12-11-2024

Electrical And
Electronics Technology
Dr. Sathish Shet K
Associate Professor
BITS Pilani Department of Electrical & Electronics Engineering
Bengaluru (Off-Campus)
Pilani|Dubai|Goa|Hyderabad

BITS Pilani
Pilani|Dubai|Goa|Hyderabad

Electrical And Electronics Technology

ENGG ZC112/ PE ZC112
LECTURE-4_5 Date :18/02/2023

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Outline for today’s lecture:

➢ Capacitance and Capacitors

➢ Capacitors in parallel
➢ Capacitors in series
➢ Electric fields
➢ Electric Field Strength and Electric Flux Density
➢ Capacitance of multi-plate Capacitor.
➢ Inductor
➢ Inductive Circuits
➢ Mutual Inductance
➢ Transients and energy stored in inductor

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Capacitance and Capacitors

A capacitor is a device which can store electric charge
for short periods of time. Like resistors, capacitors
can be connected in series and in parallel.
A capacitor is constructed of two parallel conducting
plates separated by an insulator.
Capacitance is a measure of a capacitor’s ability to
store charge on its plates. A capacitor has a
capacitance of 1 farad (F) if 1 coulomb (C) of charge
is deposited on the plates by a potential difference of
1 volt across its plates. The farad is named after
Michael Faraday, a nineteenth century English
chemist and physicist. Q=CV
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In practice, the farad is found to be inconveniently large and the

capacitance is usually expressed in microfarads (μF) or in
picofarads (pF),
Where1 μF = 10−6 F
and 1 pF = 10−12 F
Example:A capacitor having a capacitance of 80 μF is connected
across a 500 V d.c. supply. Calculate the charge.
Q = CV
∴ Charge = (80 × 10−6 ) [F] × 500 [V]
= 0.04 C = 40 mC

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Capacitors in parallel

Suppose two capacitors, having capacitances C1 and C2 farads respectively, to

be connected in parallel (as shown in Fig.) across a p.d. of V volts. The
charge on C 1 is Q1 coulombs and that on C 2 is Q 2 coulombs,
where Q1 = C1V and Q2 = C2V
If we were to replace C 1 and C 2 by a single capacitor of such capacitance C
farads that the same total charge of (Q1 + Q 2) coulombs would be
produced by the same p.d., then Q1 + Q 2 = CV.
Substituting for Q1 and Q 2 , we have C1V + C2V = CV
C = C1 + C2 farads

Hence the resultant capacitance of capacitors in parallel is the arithmetic sum of

their respective capacitances

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Capacitors in series

If V1 and V2 are the corresponding p.d.s across C1 and

C2 respectively, Then Q = C1V1 = C2V2 so that

If we were to replace C 1 and C 2 by a single capacitor of

capacitance C farads such that it would have the same
charge Q coulombs with the same p.d. of V volts, then

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But it is evident from Fig. that V = V1 + V2. Substituting for

V, V1 and V2, we have

Hence the reciprocal of the resultant capacitance of capacitors

connected in series is the sum of the reciprocals of their respective
capacitances

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Distribution of voltage across capacitors in

series
We Know

Substituting for V2 in equation , we have

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Example

Three capacitors have capacitances of 2, 4 and 8 μF respectively.

Find the total capacitance when they are connected
(a) in parallel;
(b) in series.
a) C = C1 + C2 + C3 Total capacitance = 2 + 4 + 8 = 14 μF.
b) If C is the resultant capacitance in microfarads when the
capacitors are in series, then from equation

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Example

If two capacitors having capacitances of 6 μF and 10 μF

respectively are connected in series across a 200 V supply, find
(a) the p.d. across each capacitor;
(b) the charge on each capacitor.
a) Let V1 and V2 be the p.d.s. across the 6μF and 10 μF
capacitors respectively; then,

(b) Charge on each capacitor Q = charge on C1

= 6 × 10−6 × 125 = 0.000 75 C = 750 μC

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Some points on capacitors

It follows from expression that if two similar capacitors are

connected in parallel, the capacitance is double that of one
capacitor.
But the effect of connecting two similar capacitors in parallel is
merely to double the area of each plate.
In general, we may therefore say that the capacitance of a
capacitor is proportional to the area of the plates.
On the other hand, if two similar capacitors are connected in
series, it follows from expression that the capacitance is
halved.
We have, however, doubled the thickness of the insulation
between the plates that are connected to the supply.

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Hence we may say in general that the capacitance of a capacitor

is inversely proportional to the distance between the plates;

and the above relationships may be summarized thus:

In order to clarify this relationship, we now need to consider the
space between the charged plates of a capacitor.
In this space, the charges set up electric fields.
The study of such electric fields is known as electrostatics.

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Electric Fields

The electric field is represented by electric flux lines,

which are drawn to indicate the strength of the
electric field at any point around the charged body.
The denser the lines of flux, the stronger the electric
field.

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Electric fields

The electric field strength at a point is the force acting

on a unit positive charge at that point.
Electric flux lines always extend from a positively
charged body to a negatively charged body, always
extend or terminate perpendicular to the charged
surface, and never intersect

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Electric Field Strength and Electric Flux

Density
The magnitude of the force experienced by this unit charge at any
point in a field is termed the electric field strength at that point.
Electric field strength is sometimes also known as electric stress.
It can be measured in newtons per unit charge and represented by
the symbol E.
(Since E can also represent e.m.f., we use a bold type for E when
representing electric field strength and later we will meet D
representing electric flux density.)

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It should be recalled that One Joule of work is necessary

to raise the potential of one Coulomb of charge
through one V.
When a charge moves through an electric field, the
work done against or by the electric field forces is
indicated by the change in potential of the charge.
Therefore to move a unit charge through a field so that
its potential changes by V volts requires V joules of
work.

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

The most simple field arrangement which we can investigate is that between parallel
charged plates as shown in Fig.
Let us suppose that the plates are very large and that the distance between them is very
small.
Let us also assume that there is free space between the plates.
There is a potential difference of V volts between the plates, therefore the
work in transferring 1 C of charge between the plates is V joules.
But work is the product of force and distance, and in this case the distance is d meters.
Therefore the force experienced by the charge is the electric field strength E
given by

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The electric flux density is the measure of the electric flux

passing at right angles through unit area, i.e. an area of 1m2.
It follows that if the area of the plates in the capacitor of Fig.
is A then the electric flux density D is given by

In electrostatics, the ratio of the electric flux density in a vacuum to the electric field
strength is termed the permittivity of free space and is represented by ε0. Hence,

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value of ε0. =8.854 × 10−12 F/m.

Relative Permitivity:The ratio of the capacitance of a capacitor
having a given material as dielectric to the capacitance of that
capacitor with vacuum (or air) dielectric is termed the relative
permittivity of that material and is represented by the symbol
εr

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BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Capacitance of multi-plate Capacitor

Suppose a capacitor to be made up of n parallel plates, alternate

plates being connected together as in Fig.
Let A = area of one side of each plate in square meters
d = thickness of dielectric in meters
And εr = relative permittivity of the dielectric
Figure shows a capacitor with seven plates, four
being connected to A and three to B. It will be seen
that each side of the three plates connected to B is in
contact with the dielectric, whereas only one side of
each of the outer plates is in contact with it.
Consequently, the useful surface area of
each set of plates is 6A square metres. For n plates,
the useful area of each set is (n − 1)A square metres.

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A capacitor is made with seven metal plates connected as in Fig.

and separated by sheets of mica having a thickness of 0.3 mm
and a relative permittivity of 6. The area of one side of each
plate is 500 cm2 Calculate the capacitance in microfarads.
Sol: Using expression , we have n = 7, A = 0.05 m2, d = 0.0003 m
and εr= 6.

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

A p.d. of 400 V is maintained across the terminals of the

capacitor of previous example calculate
(a) the charge;
(b) the electric field strength or potential gradient;
(c) the electric flux density in the dielectric.

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Introduction to inductors

✓ Earlier we noted that capacitors store

energy by producing an electric field
within a piece of dielectric material.
✓ Inductors also store energy, in this case
it is stored within a magnetic field.
✓ Inductors have a number of response
characteristics similar to those of the
capacitor.
BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Inductance

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.
When the e.m.f. is induced in the same circuit as that in
which the current is changing, the property is called
self inductance, L When the e.m.f. is induced in a
circuit by a change of flux due to current changing in
an adjacent circuit, the property is called mutual
inductance, M. The unit of inductance is the henry, H.

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BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

A circuit has an inductance of one henry when an e.m.f.

of one volt is induced in it by a current changing at
the rate of one ampere per second Induced e.m.f. in a
coil of N turns.

where dΦ is the change in flux in Webers, and dt is

the time taken for the flux to change in seconds (i.e.
dΦ/dt is the rate of change of flux). Induced e.m.f. in a
coil of inductance L henrys,

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where dI is the change in current in amperes and dt is the time

taken for the current to change in seconds (i.e. dIi/dt is the rate
of change of current). The minus sign in each of the above two
equations remind us of its direction (given by Lenz’s law).

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Inductors

A component called an inductor is used when the property of

inductance is required in a circuit. The basic form of an inductor is
simply a coil of wire. Factors which affect the inductance of an
inductor include:
i) The number of turns of wire – the more turns the higher the
inductance.
ii) The cross-sectional area of the coil of wire – the greater the cross-
sectional area the higher the inductance.
iii) The presence of a magnetic core – when the coil is wound on an
iron core the same current sets up a more concentrated magnetic
field and the inductance is increased
(iv) The way the turns are arranged – a short thick coil of wire has a
higher inductance than a long thin one

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Examples of practical inductors

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Inductance of a coil

If a current changing from 0 to I amperes, produces a flux change

from 0 to Φ webers, then dI = I and dΦ= Φ. Then

Energy stored: An inductor possesses an ability to store energy.

The energy stored, W, in the magnetic field of an inductor is
given by:

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Inductors in Series

Leq = L1 + L2 + L3 + ... + LN
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Inductors in Parallel

1 1 1 1
= + ++
Leq L1 L2 LN
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Mutual Inductance

If two coils A and C are placed relative to each other as in Fig. then,
when S is closed, some of the flux produced by the current in A
becomes linked with C, and the e.m.f. induced in C circulates a
momentary current through galvanometer G. Similarly when S is
opened the collapse of the flux induces an e.m.f. in the reverse
direction in C. Since a change of current in one coil is accompanied
by a change of flux linked with the other coil and therefore by an
e.m.f. induced in the latter, the two coils are said to have mutual
inductance.

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The unit of mutual inductance is the same as for self-inductance,

namely the henry.
If two circuits possess a mutual inductance of M henrys and if the
current in one circuit – termed the primary circuit – increases
by di amperes in dt seconds, e.m.f. induced in secondary
circuit is

The induced e.m.f. tends to circulate a current in the secondary circuit in such a
direction as to oppose the increase of flux due to the increase of current in the
primary circuit.

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If dφ webers is the increase of flux linked with the secondary

circuit due to the increase of di amperes in the primary, e.m.f.
induced in secondary circuit is

where N2 is the number of secondary turns.

BITS Pilani, Deemed to be University under Section 3 of UGC Act, 1956

Examples

A flux of 25 mWb links with a 1500 turn coil when a current of 3

A passes through the coil. Calculate (a) the inductance of the
coil, (b) the energy stored in the magnetic field, and (c) the
average e.m.f. induced if the current falls to zero in 150 ms.
Soln:

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When a current of 1.5 A flows in a coil the flux linking with the
coil is 90 μWb. If the coil inductance is 0.60 H, calculate the
number of turns of the coil.
Soln:

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Thank You

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