1

EM-Problems Second Semester 2018-2019

Problem-1

In the magnetic system of fig. below two sides are thicker than the other two sides. The depth of the

core is 10 cm, the relative permeability of the core, μr = 2000, the number of turns N = 500, and the

current flowing through the coil

is i = 1 A.

a- Determine the flux in the

core.

b- Determine the flux

densities in the parts of the

core.

c- Find the current i in the coil to produce a flux (ϕ = 0.012 Wb).

Solution:

𝑅𝑇ℎ𝑖𝑐𝑘 = 70∗ 10−2

2000∗4∗𝜋∗10−7∗15∗10∗10−4= 18568 [

𝐴𝑡

𝑊𝑒𝑏]

𝑅𝑇ℎ𝑖𝑛 = 80∗ 10−2

2000∗4∗𝜋∗10−7∗10∗10∗10−4= 31831 [

𝐴𝑡

𝑊𝑒𝑏]

𝑇ℎ𝑒𝑛, 𝑅𝑇ℎ𝑖𝑐𝑘 + 𝑅𝑇ℎ𝑖𝑛 = 18568 + 31831 = 50399 [𝐴𝑡

𝑊𝑒𝑏]

𝑇ℎ𝑒𝑛, 𝜙 = 500∗1

50399= 0.00992 [𝑊𝑒𝑏]

𝐵𝑇ℎ𝑖𝑐𝑘 = 0.00992

150∗ 10−4= 0.6614 [𝑇]

𝐵𝑇ℎ𝑖𝑛 = 0.00992

100∗ 10−4= 0.992 [𝑇]

𝐵𝑇ℎ𝑖𝑐𝑘 = 0.012

150∗ 10−4= 0.8 [𝑇]

𝐵𝑇ℎ𝑖𝑛 = 0.012

100∗ 10−4= 1.2 [𝑇]

𝐻𝑇ℎ𝑖𝑐𝑘 = 0.8

2000∗4∗𝜋∗ 10−7= 318.3 [

𝐴𝑡

𝑚]

𝐻𝑇ℎ𝑖𝑛 = 1.2

2000∗4∗𝜋∗ 10−7= 477.46 [

𝐴𝑡

𝑚]

𝐹 = 𝐻 ∗ 𝑙 = (318.3 ∗ 2 ∗ 35 ∗ 10−2) + (477.46 ∗ 2 ∗ 40 ∗ 10−2) = 604.8 [𝐴𝑡]

𝑇ℎ𝑒𝑛, 𝑖 = 𝐹

𝑁=

604.8

500= 1.2 [𝐴]

2

Problem-2

A circular ring of magnetic material has a mean length of 1.0 m and a cross-sectional area of 0.001

m2. A saw cut of 5 mm width is made in the ring. Calculate

the magnetizing current required to produce a flux of 1.0

mWb in the air- gab if the ring is wound uniformly with a

coil of 200 turns. Take relative permeability of the ring

material = 500 and neglect leakage and fringing.

Solution:

𝐵 = 𝜙

𝐴=

1∗10−3

0.001= 1 [

𝑊𝑏

𝑚2]

The ampere -turn for air gap is:

𝐹(𝑔) = 𝐵

𝜇𝑜 𝑙 =

1

4∗𝜋∗10−7∗ 5 ∗ 10−3 = 3979 [𝐴𝑡]

The ampere -turn for core material is:

𝐹(𝐶) = 𝐵

𝜇𝑟 𝜇𝑜 𝑙 =

1

500∗4∗𝜋∗10−7∗ 1 = 1591 [𝐴𝑡]

𝐹(𝑇𝑜𝑡𝑎𝑙) = 𝐹(𝑔) + 𝐹(𝐶) = 3979 + 1591 = 5570 [𝐴𝑡]

𝐸𝑥𝑐𝑖𝑡𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖 = 𝐹

𝑁=

5570

200= 𝟐𝟕. 𝟑𝟓 [𝑨]

Another solution

𝐴𝑠𝑠𝑢𝑚𝑒 𝕽𝒈 𝑎𝑛𝑑 𝕽𝑪 𝑎𝑟𝑒 𝑡ℎ𝑒 𝑟𝑒𝑙𝑢𝑐𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑎𝑖𝑟 𝑔𝑎𝑏 𝑎𝑛𝑑 𝑐𝑜𝑟𝑒 𝑟𝑒𝑠𝑝𝑒𝑐𝑡𝑖𝑣𝑒𝑙𝑦. 𝑠𝑜;

ℜ𝑔 = 𝑙𝑔

𝜇𝑜𝐴𝑔=

5∗ 10−3

4∗ 𝜋∗ 10−7∗0.001= 3.98 ∗ 106 [

𝐴𝑡

𝑊𝑒𝑏]

ℜ𝐶 = 𝑙𝐶

𝜇𝑟 𝜇𝑜𝐴𝐶=

1

500∗4∗ 𝜋∗ 10−7∗0.001= 1.59 ∗ 106 [

𝐴𝑡

𝑊𝑒𝑏]

The total reluctance is

ℜ = ℜ𝑔 + ℜ𝐶 = 3.98 ∗ 106 + 1.59 ∗ 106 = 5.57 ∗ 106 [𝐴𝑡

𝑊𝑒𝑏]

𝑇ℎ𝑒𝑛 𝑎𝑚𝑝𝑒𝑟 𝑡𝑢𝑟𝑛 [𝐹] 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑 𝑖𝑠 𝜙 ∗ ℜ = 0.001 ∗ 5.57 ∗ 106 = 5570 [𝐴𝑡]

𝐸𝑥𝑐𝑖𝑡𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖 = 𝐹

𝑁=

5570

200= 𝟐𝟕. 𝟑𝟓 [𝑨]

3

Problem-3

A ring of mean diameter 21 cm and cross-section 10 cm2 is made up of semi-circular sections of

cast steel and cast iron. If each joint has reluctance equal to an air gap of 0.2 mm as shown in fig.

below.

Find the amp. Turn [F] required to produce a flux of

5 * 10-4 weber in the magnetic circuit. Take μr, for

steel and iron as 825 and 165 respectively. Neglect

leakage and fringing.

Solution:

𝜙 = 5 ∗ 10−4 [𝑤𝑒𝑏] ; 𝐴 = 10 𝑐𝑚2 = 10−3 [𝑚2]

𝐵 = 𝜙

𝐴=

5∗10−4

10−3= 0.5 [

𝑊𝑏

𝑚2]

𝐻 = 𝐵

𝜇𝑜=

0.5

4∗𝜋∗ 10−7= 3.977 ∗ 105 [

𝐴𝑡

𝑚]

Air gab length is = 0.2 * 2 = 0.4 mm = 4 * 10-4 m

Then, ampere-turn [F] required = 3.977 ∗ 105 ∗ 4 ∗ 10−4 = 159 [𝐴𝑡]

𝐶𝑎𝑠𝑡 𝑠𝑡𝑒𝑒𝑙 𝑝𝑎𝑡ℎ 𝐻 = 𝐵

𝜇𝑜 𝜇𝑟=

0.5

4∗ 𝜋∗ 10−7∗825= 482 [

𝐴𝑡

𝑚]

𝐶𝑎𝑠𝑡 𝑠𝑡𝑒𝑒𝑙 𝑝𝑎𝑡ℎ 𝑙𝑒𝑛𝑔𝑡ℎ = 𝜋 𝐷

2=

𝜋∗21

2= 33 𝑐𝑚 = 0.33 𝑚

AT [F] required = H * l = 482 * 0.33 = 159 [At]

𝐶𝑎𝑠𝑡 𝑖𝑟𝑜𝑛 𝑝𝑎𝑡ℎ 𝐻 = 𝐵

𝜇𝑜 𝜇𝑟=

0.5

4∗ 𝜋∗ 10−7∗165= 2411 [

𝐴𝑡

𝑚]

𝐶𝑎𝑠𝑡 𝑖𝑟𝑜𝑛 𝑝𝑎𝑡ℎ 𝑙𝑒𝑛𝑔𝑡ℎ = 𝜋 𝐷

2=

𝜋∗21

2= 33 𝑐𝑚 = 0.33 𝑚

AT [F] required = H * l = 2411 * 0.33 = 795.6 [At]

Then the total ampere -turn [F] required = 159 + 795.6 = 1113.6 [At]

4

Problem-4

Two coils are wound on a toroidal core as shown in fig. below. The core is made of silicon sheet

steel and has a square cross section. The

coil currents are i1 = 0.28 A and i2 = 0.56

A.

a- Determine the flux density at the

mean radius of the core.

b- Assuming constant flux density

(same as at the mean radius) over

the cross section of the core,

determine the flux in the core.

c- Determine the relative permeability.

Solution:

The two mmf’s aid each other. Then,

mmf = (600*0.28) + (300*0.56) = 336 [At]

𝐻 = 336

2 𝜋∗20∗ 10−2= 267.4 [

𝐴𝑡

𝑚]

B ≈ 1.14 T from the curve

𝜙 = 𝐵 ∗ 𝐴 = 1.14 ∗ (2 ∗ 2 ∗ 10−4) = 0.000456 [𝑤𝑒𝑏]

𝜇𝑟 = 𝐵

𝜇𝑜 𝐻=

1.14

4∗ 𝜋∗ 10−7∗267.4= 3393

267.4

1.14

5

Problem-5

The magnetic circuit of fig. below provides flux in the two air gaps. The coils (N1 = 700, N2 = 200)

are connected in series and carry a current of 0.5

ampere. Neglect leakage flux, reluctance of the

iron (i.e. infinite permeability), and fringing at the

air gaps.

- Determine the flux and flux density in the

air gaps.

Solution:

mmf’s of the two coils oppose each other then,

𝐴𝑔1 = 𝐴𝑔2 = 2.5 ∗ 2.5 ∗ 10−4 = 6.25 ∗ 10 − 4 𝑚2

𝑅𝑔1 = 0.05 ∗ 10−2

4∗ 𝜋∗ 10−7 6.25∗ 10−4= 0.637 106 [

𝐴𝑡

𝑊𝑒𝑏]

𝑅𝑔2 = 0.1 ∗ 10−2

4∗ 𝜋∗ 10−7 6.25∗ 10−4= 1.274 106 [

𝐴𝑡

𝑊𝑒𝑏]

𝜙𝑔1 = 𝑁∗𝑖

ℜ=

(700−200)∗0.5

0.637∗ 106= 0.392 ∗ 10−3 [𝑊𝑒𝑏]

𝜙𝑔2 = 𝑁∗𝑖

ℜ=

(700−200)∗0.5

1.274∗ 106= 0.196 ∗ 10−3 [𝑊𝑒𝑏]

𝜙𝑡𝑜𝑡𝑎𝑙 = 𝜙1 + 𝜙2 = 0.588 ∗ 10−3 [𝑊𝑒𝑏]

𝑇ℎ𝑒𝑛, 𝐵𝑔1 = 𝜙

𝐴=

0.392∗ 10−3

6.25∗ 10−4= 0.627 [

𝑤𝑒𝑏

𝑚2]

𝐴𝑛𝑑, 𝐵𝑔2 = 𝜙

𝐴=

0.196∗ 10−3

6.25∗ 10−4= 0.3135 [

𝑤𝑒𝑏

𝑚2]

Lg1 = 0.05 cm, Lg2 = 0.1 cm, L1 = L2 = L4, = L5 = 2.5

cm, L3 = 5cm, depth of core = 2.5 cm

Red

6

Problem-6

A ferromagnetic core is shown in Figure below. Three sides of this core are of uniform width, while

the fourth side is somewhat thinner. The depth of the core (into the page) is 10 cm, and the other

dimensions are shown in the figure. There is a

200- turn coil wrapped around the left side of

the core. Assuming relative permeability µr of

2500, how much flux will be produced by a 1-A

input current?

Solution

Three sides of the core have the same cross-

sectional areas, while the fourth side has a

different area. Thus, the core can be divided into

two regions: (1) the single thinner side and (2)

the other three sides taken together. The

magnetic circuit corresponding to this core is shown

in the Figure.

The mean path length of region l is 45 cm, and the

cross-sectional area is 10 * 10 cm = 100 cm2 .

Therefore, the reluctance in the first region is

The mean path length of region 2 is 130 cm, and the cross-sectional area is 15*10 (cm = 150 cm2.

Therefore, the reluctance in the second region is

7

8

Problem-7

The Figure below shows a ferromagnetic core whose mean path length is 40 cm. There is a small gap

of 0.05 cm in the structure of the otherwise whole core. The cross-sectional area of the core is 12

cm2, the relative permeability of the core is 4000, and the coil of wire on the core has 400 turns.

Assume that fringing in the air gap increases the effective cross-sectional area of the air gap by 5

percent.

Given this information, find:

(a) the total reluctance of the flux path

(iron plus air gap)

(b) the current required to produce a flux

density of 0.5 T in the air gap

Solution

9

The effective area of the air gap is 1.05 *12 cm2 = 12.6 cm2, so the reluctance of the air gap is

Note that the air gap contributes most of the reluctance even though it is 800 times shorter

than the core.

10

Problem-8

The Figure below shows a simplified rotor and stator for a de motor. The mean path length of the

stator is 50 cm, and its cross-sectional area is 12 cm2, the mean path length of the rotor is 5 cm, and

its cross-sectional area also may be assumed to be 12 cm2, each air gap between the rotor and the

stator is 0.05 cm wide, and the cross-sectional

area of each air gap (including fringing) is 14

cm2, the iron of the core has a relative

permeability of 2000, and there are 200 turns of

wire on the core. If the current in the wire is

adjusted to be 1 A, what will the resulting flux

density in the air gaps be?

Solution

To determine the flux density in the air gap, it is necessary to first calculate the magneto motive force

applied to the core and the total reluctance of the flux path. With this information, the total flux in the

core can be found. Finally, knowing the cross-sectional area of the air gaps enables the flux density

to be calculated. The reluctance of the stator is

11

12

Problem-9

Find the relative permeability of the typical ferromagnetic material whose magnetization curve is

shown in Figure below at (a) H = 50, (b) H = 100, (c) H = 500, and (d) H=1000 A • turns/m.

Solution:

13

14

Problem-10

Calculate the line currents in the three-wire Y - Y system as shown below.

Solution:

Phase ‘a’ equivalent circuit

−=

=

=++−==

21.86.8121.816.155

0110I

8.21155.16)810()25(Z;Z

VI

Aa

T

T

ANAa jj

A2.986.811.8266.81

024II

A141.86.81

120II

AaCc

AaBb

=−=

−=

−=

−=

15

Problem-11

A balanced delta connected load having an impedance 20 - j15 is connected to a delta connected,

positive sequence generator having VAB = 3300 V. Calculate the phase currents of the load and the

line currents.

Solution:

V 0330V

87.3625 j1520Z

AB

Δ

=

−=−=

A87.15613.2120II

A13.83-13.2120II

A36.8713.238.8725

0330

Z

VI

abca

abbc

Δ

abab

=+=

=−=

=−

==

( )( )

A 87.12686.22120II

A 13.311-86.22120II

A 87.686.22

30336.8713.2

303II

AaCc

AaBb

abAa

=+=

=−=

=

−=

−=

16

Problem-12

A balanced Y - connected load with a phase impedance 40 + j25 is supplied by a balanced, positive-

sequence Δ-connected source with a line voltage of 210V. Calculate the phase currents. Use VAB as

reference.

Solution:

The load impedance, ZY and the source voltage, VAB are

V 0210V

3217.47 j2540Z

AB

Y

=

=+=

When the ∆ - connected source is transformed to a Y - connected source,

آخر الذي هو وفي تشبيكة النجمة فرق الجهد بين الخط والمحايد أ ي فرق الجهد على الطور قيمته تساوي في تشبيكة الدلتا، فولت 210هنا حول الدلتا لنجمة فأ صبح فرق الجهد بين الخط والخط ال

آخر .درجة 30وية مقسوم على جذر الثلاثة ويتأ خر عنه بزافرق الجهد بين الخط والخط ال

A 582.57120II

A 182-2.57120II

A 62-2.573247.17

03121.2

Z

VI

AaCc

AaBb

Y

anAa

=+=

=−=

=

−==

V 30-121.2

3013

0210

303

VV AB

an

=

−

=

−=

17

Problem-13

Determine the total power (P), reactive power (Q) and complex power (S) at the source and at the

load.

Solution:

Phase ‘a’ equivalent circuit Known quantities

A 21.86.81

21.816.155

0110I

ZZ

VI

A

Yline

ANA

−=

=

+=

VAR 1113Q W,1392P

j1113)VA(1392

ZI3S

LL

2

φLoad

==

+=

=

Vg =V

AN= 1100 V

ZY = 10 + j8

Zline

= 5 - j2

VAR 834.6Q W,2087P

j834.6)VA(2087

I3VS

ss

φφsource

−=−=

+−=

−=

18

Problem-14

A three-phase motor can be regarded as a balanced Y - load. A three-phase motor draws 5.6 kW

when the line voltage is 220 V and the line current is 18.2 A.

Determine the power factor of the motor

Known Quantities

• PLoad = 5600 W

• VL = 220 V

• IL = 18.2 A

Solution:

VA 6935.13

IV 3

I3VS

LL

φφ

=

=

=

0.86935.13

5600

S

Pθ cos

θ cosSP

===

=

|S| Q

P

19

Problem-15

For the Y - connected load in Figure

a) find the average power to each phase and

the total load

b) determine the reactive power to each

phase and the total reactive power

c) find the apparent power to each phase

and the total apparent power

d) find the power factor of the load

Solution:

lagging

FP

6.0

VA 6000

W3600

S

P

T

T

=

=

=

W1200

13.53cos20100

cos

=

=

= V

IIVP

W3600W120033 === PPT

VAR1600

13.53sin20100

sin

=

=

= V

IIVQ

VAR4800160033 === QQT

VA2000

20100

=

=

= IVS

VA6000200033 === SST

20

Problem-16

Each transmission line of the 3 wires, three phase system in Figure below has an impedance of

15 Ω + j 20 Ω. The system delivers a total power of 160kW at 12,000V to a balanced three-phase

load with a lagging power factor of 0.86.

a. Determine the magnitude of the line voltage EAB of the generator.

b. Find the power factor of the applied to the generator.

c. What is the efficiency of the system?

Solution:

Vø (load) = V 42.69361.73

V12000

3==LV

PT (load) = 3 V

ø I

ø cos θ

Since θ = cos-1

0.86 = 30.68o

(lagging)

= 0V V

( )( )A 94.8

86.042.69363

W160000

cos3===

V

PI T

−= 68.30A94.8I

a. total load

فرق الجهد هنا يساوي

فولت 6936.42

التيار هنا يساوي

وهو نفسه تيار الخط ل ن التشبيكة نجمة

−= 68.30A94.8I

21

، ففرق الجهد على المصدر يساوي فرق " مكونة من فرق جهد المصدر ومعاوقة خط النقل ومعاوقة الحمليعني فاز واحدهنا دائرة مكافئة فقط لطور واحد "

الحمل + الهبوط في خط النقل *** ملاحظة التيار هو نفسه تيار الخط المار في معاوقة خط النقل والخارج من المصدر الجهد على

فولت وعند حساب فرق الجهد بين 143.57يساوي هذا هو فرق جهد المصدر لكل طور

آخر يجب ضربه في جذر الثلاثة ل ن التشبيكة نجمة فيصبح فولت 12358.26الخط وال

هذه القدرة الكلية الفعالة للنظام مكونة من القدرة

المس تهلكة في الحمل والفاقد في خط النقل

22

Problem-17

A three-phase transformer has 500 primary turns and 50 secondary turns. If the supply voltage is 2.4

kV find the secondary line voltage on no-load when the windings are connected

(a) star-delta

(b) delta-star.

Solution:

23

Problem-18

For the circuit shown in Fig. below:

(a) Determine the rms value of the supply

current.

(b) Express the supply voltage v and

supply current i in the time domain.

(c) Determine the power from the supply

and the supply power factor.

(d) Draw the phasor diagram for V, I, Vr,

VL, and VC.

Solution:

24

The phasor diagram is shown in fig. below

Note: V = VR + VL + VC

هذه زاوية الطور التي تقع بين

فرق الجهد والتيار

25

Problem-19

In the three-phase balanced system shown in Figure. below, the load is Z = 8 + j6 Ω and consider

𝑽𝒂𝒃 = 𝟐𝟎𝟎∠𝟎𝒐 𝑽 (rms voltage) and phase

sequence a, b, and c.

(a) Determine the phase currents in

phasor (Iab, Ibc, and Ica).

(b) Determine the line currents in phasor

(Ia, Ib, and Ic).

(c) Determine the total three-phase

power.

(d) Draw the phasor diagrams of currents and voltages.

Solution:

26

The phasor diagram is shown in fig. below

درجة 30متأ خر عن تيار الطور بزاوية قيمتها تيار الخط اهذ

( وتساوي 30قيمتها )ثيتا + ومتأ خر عن الفولت بزاوية

درجة 66.9-( = 36.9+30)

لطور متأ خر عن الفولت تيار ا اهذ

درجة 36.9بزاوية ثيتا قيمتها

آخر آخر والفولت وال ملاحظة يمكن تطبيق ما س بق على جميع الجهود والتيارات فقط ال خذ بعين الإعتبار أ نه يوجد دائما بين التيار وال

درجة 120في نظام الثلاثي الطور

27

Problem-20

Three 1φ, 50 kVA, 2300:230 V, 60 Hz transformers are connected to form a 3φ, 4000:230 V

transformer bank.

The equivalent impedance of each transformer referred to low voltage is 0.012 + j0.016 Ω.

The 3φ transformer supplies a 3φ, 120 kVA, 230 V, 0.85 PF (lag) load.

(a) Draw a schematic diagram showing the transformer connection.

(b) Determine the transformer winding currents.

(c) Determine the primary voltage (line-to-line) required.

(d) Determine the voltage regulation.

Solution

(a)

The connection diagram is shown in Fig. below. The high-voltage windings are to be connected in

wye so that the primary can be connected to the 4000 V supply. The low voltage winding is connected

in delta to form a 230 V system for the load.

فولت ويراد ربط المحولت الثلاثة كنظام ثلاثي الطور على 230فولت والثانوي 2300هنا نظام مكون من ثلاثة محولت أ حادي الطور كل محول لديه في جهة الإبتدائي

بتدائي و ربط ملفات الإيتدائي كنجمة والثانوي كدلتا فبذلك نأ خذ في جهة الإيتدائي ، فالطريقة الصحيحة لذلك هفولت 230فولت والثانوي 4000أ ن تكون جهة الإ

فولت ل نها دلتا 230فولت وجهة الثانوي تبقى بنفس القيمة 4000 ≈ 1.732* 2300* جذر الثلاثة( 2300)

28

The primary equivalent circuit is shown in fig. below

هنا مطلوب حساب تيار الطور، فتيار الخط المغذي للحمل يساوي القدرة الظاهرية

مضروب في جذر فولت 230 مقسومة على ) فرق الجهد ولت أ مبيرف 120000وهي

الثلاثة(، وبما أ ن التشبيكة دلتا فتيار الطور يساوي تيار الخط مقسوم على جذر

أ مبير 173.92ويساوي الثلاثة

بتدائي يساوي تيار الثانوي مقسوم على نس بة التحويل وهي تيار الطور الإ

بتدائي = 10( يساوي 230\2300) أ مبير 17.39فيصبح تيار الطور الإ

بتدائي ليغذي في جهة هنا نريد حساب فرق الجهد على جهة الإ

وفي المعطيات 50.8الثانوي الحمل المطلوب بمعامل قدرة يساوي

موجودة المعاوقة للمحول منسوبة لجهة الثانوي

بت ائي دلتسهيل الحل نسبت كل القيم لجهة الإ

بتدائي هذه الدائرة المكافئة للمحول منسوبة لجهة الإ

بتدائي للمحول وهي فولت، ول ن التشبيكة نجمة ففرق الجهد 2332.4هذه قيمة فرق الجهد على ملف الطور لجهة الإ

فولت 4039.6طور مضروب في جذر الثلاثة يساوي بين الخط والآخر في جهة الإبتدائي يساوي فق الجهد على ال

𝒁𝟏 = 𝒁𝟐 ∗ 𝒂𝟐

دي الطور وعن طريق نفس المعادلةالفولت تماما مثل المحول أ حهنا حساب تنظيم ا

29

Problem-21

A 3φ, 230 V, 27 kVA, 0.9 PF (lag) load is supplied by three 10 kVA, 1330: 230 V, 60 Hz transformers

connected in Y-Δ by means of a common 3φ feeder whose impedance is 0.003 + j0.015 Ω per phase.

The transformers are supplied from a 3φ source through a 3φ feeder whose impedance is 0.8 + j5.0 Ω

per phase.

The equivalent impedance of one transformer referred to the low-voltage side is 0.12 + j0.25 Ω.

- Determine the required supply voltage if the load voltage is 230 V.

Solution

The circuit is shown in Fig. below.

The equivalent circuit of the individual transformer referred to the high-voltage side is

مكون من ثلاث محولت قدرة كل محول محول ثلاثي الطور، يتغذى من متأ خر 0.9قدرة يساوي بمعامل فولت أ مبير 27000حمل قدرته هذا السؤال يتكون من

بتدائي كنجمة فولت أ مبير 10000 عن طريق خطوط نقل ويغذي الحمل فولت 230يساوي وجهة الثانوي كدلتا 1330بفولت يساوي مرتبطة ببعضها في جهة الإ

لها معاوقة موضحة في المخطط في ال سفل والمحول نفسه يتغذى من مصدر طاقة عن طريق خطوط نقل لها معاوقة موضحة أ يضا في المخطط.

بتدائي، المطلوب حساب قيمة فرق الجهد بين الخط والآخر على ة في المعطياتملاحظة: المحول له معاوقة مكافئة منسوبة لجهة الثانوي موضح جهة الإ

بتدائيوالمحول المثالي ومعاوقة خط النقل للحملهذه الدائرة المكافئة للنظام بالكامل مكون من الحمل ومعاوقة خط ومعاوقة المحول منسوبة لجهة الإ

لى المحول النقل من المصدر اإ

30

The single-phase equivalent circuit of the system is shown Fig. below. All the impedances from the

primary side can be transferred to the secondary side and combined with the feeder impedance on

the secondary side.

The circuit is shown in fig. below

لى نجمة فتصبح التشبيكة لى نجمةلتبس يط الحل نحول الدلتا في جهة الثانوي اإ لى الآخر فيصبح فرق 230وبما أ ن الفولت على الحمل نجمة اإ فولت من الخط اإ

10تساوي 133\1330 تصبحونس بة التحويل الجديدة 133يساوي جذر الثلاثة( \فولت 230الجهد على الطور في جهة الثانوي )

ومعاوقة معاوقة المحولوأ ما ، تبقى كما هي النقل للحمل في جهة الثانويوبما أ ن معاوقة خط لجهة الثانوي حيث يوجد الحملهنا نحول جميع عناصر الدائرة المكافئة

لى جهة الثانوي حس خط نقل التغذية للمحول نس بة التحويل الجديدة بتنسب اإ

فقط لطور واحد منسوبة جميعها المعاوقة الكلية للمحول ومعاوقات خطوط النقلالدائرة الموضحة في ال سفل هي الدائرة المكافئة للنظام مكونة من الحمل مع

وفرق الجهد بين الخط والمحايد فولت 230يساوي والآخر في تشبيكة نجمة ويتضح أ ن فرق الجهد بين الخط بتشبيكة نجمة لجهة الثانوي

فوات 133( = 1.732\230يساوي)

31

الحمل مكون من فرق الجهد هنا حساب فرق الجهد الذي يغذي

على أ طراف الثانوي + الهبوط عبر المعاوقة، ومن ثم نحول هذه

بتدائي للمحول حسب لى جهة الإ وبما (10) نس بة التحويلالقيمة اإ

أ نه تشبيكة نجمة يصبح فرق الجهد بين الخط والآخر على جهة

بتدائي هي القيمة المحسوبة مضروبة في جذر الثلاثةالإ

𝑽𝑺′ = 𝟏𝟒𝟎. 𝟕 ∠𝟑. 𝟏𝒐 𝑽

𝑰𝑳 = 𝟔𝟕. 𝟔𝟕∠−𝟐𝟓. 𝟖𝒐 𝑨

𝑰𝑳 ∗ 𝑹𝑬𝑸

𝑰𝑳 ∗ 𝒋𝑿𝑬𝑸

𝑽𝑳 = 𝟏𝟑𝟑 ∠𝟎𝟎 𝑽

32

Problem-22

A 208V three phase power system is shown in Figure. It consists of an ideal 208V Y - connected three

phase generator connected to a three-phase transmission line to a - connected load. The transmission

line has an impedance of 0.06 + j0.12 Ω per phase and the load has an impedance of 12 + j9 Ω per

phase.

Solution:

This power system is shown in Figure. Since the load on this power system is Δ-connected, it must

first be converted to an equivalent Y-form. The phase impedance of the Δ-connected load is 12 + j9

Ω so the equivalent phase impedance of the corresponding Y

form is:

𝑍𝑌 = 𝑍∆

3= 4 + 𝑗3 Ω

a. The magnitude of the line current IL

b. The magnitude of the load’s line and phase voltages VL-L

and Vϕ-L

c. The real, reactive and apparent powers consumed by the load.

d. The power factor of the load

e. The real, reactive and apparent powers consumed by the transmission line.

f. The real, reactive and apparent powers supplied by the generator.

g. The generator’s power factor.

For this simple system, find

لى نجمة هنا تم تحويل الحمل الذي على شكل دلتا اإ

33

The resulting per-phase equivalent circuit of this system is shown in Figure below.

a- The line current flowing in the per-phase equivalent circuit is given by

𝐼𝑙𝑖𝑛𝑒 = 𝑉

𝑍𝑙𝑖𝑛𝑒 + 𝑍𝑙𝑜𝑎𝑑

= 120∠0𝑜 𝑉

(0.06 + 𝑗0.12 Ω) + (4 + 𝑗3 Ω)

= 120 ∠ 0𝑜 𝑉

4.06 + 𝑗 3.12=

120 ∠ 0𝑜

5.12 ∠ 37.5𝑜

= 23.4 ∠ − 37.5𝑜 𝐴

The magnitude of the line current is thus 23.4 A.

b- The phase voltage on the equivalent Y load is the voltage across one phase of the load. This

voltage is the product of the phase impedance and the phase current of the load:

𝑉𝜙𝐿, = 𝐼𝜙𝐿

, 𝑍𝜙𝐿,

= (23.4 ∠ − 37.5𝑜 𝐴) (4 + 𝑗3 Ω)

= (23.4 ∠ − 37.5𝑜 𝐴)(5 ∠36.9𝑜 Ω) = 117 ∠ − 0.6𝑜 𝑉

The original load was Δ- connected, so the phase voltage of the original load is

𝑉𝜙𝐿 = √3 (117 𝑉) = 203 𝑉

𝑉𝐿−𝐿 = 𝑉𝜙𝐿 = 203 𝑉

واحد وفرق جهد المصدر فقط لطور طوط النقللخالمعاوقة الكلية الدائرة الموضحة في ال سفل هي الدائرة المكافئة للنظام مكونة من الحمل مع

هنا نعيد حساب قيمة فرق الجهد على

حسب تشبيكة دلتا ال صلية

34

c- The real power consumed by the equivalent Y load (which is the same as the power in the

actual load) is

𝑃𝑙𝑜𝑎𝑑 = 3 𝑉𝜙 𝐼𝜙 cos Θ

= 3 (117 𝑉) (23.4 𝐴) cos 36.9𝑜 = 6571 𝑊

The reactive power consumed by the load is

𝑄𝑙𝑜𝑎𝑑 = 3 𝑉𝜙 𝐼𝜙 sin Θ

= 3 (117 𝑉) (23.4 𝐴) sin 36.9𝑜 = 4928 𝑉𝐴𝑅

The apparent power consumed by the load is

𝑆𝑙𝑜𝑎𝑑 = 3 𝑉𝜙 𝐼𝜙

= 3 (117 𝑉) (23.4 𝐴) = 8213 𝑉𝐴

d- The load power factor is

𝑃𝐹𝑙𝑜𝑎𝑑 = cos Θ = cos 36.9𝑜 = 0.8 𝑙𝑎𝑔𝑔𝑖𝑛𝑔

e- The current in the transmission is 23.4∠−37.5𝑜 𝐴, and the impedance of the line is

0.06 + j0.12 Ω or 𝟎. 𝟏𝟑𝟒∠𝟔𝟑. 𝟒𝒐 𝜴 per phase. Therefore, the real, reactive, and apparent

powers consumed in the line are

𝑃𝑙𝑖𝑛𝑒 = 3 𝐼𝜙2 𝑍 cos Θ

= 3 (23.4 𝐴)2 (0.134 Ω) cos 63.3𝑜 = 98.6 𝑊

𝑄𝑙𝑖𝑛𝑒 = 3 𝐼𝜙2 𝑍 sin Θ

= 3 (23.4 𝐴)2 (0.134 Ω) sin 63.3𝑜 = 197 𝑉𝐴𝑅

𝑆𝑙𝑖𝑛𝑒 = 3 𝐼𝜙2 𝑍

= 3 (23.4 𝐴)2 (0.134 Ω) = 220 𝑉𝐴

هنا نحسب لجميع عناصر النظام القدرة

والقدرة الظاهرية الفعالة والقدرة غير الفعالة

35

f- The real and reactive powers supplied by the generator are the sums of the powers consumed

by the line and the load:

𝑃𝑔𝑒𝑛 = 𝑃𝑙𝑖𝑛𝑒 + 𝑃𝑙𝑜𝑎𝑑

= 98.6 𝑊 + 6571 𝑊 = 6670 𝑊

𝑄𝑔𝑒𝑛 = 𝑄𝑙𝑖𝑛𝑒 + 𝑄𝑙𝑜𝑎𝑑

= 197 𝑉𝐴𝑅 + 4928 𝑉𝐴𝑅 = 5125 𝑉𝐴𝑅

The apparent power of the generator is the square root of the sum of the squares of the real and

reactive powers:

𝑆𝑔𝑒𝑛 = √(𝑃𝑔𝑒𝑛)2

+ (𝑄𝑔𝑒𝑛)2

= 8411 𝑉𝐴

g- From the power triangle, the power-factor angle (θ) is

Θ𝑔𝑒𝑛 𝑡𝑎𝑛−1 𝑄𝑔𝑒𝑛

𝑃𝑔𝑒𝑛= 𝑡𝑎𝑛−1

5125 𝑉𝐴𝑅

6670 𝑊= 37.6𝑜

Therefore, the generator's power factor is

𝑃𝐹𝑔𝑒𝑛 = cos 37.6𝑜 = 0.792 𝑙𝑎𝑔𝑔𝑖𝑛𝑔

36

Problem-23

In this example, a three-phase transformer has a its primary winding connected in wye and its

secondary winding connected in delta. The line voltage connected to the primary is 4,170 volts

and the line voltage of the secondary is 240 volts. The load is constructed of three resistors

connected in wye. Each resistor has an impedance of 2.77 ohms.

Solution:

The next step is to determine the phase voltage of the load. In a wye connection, the phase voltage is

less than the line voltage by a factor of the square root of 3 or 1.732.

VoltsEEE

E PPL

P 6.138732.1

240

3===

The phase voltage is the voltage dropped each of the resistors that form the wye connected load.

The phase current can now be determined using Ohm’s Law.

AIIZ

EI PP

PP 50

77.2

6.138===

In a wye connection, the line current and phase current are the same. Therefore, 50 amperes of

line current is necessary to supply the 50 amperes of phase current in the load. Since only one load

is connected to the transformer secondary, the line current of the secondary will be 50 amperes also.

In a delta connection the phase current is less than the line current by a factor the square root of

3 or 1.732. The phase current of the transformer secondary is 28.9 amperes.

AIII

I PPL

P 9.28732.1

50

3===

Since the load is connected directly to the output of the transformer secondary, the line voltage of

the load will be the same as the line voltage of the secondary = 240V

37

In a delta connection the phase voltage and line voltage are the same. The phase voltage of the

transformer secondary is 240 volts. At this point, all values of the connected load and secondary

winding have been determined. The next step is to determine the values that apply to the primary

winding.

In a wye connection the phase voltage is less than the line voltage by a factor of the square root of 3

or 1.732. The phase voltage of the wye connected primary is 2407.6 volts.

VEEE

E PPL

P 6.2407732.1

4170

3===

The turns ratio of the transformer can now be determined using the phase values of the primary and

secondary voltages. Since the primary voltage is higher than the secondary voltage, it has more

turns of wire. The turns ratio is, therefore, 10:1. This indicates that there are 10 turns of wire in the

primary for every 1 turn in the secondary.

10240

6.2407=== RatioTurnsRatioTurns

VoltageLower

VoltageHigherRatioTurns

A basic rule of transformers is that power in must equal power out. The volt-amperes of the

primary must equal the volt-amperes of the secondary. Since the primary has a higher voltage, it

will have less current. The current will be less by a factor equal to the turns ratio.

AIIRatioTurns

II

imaryimary

Secondary

imary PP

P

P 89.210

9.28)(Pr)(Pr

)(

)(Pr===

In a wye connection the line current is equal to the phase current. Therefore, the line current

supplying the primary of the transformer is 2.89 amperes.

PhaseLine II =

38

The calculation can be checked by determining if the input power is equal to the output power.

Use the formula for determining the apparent power in a three-phase connection to determine if the

primary and secondary values are approximately equal.

There may be some difference due to rounding off of values.

20,784VA1.73250240VA3IEVA

20,873VA1.7322.8904173

)(Secondary)(SecondaryLL)(Secondary

(Primary))(Pr)(Pr

)(Secondary)(Secondary

)(Pr)(Pr

===

=== imaryLLimary VAIEVAimaryimary

39

Problem-24

In this example a three-phase transformer contains a delta connected primary and wye connected

secondary. The line voltage supplying the primary is 13,800 volts and the secondary line voltage is

480 volts. Two loads are connected to the transformer. The first load contains three 4 ohm resistors

connected in wye. The second load contains three 6 ohms resistors connected in delta.

Solution:

Since both loads are connected to the secondary of the three-phase transformer, both have the same

line voltage as the secondary = 480V

In a wye connection the phase voltage is less than the line voltage by a factor of the square root of

3 or 1.732. The phase voltage of the wye connected load is 277.1 volts.

VEEE

E PPL

P 1.277732.1

480

3===

A voltage of 277.1 volts is applied across each of the 4 ohm resistors. The phase current of the wye

connected load can be determined using Ohm’s Law.

AIIZ

EI PP

PP 3.69

4

1.277===

In a wye connection the line current and phase current are the same. Therefore, the line current

necessary to operate the wye connected load is 69.3 amperes.

40

In a delta connection, the line voltage and phase voltage are the same. Therefore, the phase voltage

of the delta connected load is 480 volts.

A voltage of 480 volts is applied across each of the 6 ohm resistors. The phase current can be

determined using Ohm’s Law. The phase current of the delta load is 80 amperes.

AIIZ

EI PP

PP 80

6

480===

In a delta connection, the line current is greater than the phase current by a factor of the square root

of 3 or 1.732. The amount of line current necessary to supply the delta load is 138.6 amperes.

AIIII LLPL 6.138732.1803 ===

The secondary of the transformer must supply the line current for all connected loads. Therefore, the

line current supplied by the secondary is the sum of all connected load. The secondary has a line

current of 207.9 amperes

AIIIIITotalTotalLoadDeltaLoadWyeTotal LLLLL 9.2076.1383.69

)()()()()(=+=+=

In a wye connection, the phase current and the line current are equal. Therefore, the phase

current of the transformer secondary is 207.9 amperes

In a wye connection, the phase voltage is less than the line voltage by a factor of the square root of

3 or 1.732. The phase voltage of the secondary is 277.1 volts.

VEEE

E PPL

P 1.277732.1

480

3===

In a delta connection, the line voltage and phase voltage are equal.

The phase voltage of the primary is 13,800 volts

Now that the primary and secondary phase voltages are known, the turns ratio can be determined

by dividing the larger voltage by the smaller. The turns ratio is 49.8:1. This indicates that there are

49.8 turns of wire in the primary winding for every 1 turn in the secondary.

41

8.491.277

800,13=== RatioTurnsRatioTurns

VoltageLower

VoltageHigherRatioTurns

The phase current of the primary can be determined with the turns ratio. Since the primary voltage

is greater than the secondary voltage, the primary current will be less. Divide the secondary phase

current by the turns ratio to determine the primary phase current.

AIIRatioTurns

II

imaryimary

Secondary

imary PP

P

P 2.48.49

9.207)(Pr)(Pr

)(

)(Pr===

In a delta connection the line current is greater than the phase current by a factor of the square root

of 3 or 1.732. The line current necessary to supply the primary is 7.3 amperes.

AIIII LLPL 3.7732.12.43 ===

The calculation can be checked by determining if the input power is equal to the output power.

Use the formula for determining the apparent power in a three-phase connection to determine if the

primary and secondary values are approximately equal.

There may be some difference due to rounding off of values. In this calculation there is a difference

of 1643. Although this may appear to be a large difference, the two values are within 1% of each

other.

42

Problem-25

A50-kVA, 13,800/208-V, Δ-Y distribution transformer has a resistance of 1 percent and a

reactance of 7 percent per unit.

(a) What is the transformer's phase impedance referred to the high-voltage side?

(b) Calculate this transformer's voltage regulation at full load and 0,8 PF lagging, using the

calculated high-side impedance.

(c) Calculate this transfoffi1er's voltage regulation under the same conditions, using the per-unit

system.

Solution:

(a)

The high-voltage side of this transformer has a base line voltage of 13,800 V and a base apparent

power of 50 kVA. Since the primary is Δ-connected, its phase voltage is equal to its line voltage.

Therefore, its base impedance is

43

44

45

Problem-26

The equivalent circuit impedances of a 20-kVA, 8000/240 V, 60-Hz transformer are to be determined.

The open-circuit test was performed on the secondary side of the transformer (to reduce the maximum

voltage to be measured) and the short circuit test were performed on the primary side of the

transformer (to reduce the maximum current to be measured). The following data were taken:

Find the impedances of the approximate equivalent circuit referred to the primary side, and

sketch that circuit.

Solution:

The turns ratio of this transformer is a = 8000/240 = 33.3333. The power factor during the open-

circuit test is

46

Therefore, the values of the excitation branch referred to the low-voltage (secondary) side are

47

Problem-27

A transformer has 500 primary turns and 3000 secondary turns. If the primary voltage is 240 V,

determine the Secondary voltage, assuming an ideal transformer.

Solution:

For an ideal transformer, voltage ratio = turns ratio, i.e.

𝑉1

𝑉2=

𝑁1

𝑁2, ℎ𝑒𝑛𝑐𝑒

240

𝑉2=

500

3000

𝑇ℎ𝑢𝑠 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑉2 = (3000) (240)

(500)= 1440 𝑉 𝑜𝑟 1.44 𝑘𝑉

Problem-28

An ideal transformer with a turns ratio of 2:7 is fed from a 240 V supply. Determine its output voltage.

Solution:

A turns ratio of 2:7 means that the transformer has 2 turns on the primary for every 7 turns on the

secondary (i.e. a step – up transformer). Thus,

𝑁1

𝑁2=

2

7

𝐹𝑜𝑟 𝑎𝑛 𝑖𝑑𝑒𝑎𝑙 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟,𝑁1

𝑁2=

𝑉1

𝑉2 ; ℎ𝑒𝑛𝑐𝑒

2

7=

240

𝑉2

𝑇ℎ𝑢𝑠 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑉2 = (240) (7)

(2)= 840 𝑉

48

Problem-29

An ideal transformer has a turns ratio of 8:1 and the Primary current 3 A when it is supplied at 240 V.

Calculate the secondary voltage and current.

Solution:

Problem-30

An ideal transformer, connected to 240 V mains, supplies a 12 V, 150 W lamp. Calculate the

transformer turns ratio and the current taken from the supply.

Solution:

49

Problem-31

A 5-kVA single-phase transformer has a turns ratio of 10:1 and is fed from a 2.5 kV supply.

Neglecting losses, determine:

a) the full-load secondary current.

b) the minimum load resistance which can be connected a cross the secondary winding to give full

load kVA.

c) the primary current at full load kVA.

Solution:

50

Problem-32

A 2400 V/400 V single-phase transformer takes a no-load current of 0.5 A and the core losses 400 W.

Determine the values of the magnetizing and core loss components of the no-load current (Io or Ie).

Draw to scale the no-load phasor diagram for the transformer.

Solution:

The no-load phasor diagram is shown in fig. below

51

Problem-33

A transformer takes a current of 0.8 A when its primary is connected to a 240 volt, 50 Hz supply, the

secondary being on open circuit. If the power absorbed is 72 watts, determine

a) The iron loss current.

b) The power factor on no-load.

c) The magnetizing current.

Solution:

Io = 0.8 A, V = 240 V

a) Power absorbed = total core loss = 72 Watt = V1 Io cos ϕo

Hence 72 = 240 Io cos ϕo

And iron loss current, 𝐼𝑐 = 𝐼𝑜 cos Φ𝑜 = 72

240= 0.3 𝐴

𝑃𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟 𝑎𝑡 𝑛𝑜 𝑙𝑜𝑎𝑑, cos Φ𝑜 = 𝐼𝑐

𝐼𝑜=

0.3

0.8= 0.375

𝐼𝑜2 = 𝐼𝑐

2 + 𝐼𝑀2

𝐼𝑀 = √(𝐼𝑜2 − 𝐼𝑐

2)

𝐼𝑀 = √(0.82 − 0.32)

𝐼𝑀 = 0.74 𝐴

Фm

IC

IM

IO = Ie

E2

E1

52

Problem-34 A 100 kVA, 4000 V/200 V, 50 Hz single-phase transformer has100 secondary turns. Determine

a) The primary and secondary current.

b) The number of primary turns.

c) The maximum value of the flux.

Note:

The magnetic flux set up in the core of a transformer when an alternating voltage is applied to its

primary winding is also alternating and is sinusoidal.

Let ɸm be the maximum value of the flux

and f be the frequency of the supply. The

time for 1 cycle of the alternating flux is

the periodic time T, where T = 1/f seconds

The flux rises sinusoidally from zero to its

maximum value in ¼ cycle, and the time

for ¼ cycle is ¼ f seconds.

Hence the average rate of change of flux =

ɸm / (1/4 f) = 4 f ɸm Wb/S and since 1Wb/s =

1volt, the average e.m.f. induced in each turn

= 4 f ɸm volts.

As the flux varies sinusoidally, then a sinusoidal e.m.f. will be induced in each turn of both primary

and secondary windings.

For a sine wave, form factor = rms value/average value = 1.11

Hence rms value = form factor * average value = 1.11 * average value

Thus rms e.m.f. induced in each turn = 1.11 * 4 f ɸm volts = 4.44 f ɸm volts

Therefore, rms value of e.m.f. induced in primary,

E1 = 4.44 f ɸm N1 volts

And rms value of e.m.f. induced in secondary,

E2 = 4.44 f ɸm N2 volts

53

Solution:

V1 = 400V, V2 = 200 V, f = 50 Hz, N2 = 100 turns

a) Transformer rating = V1 * I1 = V2 *I2 = 100000 VA

hence primary current, 𝐼1 = 100000

𝑉1=

100000

4000= 25 𝐴

and secondary Current, 𝐼2 = 100000

𝑉2=

100000

200= 500 𝐴

from which, primary turns, 𝑁1 = (𝑉1

𝑉2) (𝑁2) = (

4000

200) (100)

b) i.e., N1 = 2000 turns

c)

𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝑚𝑎𝑥𝑚𝑖𝑚𝑢𝑚 𝑓𝑙𝑢𝑥 Φ𝑚 = 𝐸2

4.44 𝑓 𝑁2=

200

4.44 (50) (100)

(assuming E2 = V2)

𝜙𝑚 = 9.01 ∗ 10−3 𝑊𝑏 𝑜𝑟 9.01 𝑚𝑊𝑏

Where E1 = 4.44 f Φm N1

From which, 𝜙𝑚 = 𝐸1

4.44 𝑓 𝑁1=

4000

4.44 (50) (2000)

(assuming E1 = V1)

𝜙𝑚 = 9.01 ∗ 10−3 𝑊𝑏 𝑜𝑟 9.01 𝑚𝑊𝑏

54

Problem-35

A single-phase, 50 Hz transformer has 25 primary turns and 300 secondary turns. The cross-sectional

area of the core is 300 cm2. When the primary winding is connected to a 250 V supply, determine

a) The maximum value of the flux density in the core.

b) The voltage induced in the secondary winding.

Solution:

e.m.f. E1 = 4.44 f ɸm N1 volts i.e.,

250 = 4.44 (50) Φm (25)

𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑙𝑢𝑥 𝑑𝑒𝑛𝑠𝑖𝑡𝑦, 𝜙𝑚 = 250

(4.44) (50) (25) 𝑊𝑏 = 0.04505 𝑊𝑏

However,𝜙𝑚 = 𝐵𝑚 ∗ 𝐴, 𝑤ℎ𝑒𝑟𝑒 𝐵𝑚 = 𝑚𝑎𝑥𝑖𝑚𝑢𝑚 𝑓𝑙𝑢𝑥 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑜𝑟𝑒 𝑎𝑛𝑑

𝐴 = 𝑐𝑟𝑜𝑠𝑠 − 𝑠𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑟𝑒.

Hence Bm * 300 * 10-4 = 0.04505

From which, maximum flus density, 𝐵𝑚 = 0.04504

300∗ 10−4 = 1.5 𝑇

b) 𝑉1

𝑉2=

𝑁1

𝑁2, 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝑉2 = 𝑉1 (

𝑁2

𝑁1)

𝑖. 𝑒. , 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑖𝑛𝑑𝑢𝑐𝑒𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑤𝑖𝑛𝑑𝑖𝑛𝑔,

𝑉2 = (250) (300

25) = 3000 𝑉 𝑜𝑟 3 𝑘𝑉

55

Problem-36

A single-phase 500 V/100 V, 50 Hz transformer has a maximum core flux density of 1.5 T and an

effective core cross-sectional area of 50 cm2. Determine the number of primary and secondary turns.

Solution:

The e.m.f. equation for a transformer is E = 4.44 f ɸm N

And maximum flux, 𝜙𝑚 = 𝐵 ∗ 𝐴 = (1.5) (50 ∗ 10−4) = 75 ∗ 10−4 𝑊𝑏

Since E1 = 4.44 f ɸm N1

Then primary turns,𝑁1 = 𝐸1

4.44 𝑓 ∅𝑚=

500

4.44 (50) (75∗ 10−4)= 300 𝑡𝑢𝑟𝑚𝑠

Since E2 = 4.44 f ɸm N2

Then secondary turns, 𝑁2 = 𝐸2

4.44 𝑓 ∅𝑚=

500

4.44 (50) (75∗ 10−4)= 60 𝑡𝑢𝑟𝑚𝑠

56

Problem-37

A 4500 V/225 V, 50 Hz single-phase transformer is to have an approximate e.m.f. per turn of 15 V

and operate with a maximum flux of 1.4 T. Calculate

a) The number of primary and secondary turns.

b) The cross-sectional area of the core.

Solution:

a) 𝐸. 𝑚. 𝑓. 𝑝𝑒𝑟 𝑡𝑢𝑟𝑛 = 𝐸1

𝑁1=

𝐸2

𝑁2= 15

𝐻𝑒𝑛𝑐𝑒 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑡𝑢𝑟𝑛𝑠, 𝑁1 = 𝐸1

15=

4500

15= 300

𝑎𝑛𝑑 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑡𝑢𝑟𝑛𝑠, 𝑁2 = 𝐸2

15=

225

15= 15

b) E.m.f. E1 = 4.44 f ɸm N1

from which,Φ𝑚 = 𝐸1

4.44 𝑓 𝑁1=

4500

4.44 (50) (300)= 0.0676 𝑊𝑏

𝑁𝑜𝑤 𝑓𝑙𝑢𝑥 𝜙𝑚 = 𝐵𝑚 ∗ 𝐴, 𝑤ℎ𝑒𝑟𝑒 𝐴 𝑖𝑠 𝑡ℎ𝑒 𝑐𝑟𝑜𝑠𝑠 − 𝑆𝑒𝑐𝑡𝑖𝑜𝑛𝑎𝑙 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑜𝑟𝑒, ℎ𝑒𝑛𝑐𝑒

𝑎𝑟𝑒𝑎 𝐴 = 𝜙𝑚

𝐵𝑚=

0.0676

1.4= 0.0483 𝑚2 𝑜𝑟 483 𝑐𝑚2

57

Problem-38

A single-phase transformer has 2000 turns on the primary and 800 turns on the secondary. Its no load

current is 5 A at a power factor of 0.20 lagging. Assuming the volt drop in the windings is negligible,

determine the primary current and power factor when the secondary current is 100 A at a power factor

of 0.85 lagging.

Solution:

Let 𝑰𝟏′ be the component of the primary current which

provides the restoring mmf. Then

𝐼1′ 𝑁1 = 𝐼2 𝑁2

i.e.,

𝐼1′ (2000) = (100) (800)

𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ, 𝐼1′ =

(100) (800)

2000= 40 𝐴

If the power factor of the secondary is 0.85 the cos ϕ2 = 0.85, from which, ϕ2 = arccos 0.85 = 31.8o.

If the power factor on no load is 0.2, then cos ϕ0 = 0.2, and ϕ0 = arccos 0.2 = 78.5o. In the phasor

diagram shown in fig. above I2 = 100 A is shown at an angle of ϕ2 = 31.8o to V2 and 𝑰𝟏′ = 40 A is shown

in anti-phase to I2

The no load current I0 = 5 A is shown at an angle of ϕ0 = 78.5o to V1.

Current I1 is the phasor sum of 𝑰𝟏′ and I0 and by drawing to scale, I1 = 44 A and angle ϕ1= 37o

𝐵𝑦 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛, 𝐼1 cos 𝜙1 = 𝑜𝑎 + 𝑜𝑏 = 𝐼𝑜 cos 𝜙𝑜 + 𝐼1′ cos 𝜙2

= (5) ∗ (0.2) + (40) ∗ (0.85) = 𝟑𝟓. 𝟎 𝑨

𝑎𝑛𝑑 𝐼1 sin 𝜙1 = 𝑜𝑐 + 𝑜𝑑 = 𝐼𝑜 sin 𝜙𝑜 + 𝐼1′ sin 𝜙2

= (5) sin 78.5𝑜 + (40) sin 31.8 𝑜 = 𝟐𝟓. 𝟗𝟖 𝑨

𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑚𝑎𝑔𝑛𝑖𝑡𝑢𝑑𝑒 𝑜𝑓 𝐼1 = √(35.02 + 25.982) = 𝟒𝟑. 𝟓𝟗 𝑨

𝑎𝑛𝑑 tan 𝜙1 = (25.98

35.0) , 𝑓𝑟𝑜𝑚 𝑤ℎ𝑖𝑐ℎ,

𝜙1 = arctan (25.98

35.0) = 𝟑𝟔. 𝟓𝟗𝒐

𝐻𝑒𝑛𝑐𝑒 𝑡ℎ𝑒 𝑝𝑜𝑤𝑒𝑟 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑡ℎ𝑒 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 = cos 𝜙1 =

cos 36.59𝑜 = 𝟎. 𝟖 𝟎

58

Problem-39

A transformer has 600 primary turns and 150 secondary turns. The primary and secondary resistances

are 0.25 Ω and 0.01 Ω respectively and the corresponding leakage reactances are 1.0 Ω and 0.04 Ω

respectively. Determine:

a) The equivalent resistance referred to the primary winding

b) The equivalent reactance referred to the primary winding

c) The equivalent impedance referred to the primary winding

d) The phase angle of the impedance.

Solution:

𝑅𝑒𝑞 = 𝑅1 + 𝑅2 (𝑉1

𝑉1)

2

𝑅𝑒𝑞 = 0.25 + 0.01 (600

150)

2

= 0.41 Ω

𝑋𝑒𝑞 = 𝑋1 + 𝑋2 (𝑉1

𝑉1)

2

𝑅𝑒𝑞 = 1.0 + 0.04 (600

150)

2

= 1.64 Ω

𝑍𝑒 = √(𝑅𝑒2 + 𝑋𝑒

2) = 1.69 Ω

𝜙𝑒 = 𝐴𝑟𝑐𝑐𝑜𝑠 (0.41

1.69) = 75.96𝑜

Problem-40

A 5 kVA, 200V/400V, single-phase transformer has a secondary terminal voltage of 387.6 volts when

loaded. Determine the regulation of the transformer.

Solution:

𝑉𝑅 = 𝑁𝑜 𝑙𝑜𝑎𝑑 𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 − 𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙 𝑣𝑜𝑙𝑡𝑎𝑔𝑒 𝑜𝑛 𝑙𝑜𝑎𝑑

𝑛𝑜 𝑙𝑜𝑎𝑑 𝑆𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑉𝑜𝑙𝑡𝑎𝑔𝑒 ∗ 100%

𝑉𝑅 = 400 − 387.6

400 ∗ 100% = 3.1 %

59

كل عام وأنت بخيرالطلبة الأعزاء

الإختبار النهائي لن يخرج عن هذا مع بعض التوضيحات على بعض الحلول ، 1-ينات كهربائيةفي مساق ماك تم نقاشههذا كل ما

يدرس هذه الأسئلة ويستوعبها ليس لديه أي مشكلة في الدخول إلى الإختبار النهائي....من النمط من الأسئلة

ة الكتاب الذي بين يديكمامكانه حل أمثلومن يريد المزيد من الأسئلة ب

وسيتم الرد عليه بإذن الله صفحة الجامعةة باممكن يرسل لي عبر بوديه أي استفسار في النهاية من ل

لكم التوفيق والسداداللهأسأل