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 =V
YB 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 I ^ I  AOB = 30 Draw perpendicular on OC representing I .
RY BR. L

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 asVL.
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 Iphcos P(Delta)=3 Vph Iphcos

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 anemf(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. CoreType Transformer ShellType 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 the 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 emfget 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 mf 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 N2turns 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,

V1I1 V2I2
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.

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

It is given by empirical formula as,

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.

Questions asked in End-Sem Dec 2019 Examination

Q. Define (i) phase sequence (ii) balanced and unbalanced load. [3M] (Refer Q. No. 2)
Q. Derive the emf equation of 1-phase transformer. [6M] (Refer Q. No. 10)
Q. Why are steel laminations used for construction of transformer core? Sketch different types of
laminations used for core. [3M] (Refer Q. No. 09)
Q. What are losses taking place in the transformer? State the parts in which they takes place.
How to minimize these losses? [6M] (Refer Q. No. 12)
Q. Obtain the relation between phase values and line values of voltage and current in case of
balanced star connected 3-ph inductive load. Assume phase sequence RYB. Draw the
necessary phasor diagram. [8M] (Refer Q. No. 03)

28