Basic Electrical Engineering

Unit No. -1
ELECTROMAGNETISM
-----------------------------------------------------------------------------------------------------------------------------------------------------------
Review: EMF, Potential Difference, Current, temperature increases, the resistance also
Resistance increases.
Q.1 Distinguish between resistance and Resistivity: It is the property by virtue of which
resistivity and state the factors on which it opposes the flow of current.
resistance and resistivity depends.
Being property it is independent of physical
Ans: Resistance: It is defined on the actual dimensions. It can be measured by considering the
opposition to the flow of current through the
specimen of the same material.
material or substance.
It is denoted by a symbol (R) and it's unit is Factor governing the Resistivity:
ohm (Ω). The mathematical expression for 1. Temperature: As the temperature of the
resistance is,
material increases, it is found that resistivity
also increases.
Where, 2. Addition of Impurity: Resistivity also
= Resistivity of material (Ω m), changes by adding impurity in the material.
l = Length of material (m),
3. Cold Working: Resistivity also changes with
a = Cross-sectional area (m2)
the process of cold working.
Factors Governing the Resistance Value
4. Age Hardening: Due to age hardening, the
From the expression of resistance (Equation 1)
resistivity of the material also changes.
The resistance depends upon the following factors
1. Length : Resistance is directly
 Magnetic Circuit
proportional to length it means as the length of Q.2 Define the following terms related to
conductor increases, it's resistance also magnetic circuit
increases and vice-versa. (i) Magnetic flux (ii) Magnetic flux density (iii)
2. Cross-sectional area : Resistance is Magnetic field strength (iv)Reluctance (v)
inversely proportional to cross sectional area, Permeance (vi) MMF (vii) Permeability (viii)
it means as the cross-sectional area of Absolute permeability (ix)Relative
conductor increases, it's resistance decreases permeability
and vice versa. (i) Magnetic Flux ( ): The total number of lines
3. Type of material: The resistance of material of force existing in a particular magnetic field is
depends on the type of material used. called magnetic flux.
4. Temperature: As the temperature of the The unit of flux is Weber (wb) and flux is
material changes the resistance also changes. denoted by symbol ( ).
Generally, for conducting materials, as

1
(ii) Magnetic Flux Density (B): It can be defined It's unit is AT and corresponds to electromotive
as 'The flux per unit area (a) in a plane at right force (EMF) in an electric circuit.
angles to the flux is known as 'flux density'. (vii)Permeability: It is defined as ability or ease
Mathematically, ⁄ with which the magnetic flux permeates through a

The unit of flux density is , also called Tesla given medium.

(viii) Absolute permeability ( ): The ratio of
and denoted as T.
magnetic flux density (B) in a particular medium
(iii) Magnetic Field Strength/ Magnetizing
(other than vacuum or air) to magnetic field
Force/Magnetic Field Intensity (H)
strength (H) producing that flux density is called
It can be defined as 'the force experienced by a
absolute permeability of that medium.
unit N-pole (i.e. N-pole with 1 Wb of pole
strength), when placed at any point in a magnetic
field is known as magnetic field strength at that
Its unit is Henry per metre (H/m).
point.
(ix)Relative Permeability ( : Relative
For Straight conductor permeability of a material is equal to the ratio of
For Solenoid/ Toroid the flux density produced in that material to the
flux density produced in vacuum by the same
Its unit is Newton per Weber i.e. (N/Wb) or
magnetizing force (H).
Amperes per meter (A/m) or Ampere Turns per
B (Material)
meter (AT/m).
Bo(Vacuum)
(iv) Reluctance(S): The opposition offered to the
Q.3 Explain Series Magnetic Circuit
passage of magnetic flux through a material is Figure shows a composite series magnetic circuit
called it's reluctance and is analogous to consisting of three different magnetic materials
A, B, C of different permeabilities µr1, µr2 and µr3
resistance in an electric circuit.
and lengths l1, l2 and l3 and one air gap (µr = 1).
It's unit is AT/wb Each path will have its own reluctance.

(v) Permeance : It is reciprocal of reluctance and

is defined as ease or readiness with which
magnetic flux developed and is analogous to
conductance in an electric circuit.
Permeance It's unit is Wb/AT

(vi) Magneto motive Force (MMF): It is the

force required to produce flux in a magnetic The total reluctance is the sum of individual
circuit. reluctances as they are joined in series.

l
Total reluctance S T  
a

2
= reluctance of A + reluctance of B + reluctance The flux produced by the coil wound on central
of C + reluctance of air core is divided equally at point A between the two
outer parallel paths. Fig. 4.18 (b) shows the
= equivalent electrical circuit where the resistance
R
offered to the EMF. source is R || R = 2 .
as NI = Ø ST
 Flux  divides equally at point A
Total MMF=MMF for Part A, B & C + MMF of  Current I divides equally at point A
air gap The mean length of path ADB = l1 m
The mean length of path ACB = l2 m
NI=Ø (reluctance of part A, B & C+ reluctance of
The mean length of path AB = lc m
air)
AT
=Ø (ST) The reluctance of path ADB=S1 Wb
AT
= The reluctance of path ACB=S2 Wb
AT
NI AT The reluctance of path AB= Sc Wb
Also H= Or NI=H l
l m
Total m.m.f. produced=NI AT
m.m.f.
Total MMF= Flux = Reluctance m.m.f.= ·S1

For path ADBA, NI= Sc + 1 S2

For path ACBA, NI=  Sc + 2 S2
= where, ;

= Total m.m.f. = Sc + 1S1 = Sc + 2S2

Q. 5 Compare electrical and magnetic

circuits
NI=
Similarities
Q.4 Explain Parallel Magnetic Circuit
Figure shows the parallel magnetic circuit
consisting of three parallel magnetic paths ACB,
ADB and AB acted upon by the same MMF
Magnetic circuit Electric Circuit
Provides path for magnetic Provides path for electric
flux current
MMF (F) ; Amp Turn EMF (E) ; Volt

Flux (Ø); Wb Current (I); Amp

(a) S1 = S2 (b)Equivalent Electrical Circuit R1 = R2

Reluctance Resistance (

3
 1 1 small distance dx in dt seconds, across the right
Permeance = Conductance =
Reluctance Resistance
angle to the magnetic field. The area swept by the
Flux density; (wb/m )2 Current density;
(A/m2).
conductor is l dx.
Permeability Conductivity According to Faraday's law of electromagnetic
Kirchhoff's MMF and flux Kirchhoff's voltage and induction, EMF induced in the conductor is given
law is applicable to the current law is applicable to
by,
magnetic circuit. the electric circuit.
Dissimilarities
Magnetic Circuit Electric Circuit
Flux cut by conductor = Flux density x area swept
Flux does not actually flows The current actually flows
in the sense which current i.e. there is movement of by conductor = B ( l dx)
flows. electrons in electric circuit.
No magnetic insulator as Many insulators like air,
flux can pass through all the P.V.C., synthetic resin etc.
materials, even air. from which current cannot
pass.
Energy is required to create Energy must be supplied
If the conductor moves at an angle to the
the magnetic flux, but not continuously to maintain magnetic field then e.m.f. induced in the
required to maintain it. the flow of current.
Reluctance of a magnetic Resistance of an electric conductor,
circuit depends on flux (and circuit is constant and is
hence flux density). independent of the current
(or current density) as long The direction of the induced e.m.f. can be
as temperature is kept
determined by Fleming's right hand rule.
constant.
Example: D.C. Generator.

 Electromagnetic Induction 2. Statically induced EMF :

Q6. Explain the following terms: The EMF induced in a conductor when it links
1. Dynamically EMF 2. Statically induced EMF with time varying magnetic field without any
Ans: 1. Dynamically induced EMF : The e.m.f. relative physical movement with respect to
induced in a conductor due to the relative magnetic field.
physical movement with respect to steady Statically induced EMF is further divided into
magnetic field. following types:
Explanation: Self Induced EMF
The EMF induced in the primary winding of the
transformer is a good example of statically self
induced EMF
Explanation:

Consider conductor A of length 'l' mtr. as shown

below within a uniform magnetic field of B
wb/m2. Suppose the conductor moves through a

4
As shown in Fig., when the coil is carrying a 1. Self inductance
current, a magnetic field is produced through When current in the coil increases, the changing
changes. Hence the EMF is induced in the coil. magnetic field produced by the current links with
This is known as self induced EMF. The induced coil, hence according to Faraday's law's an EMF is
EMF always opposes the cause producing it. The induced in the coil. The EMF induced in the coil
EMF induced is given as opposes the cause producing it i.e. it opposes
increase in current in the coil. When current in the
coil decreases, the changing magnetic field again
Mutually Induced EMF
induces EMF in the coil which opposes decrease
EMF induced in a conductor when it links with
in current in the coil.
time varying magnetic field created by some
This property of the coil which opposes change in
other coil.
current through it is called as self inductance or
Example : EMF induced in the secondary winding
inductance of the coil.
of the transformer.
Explanation: Henry

2. Mutual inductance
A coil possesses an inductance whenever the flux
linking with it is changed. If the flux produced by
some another coil get linked with coil then the
inductance possessed by the coil is called as
Let us consider the two coils A and B placed mutual inductance.
adjacent to each other as shown in Figure. A part The mutual inductance is defined as it is flux
of flux produced by coil A links the coil B. If the linkage to one coil with respect to change in
current flowing through A changes, the flux current in other coil. It is denoted by M and
produced by coil B also changes. Hence the flux measured in Henry.
linking to the coil B also changes, thus EMF Henry
is induced in the coil B. The EMF induced in the
Q8 Derive Expression for following terms
coil B is called as mutually induced EMF. The
1. Self inductance 2. Mutual inductance
magnitude of mutually-induced EMF is given by
Self Inductance
(EMF induced

in second coil due to current change in first coil)

(EMF induced

in first coil due to current change in second coil)

A coil possesses an inductance whenever the flux
Q7. Define the following terms
linking with it is changed. If the own flux link
1. Self inductance 2. Mutual inductance

5
with the coil then the inductance possessed by the As 21 and
coil is called as self inductance. 1 i1
As shown in above figure, current I is responsible
2i1
for producing flux . Therefore, 2=K i1
i 2
K= I = constant
 = Ki 1
 2
K= I = constant 2= I · i1
1
The flux  can be written as,
d2 2 di1
= Differentiating, dt = I dt …(1)
1

 When 21 links with coil B, according to

= I i
Faraday's law a mutually induced e.m.f. e21 is
d  di induced in coil B given by :
Differentiating, dt = I dt …(1)
d2
When current i in the coil is changed EMF e21=– N2 dt
induced in the coil is given by,
–N22 di1
d From (1 )e21= I
E = – N dt 1 dt
di1
 di e21=– M dt …(A)
From equation (1), e= – N I dt
This is magnitude of mutually induced e.m.f.
di
e = –L
dt
…(A) N22
M= I
1
N
where, L = I …(B) =Mutual inductance
As 2 is part of 1
L  Self inductance of the coil
2 =K1 1
Mutual Inductance K1 indicate amount of flux linking with coil B
If the flux produced by some another coil get N2 K1 1
linked with coil then the inductance possessed by M= …(2)
I1
the coil is called as mutual inductance. Q 9. State the factors on which self
The mutual inductance is defined as it is flux inductance and mutual inductance depends.
linkage to one coil with respect to change in
The self and mutual inductance's are given as…
current in other coil. It is denoted by M and
measured in Henry. ,

1. Self and mutual inductance's are directly

proportional to number of turns of the coil.
2. Directly proportional to cross sectional area
of magnetic circuit.
3. Inversely proportional to length of magnetic
circuit.
Let, 1 : Flux produced by current I1 again
4. It is directly proportional to relative
called as self flux of coil A.
permeability (µr) of core. Coils having
2 : Part of 1 linking with coil B,
magnetic material are a core possesses large
again called as mutual flux.
inductance where as coils having non
6
 magnetic material as a core like air possesses From above equation coefficient of coupling is
less inductance. defined as it is a ratio of actual mutual inductance
5. As µr varies with flux density, the inductance
between the two coils to maximum possible mutual
varies with respect to flux density.
inductance between two coils.
Q10. Derive expression for coefficient of
Q11. Derive expression for energy stored in
coupling.
inductor per unit volume.

When the coil of inductance 'L' Henry is

The mutually induced EMF in coil B due to
current I1 is.. connected across supply, the lines of forces are
---(1) created.
The mutually induced EMF in coil A due to Due to the lines of force linking to the coil, EMF
current I2 is.. is induced in the coil. It is given as
---(2) di
Hence, the mutual inductance between the two e = – L dt
coils is given by ……… But e = –v (as induced e.m.f. is always opposed
---(3)
by the cause producing it as per Lenz's law)
---(4) di di
– v = – L dt , v = L dt … (1)
Multiplying equation (1) and (2)
Multiplying both sides of equation (1) by i dt we
have
v i dt = L i di

where, But v i dt be the electrical energy supplied to the

coil by source.
Taking square root
Total energy supplied by the source to the coil
M √
when current varies from 0 to I
Whenever there is 100% flux linkage between two
Energy stored =∫ =∫
coils, the mutual inductance between the two
coils is said to be maximum. ( ) [ ] Joule

If √ Energy stored per unit volume

Then coefficient of coupling Energy stored= (As )
---(5) Also
√

Therefore Energy stored=

7
( where a l is the volume) magnetizing coil of 200 turns produces a total flux
of 1.2 mWb in the iron. Calculate: i) Flux density
Energy stored per unit volume = Joule
in the iron ii) Absolute and relative permeability
List of Formulas of iron iii) Reluctance of the circuit
[1 Wb/m2, 0.002 H/m, 1590, 3.33x106 AT/Wb]

3. A mild steel ring of 30 cm mean circumference
 ;
has a cross sectional area of 6 cm2 and has a
 ; winding of 500 turns on it. The ring is cut through
 ; at a point so as to provide an air gap of 1mm in
the magnetic circuit. It is found that current of 4
 Ampere in the winding produces a flux density of
1T in the air gap. Find (i) Relative permeability of
 e =-N Volt; mild steel (ii) L of winding. [197.5, 0.075 H]
 e = ; 4. A closed magnetic circuit is composed of two
 sections. Section A has a length of 40 cm and
cross sectional area of 10mm2. Section B has a
 ; length of 50 cm and cross sectional area of
14mm2. Both the sections are made up of same
 material having permeability of 650. A coil with
400 turns is wound over one of the section. Find
 ; the current required in the coil so as to develop a
flux density of 1.4 Tesla in section B. [4.54 Amp]

5. A ring has a diameter of 21 cm and across
 √
sectional area of 10 cm2.The ring is made up of
 Energy Stored= Joule; semicircular sections of cast iron and cast steel
with each joint having reluctance equal to an air
 Energy Stored per unit volume = Joule gap of 0.2mm. Find the Ampere turns required to
produce a flux of 8 x 10-4Web. The relative
permeability of cast steel and cast iron are 800 and
Numericals for Practise: 166 respectively. [1783]
1. A coil of 500 turns and resistance 20 Ohm is
wound uniformly on an iron ring of mean 6. A rectangular iron core is shown in the fig. It
circumference 50 cm and cross-sectional area has a mean length of magnetic path of 100 cm,
4cm2. It is connected to a 24 V d. c. supply. Under cross section of (2cm x 2cm), relative
these conditions, the relative permeability of iron permeability of 1400 and an air-gap of 5 mm cut
is 800. Calculate the values of (i) Magnetomotive in core. The three coils carried by the core have
force of the coil (ii) Magnetising force (iii) Total number of turns N1 = 335, N2 = 600 and N3= 600;
flux in iron (iv) Reluctance of the ring [600AT, and the respective currents are 1.6 A, 4 A and 3 A.
1200 AT/m, 0.483mWb, 1.24x106 AT/Wb] The direction of currents are as shown. Find the
flux in the air-gap. [5.63µWb]
2. A magnetic circuit consists of an iron ring of
mean circumference 80cm with cross-sectional
area 12cm2 throughout. A current of 2A in

8
 13. Calculate the inductance of a ring shaped coil
having a mean diameter of 200 mm wound on a
wooden core of diameter 20 mm. The winding is
evenly wound and contains 500 turns.
If the wooden core is replaced by an iron core
which has relative permeability of 600 when the
current is 5 A, calculate the new value of
inductance. [157 mH , 94.2 mH]

14. Two identical coils P and Q each with 1500

2
7. An iron ring of 40 cm diameter and 7 cm cross turns, are placed in parallel planes near to each
section has an air gap of 2mm.it is uniformly other, so that 70% of the flux produced by current
in coil P links with coil Q. If a current of 4 A is
wound with 750 turns of wire and carries a current
passed through any one coil, it produces a flux of
of 3 Ampere. The iron takes 60 % of the total 0.04 mWb linking with itself. Find the self-
mmf. Find the (i) Total mmf ii) Flux iii) inductances of the two coils, the mutual
Reluctance iv) Flux density [2250AT, inductance and coefficient of coupling between
6 them. [15 mH, 15 mH , 10.5 mH , .7 V]
0.395mWeb, 5.696 x10 AT/ Web, 0564 T]
8. A 100 cm long straight conductor carrying 50 A
15. Two coils A and B in a magnetic circuit have
lies perpendicular to a uniform field of 1 Web/m2.
600 and 500 turns respectively. A current of 8
Find i) force on the conductor ii) power required Amp in coil A produces a flux of 0.04 wb. If the
to move the conductor at a uniform speed of 5 co-efficient of coupling is 0.2. Calculate (i) The
m/s. [50 N , 250 Watt] self inductance of coil A when B is open circuited,
(ii) Flux linkage with coil B, (iii) Mutual
9. A conductor carrying a current of 100 A at a inductance, (iv) Emf induced in B when flux
right angle to the magnetic field has a density of changes from zero to full value in 0.02 sec.
0.5 Tesla. Calculate the force on the conductor per [ 3 H , 4 Web-T ,0.5 H , -200 V]
meter length. Also find emf generated by a
16. Two coils A and B have self-inductances of
conductor in 1 second when the flux of 0.5 Web is 120 mH and 300 mH respectively. A current of 1
cut at a uniform rate. [50 N/m, 0.5 V] A through coil 'A' produces flux linkage of 100
µWb turns in coil 'B'. Calculate, (i)Mutual
10. A magnetic flux of 900 µWb passing through inductance between the coils.(ii) Average e.m.f.
a coil of 1000 turns is reversed in 0.2 Sec. induced in coil 'B' if current of 1A in coil 'A' is
Calculate average value of emf induced. [9 V] reversed at a uniform rate in 0.1 sec. Also find the
coefficient of coupling. [100 H, 0.2 mV, 0.527]
11. Calculate the inductance of a toroidal coil of
100 turns wound uniformly on a nonmagnetic core 17. A conductor of length 10 cm carrying 5 A is
placed in a uniform magnetic field of flux density
of mean diameter 140 mm and cross section area
1.25 tesla. Find the force acting on the conductor
of 750 mm2 [21.4 Micro-Henry] if it is placed (i) along the lines of magnetic flux,
(ii) perpendicular to the lines of flux, and (iii) at
12. Calculate the inductance of a toroid 25 cm 30° to the flux. [0, 0.625 N, 0.3125 N ]
mean diameter and cross section area 6.25 cm2
wound uniformly with 1000 turns of a wire. Also
calculate the emf induced when a current
increasing at the rate of 200 A/S flows in the
winding. [1mH , - 0.2 V]

9
 AC Fundamentals
Define the terms as related to alternating quantities. (1) Instantaneous value (2) Waveform (3)
Cycle (4) Periodic Time (5) Frequency (6) Amplitude
Instantaneous Value
Q.11: Define RMS Value and derive the
The value of an alternating quantity at a particular
expression for RMS value of sinusoidal
instant is known as its instantaneous value, e.g. i1, alternating current in terms of its peak value.
i2 and i3 shown in Fig. are the instantaneous values The RMS value of an alternating current is given by
of alternating current at different instances. that value of direct current which, when flowing
through a given circuit for a given time produces
Waveform
the same amount of heat as produced by alternating
The graph of instantaneous values of an alternating
quantity against time is called as its waveform. when flowing through the same circuit for the same
Figure shows the waveform of alternating current. time.
Alternating current can be expressed analytically as,
i=Im sin where, =t
Taking square of current,
i2=Im2sin2 


Fig. Waveform of Alternating Current

Cycle
Each repetition of a complete set of changes
undergone by the alternating quantity is called
cycle. These repetitions recur at equal intervals of
time.
Wave form for i2 is plotted for an half cycle. To find
Time Period (T)
the area under the curve (i2), an interval of d is
Time Period is the time taken by the alternating
consider at a distance  from origin.
quantity to complete one cycle. π

Area of squared curve over half cycle=  2

 i · d
Frequency (f) 0
The number of cycles completed per second by an The length of base is π. Therefore, mean value of
alternating quantity is known as frequency. square of current over half cycle,
1 Area of squared curve over half cycle
f  Hz = Length of base over half cycle
T


i
Amplitude 2
d 
The maximum value attained by an alternating 1
  i d
0 2
quantity during its positive or negative half cycle is   0
called as amplitude or peak value. 
1
 I sin 2 d
2

 0
m

2 
I 1  cos 
  d
m

 0 2
2
 sin 2 
 I  
m

Figure shows the amplitude of A.C. voltage. 2  2 

  sin 2  
2
It is denoted by Vm or I m , where Vm represents peak I
 m   0  


value of voltage and I m represents peak value of the 2   2  0 
current.
 2 Form factor
 I  m
The ratio of r.m.s value to average value of an
2
2 alternating quantity is called as form factor. Thus,
mean square value  I m for sinusoidal voltage and current :
2 R.M.S. value
Im Form factor, Kf = Average value
Root mean square value  Irms  I 
2 0.707  Maximum value
= = 1.11
Hence, root mean square i.e. r.m.s. value can be 0.637  Maximum value
calculated as, Irms=0.707 Im
Peak factor
Q.12: Define Average value and derive the The peak factor of an alternating quaintly is defined
expression for average value of sinusoidal as ratio of maximum value to the r.m.s. value. This
alternating current in terms of its peak value. factor may also be called as crest factor or
Average value of an alternating quantity is defined amplitude factor.
as the sum of all the instantaneous values over a For sinusoidal voltage and current,
given interval divided by the number of intervals of Maximum value
Peak factor, Kp= R.M.S. value
that interval.
Alternating current can be expressed analytically as, Maximum value
= =1.414
i  I m sin  0.707  Maximum value

Q.14: What is phase of an electrical quantity?

Explain the term in phase and out of phase of an
electrical quantity.
Phase
The fraction of the time period (T) that elapses in
achieving certain instantaneous value is known as
Figure shows one half cycles for instantaneous phase of that alternating quantity. For example, in
current. Here to find area under the curve, consider Fig. phase of the alternating quantity at point S is
an interval of d at a distance  from the origin. T/4 seconds or when expressed in terms of angle, it
Average value of current over a half cycle, is π/2 radians.
Area under the curve over half cycle
Iav= Length of base over half cycle

1
I avg 
  id
0


 i d
I avg  0

 Inphase Quantities
1
I avg 
 I
0
m sin  d

I avg 
Im
 cos  o

2I m
I avg   0.637 I m

Thus average value of sine wave is equal to 0.637
times the peak value. When two alternating quantities of the same
frequency attain their, corresponding values (e.g.
Q.13: Define form factor and peak factor
zero, positive maximum, etc.) simultaneously, they A is represented by the equation,
are said to be in phase with each other. eA  EmA sin   EmA sint 
The two alternating e.m.f.’s considered can be
then, the e.m.f. in the coil B will be represented by,
represented by the following equations:
eB  EmB sin(   )  EmB sin t   
eA  EmA sin   EmA sint 
Similarly, lagging alternating quantity is that which
eA  EmB sin   EmB sint  attains its zero or maximum value latter than the
Out of phase Quantities other quantity.
Two alternating quantities of the same frequency
which attain their corresponding values at different
instants are said to be out of those.
Leading alternating quantity as one which attains
its zero or maximum value earlier as compared with
the other quantity.
A is represented by the equation,
eA  EmA sin   EmA sint 
then, the e.m.f. in the coil B will be represented by,
eB  EmB sin(   )  EmB sin t   
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Unit 3: Single Phase AC Circuit (Q No. 1 and Q. No. 2)

Pure R, Pure L and Pure C AC Circuit
1. Prove that the voltage and current are in phase in case of purely resistive circuit. Also
find power consumed in circuit.
Ans: Consider a circuit which consists of pure resistance R only as shown in Fig.

Fig. Purely Resistive Circuit

Let, the applied alternating voltage be given by equation,

v = Vm sin t … (1)

As, voltage v is applied in a close loop so an alternating current will be set up in the circuit.
At any instant, the value of current is given by Ohm's law as,
v V sin wt Vm
i=R = m = R sin t
R
The current will be maximum when sin t becomes unity, i.e. I m  (Vm R) ,

i = (Vm R) sin t =Im sin t … (2)

Fig. Voltage and Current Waveforms & Phasor diagram

So in case of Resistive circuit, voltage and current obtains their zero and maximum value
at same instant. Hence they are in phase as shown in Fig.

Power
In a.c. circuits, power at any instant is given by the product of instantaneous voltage and
instantaneous current.
Instantaneous power, p = v  i

Where, for pure resistance, v = Vm sin t and i = Im sin t

1
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

p = Vm sin t  Im sin t
1 – cos 2 t
= Vm Im sin2 t = Vm Im  
 2 
Vm Im Vm Im
= 2 – 2 cos 2t

Thus, instantaneous power consists of a constant part (Vm I m ) 2 and a fluctuating part

[(Vm I m ) 2 ]Cos2t . The fluctuating part is a cosine curve of frequency double that of

voltage and current. For one complete cycle, average value of [ (Vm I m ) 2 ] Cos2t is zero.

Fig : Voltage, current and power waveforms

Average Power
Average power over a cycle,
 VmIm   VmIm  VmIm
P = Average of   – Average of  cos 2 (t) = –0
 2   2  2
Vm Im
P=  =V  I Watt
2 2

2. Prove that the current lags behind applied voltage by 900 in case of purely inductive
circuit. Also find power consumed in circuit.
Ans: Consider a circuit which consists of pure Inductance, L only as shown in Fig.

Fig. Purely Inductive Circuit

Wherever, an alternating voltage of v volt is applied to a purely inductive coil of inductance
L Henry, an alternating current starts flowing in the circuit. This current produces an

2
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

alternating (changing) magnetic field which links with the same coil to produce an emf of
self-induction which is given by,

 di
e=– L dt
 
As the circuit contains only L, therefore emf of self-induction will always oppose the
applied voltage as per Lenz's law,

 di di
 v = – e = – – L dt = L dt , Where v = Vm sin t ----(1)
 
di
Vm sin t = L ·
dt

L · di = Vm sin t ·dt

Vm
di = L sin t · dt
Integrating on both sides,
Vm
i= ∫ L sin t · dt
Vm  cos t
i= L – 
  
Vm
i= sin (t – π/2) … (2)
L
When sin(t – π/2) becomes unity, the current attains maximum value which is given by,
Vm
Im = where L is called as inductive reactance, XL.
L
i = Im sin (t – π/2) or i= Im sin (t – 90o) …(3)

Fig. Voltage and Current Waveforms & Phasor diagram

The current obtains zero value 90° after voltage. Thus, the current always lags behind
voltage by 90° as shown in figure.
Power
Instantaneous power, p = v · i
=Vm sin t · Im sin (t – π/2)=Vm Im sin t · (– cos t)
3
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Vm Im
= – Vm Im · sin t · cos t= – sin 2t
2

Fig. Voltage, current and power waveforms

Average Power
Average power over one complete cycle,
 Vm Im 
P = Average of – 2 sin 2t = 0
 
3 If a sinusoidal voltage of v = Vm sin ωt is applied across purely capacitive circuit,
derive the expression for current drawn and power consumed.
Ans: Consider a circuit which consists of pure Capacitance, C only as shown in Fig.

Fig. Circuit Diagram

When a pure capacitor is connected across an alternating voltage of v capacitor is charged

in one direction then in the opposite direction. Thus instantaneous charge q stored on the
plates of capacitor depends on that instant (time t). The instantaneous charge q on the
capacitor of capacitance value C is given by,
q=C·v ,
where v= Vm sint …..(1)

q= C · Vm sint

Let dq is the small charge which stored on a capacitor plate, in small time interval dt
second, when the instantaneous value of current is
dq d d
i = dt = dt (C · Vm sint) = C · Vm dt (sin t)

4
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

= C · Vm (cos t · ) = (C) · Vm · cos t

Vm Vm
= · cos t = · sin (t + π/2) .....(2)
1/(C) 1/(C)

Current reaches its maximum value Im, when sin (t + π/2) = 1

Vm
Im = = Vm · (C)
(1/C)
where term (1/C) is called as capacitive reactance XC
Vm
Substituting = Im in equation (2),
1/C

i = Im · sin (t + π/2) …..(3)

Fig. Voltage and Current Waveforms & Phasor diagram

The current obtains zero value 90° before voltage. Thus the current always leads ahead
the applied voltage by 90°.
Power
Instantaneous power, p = v · i
p= Vm sint · Im sin(t + π/2)= Vm Im sint · cos t

Vm Im
p= 2 sin 2t

Thus instantaneous power wave has double frequency as that of applied voltage and current
as shown in Fig.

Fig. Voltage, current and power waveforms

5
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Average Power

Vm Im 
Average power over a cycle = Average of  2 sin 2t = 0
 
4. Show that the average power taken by pure Inductor or pure Capacitor over a cycle
is zero.
Ans: For Pure Inductive circuit
Instantaneous power, p = v · i

p=Vm sin t · Im sin (t – π/2)=Vm Im sin t · (– cos t)

Vm Im
p= – Vm Im · sin t · cos t= – 2 sin 2t

Average power over one complete cycle,

2
 Vm Im   Vm Im 
P = Average of – 2

sin 2t =
 
0
–
 2 sin 2 t  =0


For Pure Capacitive circuit

Instantaneous power, p = v · i

p=Vm sin t · Im sin (t + π/2)=Vm Im sin t · ( cos t)

Vm Im
p= Vm Im · sin t · cos t= 2 sin 2t

Average power over one complete cycle,

2
 Vm Im   Vm Im 
P = Average of  2

sin 2t =
 0

 2
sin 2 t =0


5. What is Inductive and Capacitive reactance? Explain their dependency on frequency.

Ans: Inductive Reactance: It is the opposition made by Inductor to flow of an alternating
current.
The opposition is expressed as XL (Inductive
reactance) and is given by, X L  V I
Where, X L  2fL Ohm .
X L  f , When value of L (Self Inductance) is
constant.
6
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

The variation of XL with respect to change in frequency is

shown in figure. The higher the frequency, the greater the
inductive reactance; the lower the frequency, the less the
inductive reactance for a given inductor.

Capacitive Reactance: It is the opposition made by Capacitor to flow of an alternating

current.
The opposition is expressed as XC (Capacitive reactance)
and is given by, X C  V I
Where, X C  1 (2f C ) Ohm .
X C  1 f , When value of C (Capacitance) is constant.

The variation of XC with respect to change in frequency is

shown in figure. The higher the frequency, the less the
Capacitive reactance; the lower the frequency, the higher
the Capacitive reactance for a given Capacitor.

R-L, R-C and R-L-C series AC Circuit

6. If v=Vm sin ωt is applied across single phase circuit and current flowing through the
circuit is i=Im sin (ωt-Ф). Derive the expression for average power consumed in the
circuit. Draw voltage, current and power waveforms.
Ans: Let us consider the R-L circuit as shown in fig. Let v = Vm sin t

Fig. Circuit diagram

Now the current flowing through the circuit at any instant is i = Im (t – )

The instantaneous power is given as,

p = v.i
= Vm sin t . Im sin (t – ) = Vm Im sin t . sin (t – )

7
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Vm Im VmIm VmIm
= 2 [cos  – cos (2 t – ) ] = 2 cos  – 2 cos (2t – )
In the above expression first term is constant and the second term having the double of the
supply frequency.

Fig. Voltage, current and power waveforms

Average Power
VmIm VmIm
P = Average of [ 2 cos  – 2 cos (2t – )]

Hence the average power consumed over a cycle by second term is zero.

VmIm Vrms Irms

P = 2 cos  = . cos 
2 2
P = V. I. cos  Watt

7. If v=Vm sin ωt is applied across single phase circuit and current flowing through the
circuit is i=Im sin (ωt+Φ). Derive the expression for average power consumed in the
circuit. Draw voltage, current and power waveforms.
Ans: Let v = Vm sin t

Fig . Circuit diagram

The current flowing through the circuit at any instant is i = Im sin (t + )

The instantaneous power is given as,

p = v.i

8
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

= Vm sin t . Im sin (t + ) = Vm Im sin t . sin (t + )

Vm Im VmIm VmIm
= 2 [cos  – cos (2 t + ) ] = 2 cos  – 2 cos (2t + )

In the above expression first terms is constant and the second term having double of the
supply frequency, hence the average power consumed by the second term is zero.

Fig. Voltage, current and power waveforms

Average Power
VmIm VmIm
P = Average of [ 2 cos  – 2 cos (2t +)]

Hence the average power consumed over a cycle by second term is zero.
VmIm Vrms Irms
P= 2 cos  = . cos 
2 2
P = V. I. cos  Watt

8. What is Impedance? Draw Impedance Triangle for R-L and R-C Series circuit
Ans: Impedance: It is the opposition made by entire circuit to flow of an alternating current.
It is denoted by Z and its unit is ohm.
Z  V I , Z may be R-L/ R-C/ R-L-C combination

R-L Series Circuit R-C Series Circuit

Impedance Triangle

Z=R+jXL=Z∠Ф Z=R-jXC=Z∠-Ф

9
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

9. Explain concept of active, reactive and apparent power. Draw the power triangle for
R-L series or R-C series circuit.
Ans: Active Power or Real Power (P)
Active power is the power which is actually consumed in the circuit. It is the product of
RMS value of voltage and current. It is measured in watts or kW and its symbol is P.
P = V I cos Ф Watt

Reactive Power (Q)

Reactive power is not actually consumed by the circuit rather this power is taken and
returned back to the source by components like inductance or capacitance. Reactive power
is product of RMS value of voltage and quadrature component of current. Its symbol is Q
and unit is VAr or kVAr.
Q = V I sin Ф VAr

Apparent Power (S)

Apparent power is the total power available at source. This power is the product of RMS
value of voltage and current. Its symbol is S and unit is volt ampere or VA.
S = V I VA

Apparent power, S= (Real power)2 + (Reactive power)2 = P2 + Q2

R-L Series Circuit R-C Series Circuit

Power Triangle

Complex Form S=P+jQ=S∠Ф S=P-jQ=S∠-Ф

10. What is complex power? Explain its physical significance.

Complex power, S (in VA) is the product of the RMS voltage
phasor and the complex conjugate of the RMS current phasor.
As a complex quantity, its real part is real power P and its
imaginary part is reactive power Q.

S  V I *  P  jQ

10
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

For RL series circuit, position of voltage and current phasors

are shown in fig. Voltage and current in complex form…
V V1 and I  I2

S  V I *  V1 x I  2  V I 1  2 =P+jQ

Real Part, P= VI cos (Ф1-Ф2)
Imaginary Part Q= VI sin (Ф1- Ф2)

When S lies in the first quadrant, the load is inductive in nature

with lagging power factor.
Where ,Ф1- Ф2 >0
When S lies in the fourth quadrant, the load is capacitive in
nature with leading power factor.
Where, Ф1- Ф2 <0

11. Draw the phasor diagram for following condition

(a) XL > XC (b) XL < XC (c) XL = XC for R-L-C series circuit.
VL > Vc VL < VC VL = Vc

R-L-C series Resonance Circuit

12. Derive the condition for series resonance in series R-L-C circuit.
Ans: Consider a series circuit of resistance R, inductance L and capacitance

11
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

The total impedance of the circuit is given by Z= R2 + (XL – XC)2

Now, if the supply voltage across the circuit is maintained constant and frequency is
gradually increased from zero to a high value then inductive reactance, (XL=2 π fL), starts

 1 
increasing from zero while capacitive reactance XC = 2πfC decreases from its infinitely
 
large value as At a certain frequency fr, the two reactance’s becomes numerically equal i.e.
at this frequency, XL = XC

Let, fr be the resonant frequency. Then at this frequency,

XL= XC
1
2 π fr · L = 2π f · C
r
2 1
fr =
4π2 LC
1
fr = Hz
2π LC

13. State any four characteristics of series resonance. Show the variation of XL, XC, Z and I
against frequency.
Ans: i) Inductive reactance (XL) = Capacitive Reactance (XC)
ii) Impedance, Z=R + j (XL-XC)= R+j0=R. Hence Z reaches to minimum value
iii) Current I=V/Z=V/R. Hence the current reaches to maximum value
iv) As Z=R, the circuit behaves like pure resistive circuit. Hence pf (CosФ) becomes 1

Fig. Variation of XL, XC, Z and I with frequency

12
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Parallel AC Circuit
14. Define the terms (i) Admittance (ii) Conductance (iii) Susceptance.
Ans: (i) Admittance(Y): Admittance is defined as the reciprocal of the impedance. It is
denoted by Y and is measured in unit Siemens or Mho.
1
Y = Z Siemens

Conductance (G) : It is defined as the ratio of the resistance to the square of the
impedance. It is measured in unit Siemens or Mho.
R
G = 2 Siemens
Z
Susceptance (B) : It is the ratio of the reactance to the square of the impedance. It is
measured in the unit Siemens or Mho.
X
B = 2 Siemens
Z
14. What is admittance? What are its two components?
Admittance is defined as the reciprocal of the impedance. It is denoted by Y and is measured
in unit Siemens or Mho. Consider a circuit shown in the Fig. The total current is phasor sum
individual branch currents.
– – – –
I = I 1 + I 2 + I 3 and V = I Z
– – –
– V V V
I =– +– +–
Z1 Z2 Z3

–
V –1 1 1
– = V  – + – + – 
Z Z1 Z2 Z3
1 1 1 1
– =  – + – + – 
Z Z1 Z2 Z3
– – – –
Y =Y 1 + Y 2 + Y 3

Where, Y is the total admittance of the circuit.

Components of Admittance:
Consider an impedance Z is given by,

13
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Z = R  jX
Where +ve sign is for inductive reactance and – ve sign is for capacitive reactance,
1 1
Admittance, Y = Z =
R  jX
Rationalizing the above expression, we get,

1 – jX
R+ R+– jX R X R – X
Y=  – = 2 = 2 – j
+ = 2 + j 2
R  jX R + jX R +X 2 2
(R + X ) R + X2
2 Z Z

– j, R X
Y=G+ where, G = 2 is conductance and B = 2 is Susceptance
Z Z
15. Draw the admittance triangle for inductive circuit and capacitive circuit.
The triangle in which sides are representing Conductance, Susceptance and Admittance of
the circuit is known as admittance triangle.
R-L Series Circuit R-C Series Circuit

Admittance Triangle

Complex Form – –
Y = Y  – G – j BL Y = Y  G + j BC

14
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Unit 4:
Polyphase AC Circuit and Single Phase Transformer (Q No. 3 and Q. No. 4)
Part A: Polyphase AC Circuit
1. State the advantages of three phase ac system over single phase ac system.
Ans: There are several reasons why three-phase system is superior to single-phase system.
(i) The rating of three-phase motor and three-phase transformer are about 150% greater
than single-phase motor or transformer with a similar frame size.
(ii) The power delivered by a single-phase system pulsates. The power falls to zero, three
times during each cycle. The power delivered by a three-phase circuit pulsates also, but
it never falls to zero. So in three-phase system, power delivered to the load is same at
any instant. This produces superior operating characteristics for three-phase system.
(iii) To transmit certain amount of power at a given voltage over a given distance, three-
phase transmission line requires less amount of copper than single-phase line. This
reduces the cost of material required, hence, becomes economical.
(iv) Power factor of three-phase motor is greater than single-phase motor for same rating.
(v) Three-phase motors are self-starting, as the magnetic field produced by three-phase
supply is rotating. But the magnetic field produced by single-phase system is pulsating,
so most of the single-phase motors are not self-starting.
2. Define the following terms :
(i)Symmetrical System (ii) Phase sequence (iii) balanced load
Ans: (i) Symmetrical System: A three-phase system is said to be symmetrical when voltages of
same frequency in different phases are equal in magnitude and displaced from one another by
equal phase angles.

(ii) Phase sequence: A sequence in which three voltages will achieve their positive maximum
values is called phase sequence.

(iii) Balanced load: The load is said to be balanced when loads in each phase are equal in
magnitude and identical in nature.

3. Derive the relation between line and phase values of currents and voltages for balanced
three phase star connected (Resistive/ Inductive/ Capacitive) load.

15
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Ans: Consider the balanced star-connected load.

Fig. Circuit diagram

Line voltages, VL = VRY = VYB = VBR

Line currents, IL= IR = IY = IB
Phase voltages, Vph =VRN = VYN = VBN
Phase currents, Iph= IR = IY = IB
As path for line current, IL and phase current, Iph is same, IL = Iph
To derive relation between VL and Vph, let us consider line voltage VL = VRY.
—— —— ——
V RY =V RN + V NY
—— ——
as V NY = V YN
—— —— ——
Hence, V RY =V RN  V YN  (i)
—— —— ——
Similarly, V YB =V YN  V BN  (ii)
—— —— ——
V BR = V BN  V RN  (iii)
The phasor diagram will give relation between line voltage and phase voltage.

Fig. Phasor diagram for resistive load

As shown in Fig. take phase voltage VRN as reference. The three-phase voltages are displaced
by 120 from each other.
—— —— —— ——
The phasor VRY line voltage is addition of V RN and  V YN , to get  V YN ,V YN is reversed.

16
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

VL
The perpendicular is drawn from point A on phasor OC representing VL. OB = BC = 2
Angle between VRN and – VYN is 60.
So  AOB = 30 (OC bisects V ^ V )
RN YN

VRY
OB 2
From  AOB, cos 30 = OA = V
RN

VL
3 2
2 = Vph

VL = 3 Vph
Thus, line voltage is 3 times the phase voltage and line current and phase currents are same.

Phasor diagram for Inductive load

Phasor diagram for Capacitive load

4. Derive the relation between line and phase values of currents and voltages for balanced
three phase Delta connected (Resistive/ Inductive/ Capacitive) load.

17
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Consider the balanced delta-connected load.

Fig. Circuit diagram

Line voltages, VL = VRY = VYB = VBR

Line currents, IL = IR = IY = IB
Phase voltages, Vph=VRY = VYN = VBR
Phase currents, Iph=IRY = IYB = IBR
As seen earlier, VL = Vph for delta-connected load. To derive relation between IL and Iph,
apply KCL at the node R of the load as shown in Fig.
 current entering =  current leaving the node R
—— —— ——
IR + I BR = IRY
—— —— ——
IR = IRY  IBR  (i)
Similarly, at node Y and node B, we get
—— ——
IY = – IRY  (ii)
—— —— ——
IB + I BR =  IYB  (iii)
The phasor diagram will give relation between line current and phase current.

Fig. Phase diagram for resistive load

As shown in Fig., take phase voltage VRY as reference. Three-phase voltages are displaced by
120 from each other.

18
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Consider resistive load. Draw Iph in phase with Vph. Angle between IRY and  IBR is 60. OC
will bisect IRY ^ IBR.  AOB = 30 Draw perpendicular on OC representing IL.
IR IL
OB =OC = 2 = 2
IR IL
OB 2 2
cos 30 = OA = I = I
RY ph

IL
3 2
=
2 Iph

IL = 3 Iph
Line voltage VL appears across load. Hence the voltage across load, Vph is same as VL.
VL = Vph
Phasor diagram for Inductive load

Phasor diagram for Capacitive load

19
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

5. State the relations between line and phase values of voltages and currents in case of star
and delta connected three phase system.
Voltage Relation Current Relation
VL(Line Voltage) & IL(Line Current) &
Vph (Phase voltage) Iph=(Phase Current)
Star Connection VL  3 Vph I L  I ph

Delta Connection VL Vph I L  3 I ph

6. Prove that power taken by three phase delta connected balanced load is always three
times to power taken by three phase star connected balanced load.

For Star VL  3 Vph , I L  I ph For Delta VL Vph , I L  3 I ph

P(star) =3 Vph Iph cos  P(Delta) =3 Vph Iph cos 

VL Vph Vph
=3  Z  cos  =3 VL Z  cos 
3 ph ph

VL VL VL
=3  cos  =3VLZ cos 
3 3 Zph ph

VL2 VL2
= cos  =3 Z cos 
Zph ph

20
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Single Phase Transformer

7. Define transformer and explain its working principle
Transformer can be defined as the static device which transfers electrical energy from one
alternating current circuit to another circuit with desired change in voltage or current without
change in frequency.

Figure: Basic Transformer

Consider the two coils with N1 and N2 be the number of turns say 1 and 2 wound on simple
magnetic circuit. These coils are isolated from each other, and there is no electrical connection
between them. The coil which is connected across the supply voltage is called as primary
winding and the coil which delivers energy to the load is called as secondary winding. When
the supply voltage (V1) is applied across the coil 1, the current (I1) starts flowing through it.
This alternating current produces an alternating flux (Ф) in the magnetic core, which links the
N1 turns of coil 1 and induces an emf (E1) in it, by self-induction. Assuming it is an ideal
transformer, all the flux produced by coil 1 links the turns of coil 2. Thus, induces an emf (E2)
in coil 2 due to principle of mutual induction. As the coil 2 is connected to load, the alternating
current (I2) starts flowing through it. Thus the energy will be delivered to the load.

8. Compare core type and shell type transformer

Sr. No. Core Type Transformer Shell Type Transformer

It has double magnetic circuit.

1. It has single magnetic circuit.

21
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

2. Windings used in core type trans- Sandwich type windings are used.
former are cylindrical in form.
3. Core is surrounded by the winding. The windings are surrounded by the
core.
4. It is easy for repair and maintenance. It is difficult for repair and main-
tenance.
5. Natural cooling is good. Natural cooling is poor.

9. Explain different types of laminations used in transformer.

The magnetic core of the transformer is made up of laminations with a thickness of 0.35mm
to 0.5 mm to form frame required for Core type as well as shell type transformer. Laminated
magnetic core is used to reduce eddy current losses. The laminations are cut in the form of a
strip of T’s, U’s, L’s, I’s, E’s and I’s as shown in the figure.

Typical combination of laminations used to form a magnetic core of core type and shell type
transformer is shown below.

10. Derive the EMF equation of single phase transformer

When transformer primary winding is connected across the alternating current supply, the
current starts flowing through it. This alternating current produces alternating flux which links
with primary and secondary winding and induces an emf of E1 and E2 in it respectively.
Magnitude of E1 and E2 can be derived using following method
Let us consider the flux waveform, as shown below

22
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

According to the Faraday's law of electromagnetic induction the average emf get induced in
each turn.
d
Average emf induced in each turn = dt
where, d : be the change in flux and dt : be the time required for change in flux
Now, considering quarter cycle of the flux waveform.
d : m - 0 and dt : T/4
 Substituting this in above equation, average emf induced in each turn,
d m – 0 4 m
=
dt T/4 = T
1
But,Time period, T= f

d 4 m
 dt = 1/f = 4 mf
But the flux considered very sinusoidally with time, the emf induced is also sinusoidal in
nature.
RMS value
For pure sine wave: Hence, Form Factor = Average value = 1.11
 RMS value of emf induced in each turn,
= Average value  1.11
= 4 m f  1.11 = 4.44 m f volt
Total emf induced in primary winding with N1 number of turns
E1 = 4.44 m f N1 volt.
Similarly, emf in induced in the secondary winding with N2 turns due to mutual induction.
E2 = 4.44 m f N2 volt

11. What is KVA rating of transformer? Explain, why rating of transformer is expressed in
KVA.
KVA rating: It is the output given by transformer at rated voltage and rated frequency under
usual service conditions without exceeding the standard limits of temperature rise.
If I1 and I2 be the rated full load current and V1, V2 be the rated primary and secondary
voltages.

23
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Then kVA rating of transformer,

V1 I1 V2 I2
kVA rating = 1000 = 1000

The transformer rating is always expressed in kVA because:

The transformer is designed for a particular value of operating voltage and current for each of
the winding not for particular value of output power. The load connected across the secondary
side of the transformer may be lagging, leading or unity. Thus, for the same operating voltage
and current the out power can be different at different loading conditions. Hence, only the
operating voltage and current are specified for transformer. This operating voltage and current
are called as rated voltage and rated current of particular winding. Hence, product of rated
voltage and rated current is called as 'Volt-Ampere' rating of transformer. In large transformer
it is expressed in kVA i.e. Volt-Ampere divided by 1000.
12. What are the different types of losses taking place in the transformer? How these losses
are minimized. State the parts in which it takes place.
Since, the transformer is a static device and not a rotating machine, therefore friction and
windage losses are not present. The losses which takes place in a transformer are of two types
(i) Iron losses or core losses (constant losses), [takes place in transformer magnetic core]
(ii) Copper losses (variable losses). [Takes place in transformer copper windings]
(i) Iron Losses
These losses occurs due to the alternating flux in the transformer core. These losses consist
of : (a) Hysteresis loss, (b) Eddy current loss.
These losses remains constant at any load condition.
(a) Hysteresis Loss : When transformer core is subjected to a magnetic field, the molecules
in the material are forced to get aligned in the direction of applied magnetic field. If the applied
magnetic field is alternating in nature then, the molecules are forced to change the directions
with the same frequency of applied magnetic field. But the molecules are very much reluctant
to change their direction.
Hence, some energy is required in order to change their direction as per the applied
alternating magnetic field. This loss of energy is called as hysteresis loss. It is dissipated in
the form of heat.
It is given by empirical formula as,

24
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Hysteresis loss, Ph =  Bmax f v watt

1.6

where,  : stenmitz constant, Bmax: maximum flux density in the core.

f : frequency of alternating flux v : be the volume of core material.
(b) Eddy Current Losses: Due to the linking of alternating flux to transformer core, emf get
induced in the transformer core. It gives rise to circulating, current in the core. These
circulating currents are called as eddy currents. Now the every path of circulating current in
the core has some resistance which causes the loss of energy. The total loss of energy due to
the total eddy current is called as eddy current loss. It is also dissipated in the from of heat. It
is also given by an empirical formula.
2
Eddy current loss Pe=Ke Bmax f2 t2 v watt
Where, Ke : constant depends on the resistivity of core material
Bmax: maximum flux density, f : frequency of alternating flux
t : thickness of the lamination of the core, v : volume of core material
The flux density in the core remains practically constant from no load to full load as well as
supply frequency also remains constant therefore iron losses are also called as constant losses.
(i) Copper Losses
These losses occurs in the primary and secondary windings due to resistance of primary and
secondary winding.
Let I1 and I2 : the primary and secondary current.
R1 and R2 : the primary and secondary winding resistance.
2 2
Hence, Total copper loss = I1 R1 + I2 R2

13. With neat circuit diagram explain the direct loading test on single phase transformer
for finding the voltage regulation and efficiency.

Fig. Direct Loading Arrangement on Single Phase Transformer

25
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

Theory: The efficiency and regulation of transformer can be found by direct loading method.
The circuit diagram for direct loading method is as shown in Fig.

For finding the efficiency and regulation of transformer, the primary winding terminals are
connected across supply and a variable load is connected directly across secondary terminals
as shown above. The wattmeters i.e. W1 and W2 are inserted in the circuit diagram in order to
measure the power input and power output of the transformer. Ammeters and voltmeters are
used for measurement of current and voltage in the circuit. The load on transformer is varied
step by step and the readings are noted down.

This test is useful only for small transformer and not for large ratings transformer, because of
the non-availability of the load. The results obtained by this test are very accurate as the
transformer is directly loaded for a particular load.

Procedure:
(1) Make the connections as per the circuit diagram.
(2) At start switch off the load.
(3) Switch on the supply and slowly increase the voltage with the help of auto transformer.
(4) Adjust the rated voltage of transformer.
(5) Now slowly increase the load on secondary and note down the readings of ammeter,
voltmeter and wattmeter.
(6) Load the transformer up to the rated capacity of transformer or 25% more than the rated
capacity.
Observation Table :
Sr. No. I1 V1 W1 I2 V2 W2
1.
2.

Formulae :
W2
Efficiency : Efficiency of the transformer can be calculated as, ... %  = W  100
1

Voltage Regulation :
Voltage regulation of a transformer can be calculated as,
E2 – V2
% regulation =  100
E2

26
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

When the secondary is open the secondary voltage V2 = E2.

Hence, at start when the load is switched off regulation of transformer is zero. The subsequent
regulations are calculated by using the above formulae.
Graphs:
From the results obtained, curves are plotted for efficiency
and regulation against load current or output power as
shown in fig.
The efficiency and regulation at any desired load can be
found from these curves.

14. Write a short note on Autotransformer.

An auto transformer is one in which single winding is used as primary and secondary winding.
It can be used as step up or step down transformer. The step down and step up transformers
are as shown in Fig. (a) and (b).

(a) Step Down transformer (b) Step up transformer

As shown in Fig. (a), the winding XZ forms the primary winding of the transformer having
N1 number of turns. The winding YZ forms the secondary winding having N2 number of turns.
Similarly, in Fig. (b) the portion XZ forms the secondary and YZ forms the primary winding.
If the transformer losses are neglected, then the same relationship holds good as in two
winding transformer.
V2 N2 I1
i.e. K = = =
V1 N1 I2
Advantages:
(1) Copper required in case of auto transformer is always less than the two winding
transformer, it is always cheaper.
(2) For same rating, weight of auto transformer is less than two winding transformer.
(3) The copper losses taking place in a transformer are less.

27
 SCOE-BEE End-Sem Notes | Dr. A. N. Sarwade

(4) Due to less copper loss, efficiency of the transformer is higher than that of two winding
transformer.
(5) Auto transformer has better voltage regulation than that of two winding transformer.
Disadvantages:
(1) There is always risk of electric shock, as the primary and secondary are not
electrically separated.
(2) In case of step down auto transformer, if the common part gets opened due to any
fault, the high voltage on primary side will damage the measuring instrument
(typically voltmeter) connected on secondary side.
Applications:
(1) It can be used as starter for squirrel cage induction motor.
(2) It can be used as booster to raise the voltage in A.C. feeders.
(3) It can be used in industry as furnace transformers for getting required voltage.
(4) It can be used as dimmer for dimming the light.

28