conductor radius R.
Inductance of a pair of concentric cylinders
The cross-section of a concentric (or coaxial) cable is shown in
Figure 40.18. Let a current of I amperes flow in one direction in the
core and a current of I amperes flow in the opposite direction in the outer
sheath conductor.
Consider an element of width υr at radius r, and let the radii of the
inner and outer conductor be a and b respectively as shown. The magnetic
field strength at radius r,
H
r
D
Ni
I
D
1I
2r
D
I
2r
a
b
r
δr
Figure 40.18 Cross-section of
a concentric cable
The flux density at radius r, B
r
D
0
r
H
r
D
0
r
I
2r
For a 1 m length of the cable, the flux ) within the element of width υr
is given by
D B
r
A D
0
r
I
2r
υr ð 1 D
0
r
I
2r
dr webers
This flux links the loop of the cable formed by the core and the
outer sheath. Thus the flux linkage per metre length of the cable is
0
r
I/2rυr weber turns, and
744 Electrical Circuit Theory and Technology
total flux linkages per metre D
b
a
0
r
I
2r
dr D
0
r
I
2
b
a
1
r
dr
D
0
r
I
2
[lnr]
b
a
D
0
r
I
2
[lnb ln a]
D
0
r
I
2
ln
b
a
weber turns
Thus inductance per metre length D flux linkages per ampere
D
m
0
m
r
2p
ln
b
a
henry=metre 40.19
At low frequencies the inductance due to the internal linkages is added
to this result.
Hence the total inductance per metre at low frequency is given by
L D
1
4
m
0
m
r
2p
C
m
0
m
r
2p
ln
b
a
henry=metre 40.20
or
L =
m
2p
1
4
Y ln
b
a
henry=metre
40.21
Problem 16. A coaxial cable has an inner core of radius 1.0
mm and an outer sheath of internal radius 4.0 mm. Determine the
inductance of the cable per metre length.
Assume that the relative permeability is unity.
From equation (40.21),
inductance L D
2
1
4
C ln
b
a
H/m
D
0
r
2
1
4
C ln
4.0
1.0
D
4 ð 10
7
1
2
0.25 C ln4
D 3.27
× 10
−7
H=m or 0.327 mH=m
Problem 17. A concentric cable has a core diameter of 10 mm.
The inductance of the cable is 4 ð 10
7
H/m. Ignoring inductance
due to internal linkages, determine the diameter of the sheath.
Assume that the relative permeability is 1.
From equation (40.19),
inductance per metre length D
0
r
2
ln
b
a
Field theory 745
where b D sheath radius and a D core radius. Hence
4 ð 10
7
D
4 ð 10
7
1
2
ln
b
5
from which 2 D ln
b
5
and e
2
D
b
5
Hence radius b D 5e
2
D 36.95 mm
Thus the diameter of the sheath is 2 ð 36.95 D 73.9mm
Problem 18. A coaxial cable 7.5 km long has a core 10 mm
diameter and a sheath 25 mm diameter, the sheath having
negligible thickness. Determine for the cable (a) the inductance,
assuming nonmagnetic materials, and (b) the capacitance, assuming
a dielectric of relative permittivity 3.
(a) From equation (40.21),
inductance per metre length D
2
1
4
C ln
b
a
D
0
r
2
1
4
C ln
12.5
5
D
4 ð 10
7
1
2
0.25 C ln2.5
D 2.33 ð 10
7
H/m
Since the cable is 7500 m long,
the inductance D 7500 ð 2.33 ð 10
7
D 1.75 mH
(b) From equation (40.7),
capacitance, C D
2ε
0
ε
r
lnb/a
D
28.85 ð 10
12
3
ln12.5/5
D 182.06 pF/m
Since the cable is 7500 m long,
the capacitance D 7500 ð 182.06 ð 10
12
D 1.365 mF
Further problems on the inductance of concentric cables may be found in
Section 40.9, problems 24 to 27, page 756.
746 Electrical Circuit Theory and Technology
40.7 Inductance of an
isolated twin line
Consider two isolated, long, parallel, straight conductors A and B, each
of radius a metres, spaced D metres apart. Let the current in each be
I amperes but flowing in opposite directions. Distance D is assumed to
be much greater than radius a. The magnetic field associated with the
conductors is as shown in Figure 40.19. There is a force of repulsion
between conductors A and B.
Figure 40.19 Figure 40.20
It is easier to analyse the field by initially considering each conductor
alone (as in Section 40.3). At any radius r from conductor A (see
Figure 40.20),
magnetic field strength, H
r
D
Ni
l
D
I
2r
ampere/metre
and flux density, B
r
D
0
r
H
r
D
0
r
I
2r
tesla
The total flux i
n1moftheconductor,
D B
r
A D
0
r
I
2r
υr ð 1 D
0
r
I
2r
υr webers
Since this flux links conductor A once, the linkages with conductor A
due to this flux D
0
r
I
2r
υr weber turns.
There is, in fact, no limit to the distance from conductor A at which a
magnetic field may be experienced. However, let R be a very large radius
at which the magnetic field strength may be regarded as zero. Then the
total linkages with conductor A due to current in conductor A is given by
Field theory 747
R
a
0
r
I
2r
dr D
0
r
I
2
R
a
dr
r
D
0
r
I
2
[lnr]
R
a
D
0
r
I
2
[lnR lna] D
0
r
I
2
ln
R
a
Similarly, the total linkages with conductor B due to the current in A
D
R
D
0
r
I
2r
dr D
0
r
I
2
ln
R
D
Now consider conductor B alone, carrying a current of I amperes. By
similar reasoning to above, total linkages with
conductor B due to the current in B D
0
r
I
2
ln
R
a
and total linkages with conductor A due to the current in B
D
0
r
I
2
ln
R
D
Hence total linkages with conductor A
D
0
r
I
2
ln
R
a
C
0
r
I
2
ln
R
D
D
0
r
I
2
ln
R
a
ln
R
D
D
0
r
I
2
ln
R/a
R/D
D
0
r
I
2
ln
D
a
weber-turns/metre
Similarly, total linkages with conductor B
D
0
r
I
2
ln
D
a
weber-turns metre
Fora1m
length of the two conductors,
total inductance D flux linkages per ampere
D 2
0
r
2
ln
D
a
henry/metre
i.e., total inductance
=
m
0
m
r
p
ln
D
a
henry=metre 40.22
Equation (40.22) does not take into consideration the internal linkages of
each line.