Dec 2009

1a
i High starting torque
ii Low starting current

b)
synchronou sspeed − Rotorspeed
i Percentage slip (% S = ×100
synchronou sspeed
At standstill, the synchronous speed and the rotor speed are the same
% slip is 100%

Rotorgross − output − in − watts

ii Torque produced at synchronous speed = = Zero (0)
2π × synchronou sspeed

c)
Motor output=30KW; Motor speed = 0.9 synchronous speed ; Rotational losses = 1.4KW

slip × rotorgross − output

=
i Rotor copper loss 1 − slip
rotorgross − poweroutpu t = motoroutpo wer + rotational losses

synchspeed − motorspeed N − 0.9 N S

%slip = × 100 = S = 0.1 × 100 = 10%
synchspeed NS

0.1 × (30000 + 1400 )

Rotor copper loss = = 3488 .9W = 3.4889 KW
0.9

ii Operating efficiency
output − power 30000
= ×100 = ×100 = 78 .4%
output − power + losses 30000 + (3488 .9 + 3400 +1400 )

2a
Armature reaction is due to the magnetic flux set-up by the armature current which distorts the main
flux distribution as show above

b)

i) Sparking at the commutator contact

ii) Wearing of carbon brush
Resulting in increase in motor speed and reduction in generator terminal voltage

c)

i Compensating winding ( interpoles ) ; A set of compensating windings is introduced or slotted

in the field winding arrangement to compensate for the loss due the distortion caused by the
armature reaction on the flux distribution.

ii Moving Brush Gear ; The movable or adjustable brush gear is used for adjusting the brush
contacts for proper and firm connection with the commutator to avoid sparking.

3a

KW Active − power 560000

KVA = = = = 903 .2 KVA
COS φ Power − factor 0.62

With the KVA Value stays the same, the power in KW @ .094 lagging

kw = kva ×cos φ = 903 .2 ×0.94 = 849 .03 kw

The additional power that the transformer can generate @ 0.94 lagging = 849.03- 560=
289.03kw

b)

Cos φ = 0.7 φ = 45 .6 0 ; cos φ / = 0.96 φ / =16 .3 0

The horizontal component of load current (I) @ 0.7 pf lagging = 68 × 0.7 = 47 .6 A

The vertical component of load current (I) @ 0.7 pf lagging = 68 × sin 45 .6 = 48 .6 A

The horizontal component of load current ( I ′ ) @improved pf of 0.96lagging = I ′ × 0.96

Since the horizontal component of load current before improvement is equal to the horizontal
component of load current after improvement. It implies that;

I cos φ = I ′ cos φ′
47 .6
47.6 A = I ′ × 0.96 I′ = = 49 .6 A
0.96

The new supply current= 49.6A

(ii) KVAR rating of the capacitor required fro this improvement is given by

230 × 47 .6
KVAR = KW (Tan φ − Tan φ′) = (Tan 45 .6 0 − Tan 16 .3 0 ) = 7.978 KVAR ≅ 8.0 KVAR
1000

OR

Vertical component of the load current (I ) after improvement

= I ′sin φ′ = 49 .6 ×sin 16 .3 = 13 .92 A

The capacitor current required to be injected to achieve the improvement is given by

I C = I sin φ − I ′ sin φ ′ = 48 .6 −13 .92 = 34 .70 A

For single- phase,

VP I C 230 × 34 .70
The KVAR rating = = = 7.981 KVAR ≅ 8.0 KVAR
1000 1000

c)

Methods of power factor improvement

i) Using of static capacitor for individual load

 ii) Using of synchronous motor for overall load

4a

b)

let
P1 = Re ading − of − wattmeter − one
P2 = Re ading − of − wattmeter − two
φ = phase − angle

− 1
P2 − P1 P2 = 3.8 KW
φ = tan 3( )
P2 + P1 P1 =1.2 KW

3.8 −1.2
φ = tan −1 3 ( ) = 42 0
3.8 + 1.2

Power factor is the cosine of the phase angle= cos42 = 0.74

When the connection to the current coil of wattmeter is reversed, it gives a negative reading. Thus,
P1 = −1.2 KW
P2 − (−P1 ) 5
φ = tan −1 3 ( ) = tan −1 3 ( ) = 73 .30
P2 − P1 6
Power − factor = cos 73 .3 = 0.29 ≅ 0.3

5a

power − output
Efficiency = ×100
power − input

i Power output in kw @ unity power factor @ half-load=

 1 1 1 2
kva cos φ = ×60 ×10 3 ×1 The copper loss = ( ) × 916 = 229 w
2 2 2
= 30 kw

30 ×10 3
Efficiency @ unity power factor @ half-load = ×100 = 97 .3%
30 ×10 3 + (229 + 615 )

ii Power output in kw @ half-load @ 0.65 power factor = 30 ×10 3 × 0.65 = 19 .5kw

Copper loss remain the same since the load is still half.( i.e 229w )

19 .5 ×10 3
Efficiency @ 0.65 power factor @ half-load = ×100 = 95 .9%
19 .5 ×10 3 + ( 229 + 615 )

b)
Efficiency of the transformer is maximum when the variable losses is equal to constant losses i.e
copper loss = iron loss = 615w

c)

I P = primary − current
I S = sec ondary − current
VS = sec ondary − voltage
VP = primary − voltage
N P = primary − winding
N S = sec ondary − winding
( I S − I P ) = current − in − the − common − part − of − the − winding

6a
In an IT network, the distribution system has no connection to the earth at all, or it has only high
impedance connection. In such systems an insulation monitoring device is used to monitor the
impedance.

Exposed conductive part- This is the metal work of an electrical appliance or the trunking and
conduit of an electrical system which can be touched because they are not normally live, but which
may become live under fault condition.

c)

Advantages of mineral insulated, metal sheathed cables compared with pvc insulated and sheathed
cables

i It is mechanically robust and resistance to impact

ii It is water proof and resistance to ultraviolet light and many corrosive elements
iii It does not initiate an explosion even during circuit fault conditions
iv It does not contribute fuel or hazardous combustion products to a fire and cannot propagate a fire
within a building.

7a

The term split-phase as its applied to single-pase induction motor is the connection of an auxiliary
winding of smaller gauge wire in parallel with the main winding so that the supply current to the
motor is shared between these windings for the purpose of creation of rotating magnetic field.

b)

A single- phase induction motors are not self-starting because for a rotating magnetic flux to be
produced, two anti-phase current must be involved.In split-phase induction motor the anti-phase
currents are produced by connecting an auxiliary winding of smaller gauge in series with reactor
both connected in parallel with the main winding as show below.
 d)

frequency 50
synchspeed = = = 25 rev / s
pair − of − poles 2

slipspeed = Synchspeed − Rotorspeed

synchspeed − rotorspeed
% slip = ×100
synchspeed
25 − N r
0.06 =
25
N r = 23 .5rev / s

slipspeed = 25 − 23 .5 = 1.5rev / s

8a

I = load − current
I f = field − current
I a = armature − current
Ra = armature − resis tan ce
 R sh = field − resis tan ce
Vt = ter min al − voltage
E g = generated − e.m. f

E g =Vt + I a Ra
Ia = I f +I 12000
I = = 52 .2 A
230
230
If = = 2A
115

I a = 52 .2 + 2 = 54 .2
E g = 230 + 54 .2 × 0.3 = 246 .26V

2
ii armature copper loss = I a Ra = 54.2 2 × 0.3 = 881.3w

iii Output power of the generator (VI) = 12000w

The input power to the generator = Output power of the motor driving it
= Output power of the generator + total losses

2 2
Total − losses = I a Ra + I f Rsh + rotational losses
= 881 .3 + 460 + 20 = 1361 .3w
Inputpower = 12000 +1361 .3 = 13361 .3w
Mechanical power − of − the − motor = Tω
= 13361 .3W
13361 .3
Torque = = 151 .9 Nm
840
2π ( )
60

b)

I = load − current
I f = field − current
I a = armature − current
Ra = armature − resis tan ce
E g =Vt − I a Ra R sh = field − resis tan ce
I a = I −I f Vt = ter min al − voltage
E b = back − e.m. f

9a

b)

Induction relay have inverse- definite minimum time (IDMT) time- current characteristics in which
the time varies inversely with current at low fault currents, but attains a constant minimum value at
higher currents. This constant minimum value depends upon the adjustments. Further adjustment is
possible by means of tapping on the relay winding.

c)

FACTORS

i Where automatic and remote protection system is required

ii Where rapid action to restore supplies following a fault is necessary.
 10 a

i )Magnetising component of the no- load current( I

m ) = no − loadcurren t × sin e − of − the − no − load − phaseangle

I m = I O sin φo cos φo = 0.26

I O =no-load current= 3A
φo = 74 .93 0 I m = 3 × sin 74 .93 = 2.9 A

ii Power component of the no-load current I c = I O cos φo = 3 × 0.26 = 0.78 A

b)
′
I p cos φp = I O cos φo + I p cos φs
′
I p sin φp = I O sin φo + I p sin φs

′
I p V p = I sV s
′ I V 70 ×115
Ip = s s = = 35 A
Vp 230
cos φs = 0.76
φs = 40 .5 0
I p cos φp = 0.78 + (35 × 0.76 ) = 27 .38 A
I p sin φp = 2.9 + (35 × sin 40 .5) = 25 .6 A
Pr imary − current ( I p ) = 27 .38 2 + 25 .6 2 = 37 .5 A
I p = primary − current
I s = sec ondary − current
′
I p = primary − current − due − to − load
V p = primary − voltage
Vs = sec ondary − voltage
φp = phase − angle − of − primary − current − due − to − load

( I p cos φ p + I p sin φ p )
2 2
Primary current I P =
 25 .6
tan φ p =
27 .38
φp = 43 .08 0
primary − power − factor = cos φ p = cos 43 .08 = 0.73

June 2009

b)

Input − power = 3V L I L cos φ

3.5 ×10 3 = 3 × 400 × I L × 0.65
3.5 ×10 3
IL = = 7.8 A
3 × 400 × 0.65

Reading of each wattmeter

Let w 1 = reading − of − wattmeter − one
w 2 = reading − of − wattmeter − two
w1 = VL I L cos( 30 0 + φ ) = 400 × 7.8 cos( 49 .5 + 30 0 ) = 568 .8w
w2 = V L I L cos( 30 0 − φ ) = 400 × 7.8 cos( 49 .5 − 30 0 ) = 2941 .0 w

2a

For open-circuit test, rated voltage is applied at the primary side while the secondary side is open-
circuited. The power measured or recorded by the wattmeter is the iron-losses

For short- circuit test, small voltage of the range 10-12V is applied at the primary side while the
secondary side is short-circuited, the power measured or recorded by the wattmeter is the copper
losses.

b)

Output power of the transformer in KW at 0.8 power factor lagging at full-load

= 150 ×10 3 × 0.8 = 120000 w = 120 kw
Total − losses = copperloss − and − ironlosses
= 3500 + 2000 = 5500 w
Power − output Power − output
Efficiency = ×100 = ×100
Power − input Power − output +Totallosse s
120000
= ×100 = 95 .6%
120000 + 5500
 c)

The e.m.f induced in the primary winding of a transformer E p = 4.44 ϕfN p

The e.m.f induce in the secondary winding of the transformer E s = 4.44 ϕfN s
For an ideal transformer, the power input is equal to the power
powerinput = poweroutpu t , output
V p I p =Vs I s

Vp Np Is Ep V p = primary − voltage
= = =
Vs Ns Ip Es Vs = sec ondary − voltage
I p = pimary − current
I s = sec ondary − current
N p = primary − turns
N s = sec ondary − turns

3a

A single-phase induction motors are not self starting because for a rotating magnetic flux to be
produced, two anti- phase currents must be involved. In a single-phase induction motor, the anti-
phase currents are produced by the connecting an auxiliary winding in series with a reactor both
connected in parallel with the main winding as shown below.

When the motor has come up to about 70% to 75% of synchronous speed, the auxiliary winding may
be opened by a centrifugal switch, and the motor will continue to operate as a single-phase motor.

b)

i Method of starting is by connecting an auxiliary winding of smaller-gauge wire in parallel with

the main winding.
ii The direction of rotation may be reversed by reversing the connection to the auxiliary winding.

c)

4a
See the attached graph

b)
Vt 220
If = = = 2.75 A
RSH 80
E g = Vt + I a Ra
P 11000
Ia = I f +I I = = = 50 A
Vt 220
I a = 2.75 + 50 = 52 .75 A

E g − Vt 228 − 220
Ra = = = 0.2Ω
Ia 52 .75

5a

i Excessive bill
ii Poor power efficiency

b)

Total kw = 200+ 25 = 225kw

200 25
Total kva = + = 275 kva
0 .8 1
 kw
The combined power factor before improvement= = 0.8lagging
kva

The kvar rating of capacitors required to improve the power factor of the combined load to 0.95
lagging
= 225 (tan φ − tan φ ′) = 225 (tan 34 .92 − tan 18 .2) = 83 .1k var

ii With the capacitors in circuit and the motor load switched off and the lighting and heating load
reduced to 10kw

k var − rating − of − the −capacitor

Tan φ′′=
kw −of − the − lighting −and − heating −load
83 .1
= =8.31
10
φ=83 .10

Overall power factor = cos 83.2 = 0.12 leading

6a

In oil break circuit breaker, the contact are separated under the whole of the oil in the tank. There is
no special arc control system other than increasing length caused by separation of contacts. Arc
extinction occurs when a critical gap is reached between the contacts.

ii Plain air break circuit breaker cool the gases to naturally deionized them, causing arc interruption.
The arc can be stretched. Its resistance can be increase by increasing its length. The increase in
resistance is significant so that the current and voltage are brought into phase.

Advantages of oil over air

1 Oil is very good insulator

2 Has a high dielectric strength

Disadvantages of oil over air

1 There is no special control over arc other than increase in length by separating the moving
contacts. And for successful interruption long arc length is required.
2 There is risk of fire.

c)
7a

b)

The types of three- phase load for which the two-wattmeter method of total power measurement is
Balanced three-phase loads.

c)

The earth electrode under test is driven to the ground at the same ground level with two steel rods
placed at different distance from the earth electrode under test. The distance of the two steel rods
from the earth electrode under test is properly chosen to avoid the overlapping of the resistance area .
The earth tester is connected as shown above and operated accordingly and the resistance value
display by the earth tester is the resistance of the earth electrode under test.

8a
 synchspeed − motorspeed
per − unit − slip =
synchspeed
120 × frequency 120 × 50
synchspeed = = = 1500 rev / min
number − of − poles 4
1500 −1440
per − unit − slip = = 0.04
1500
slip × rotorgross − outputpowe r
ii ) Rotor − copperloss =
1 − slip
Rotorgross − outputpowe r = motor − outputpowe r + rotational losses
= 25000 + 800 = 25800 w
0.04 × 25800
Rotor − copperloss = = 1076 w
1 − 0.04
power − output
Efficiency = ×100
power − output + losses
25000
0.86 =
25000 + losses
losses = 29069 .8 − 25000 = 4069 .8w
losses = statorcopp erloss + rotorcoppe rloss + rotational losses
statorcopp erloss = 4069 .8 − ( rotorcoppe rloss + rotational losses )
= 4069 .8 − (1076 + 800 ) = 2193 .8w = 2.194 kw
b)

This type of starter circuit uses an autotransformer to apply reduced voltage across the windings of
the motor during start-up .Three autotransformers are connected in the star configuration and taps
are selected to provide an adequate starting current for the motor. After a certain time lapse, full
voltage is applied to the motor bypassing the autotransformers.

9a
primary − resis tan ce ( R p ) = 0.8Ω
sec ondary − resis tan ce ( R s ) = 0.1Ω
primary − leakagerea c tan ce ( X p ) = 1.6Ω
sec ondary − leakagerea c tan ce ( X s ) = 0.06 Ω
primary − voltage (V p ) = 1000 V
sec ondary − voltage (V s ) = 250 V
Transforme r − rated − capacity = 12 KVA
′ V 250 2
Re sis tan ce − referred − to − the − sec ondary ( R p ) = R p ( s ) 2 = 0.8 × ( ) = 0.05 Ω
Vp 1000
′ V 250 2
Leakagerea c tan ce − referred − to − the − sec ondary ( X p ) = X p ( s ) 2 = 1.6( ) = 0.1Ω
Vp 1000
Equivalent − resis tan ce − referred − to − the − sec ondary =
′
Re = R s + R p = 0.1 + 0.05 = 0.15 Ω
Equivalent − leakagerea c tan ce − referred − to − the − sec ondary =
′
X e = X s + X p = 0.06 + 0.1 = 0.16 Ω

b)
I ( Re cos φ + X e sin φ)
Voltage drop @ 0.6 power factor =

12 ×10 3
Is = = 48 A
250
voltage − drop @ 0.6 pf = 48 (0.15 × 0.6 + 0.16 sin 53 .1) = 10 .7V
1
ii − per − unit − regulation @ 0.96 pf = I S ( Re × 0.96 + X e sin 16 .3) ×
Vs
48 (0.15 × 0.96 + 0.16 × 0.28 )
= = 0.04
250
10 a ) E b = Vt − I a Ra
E b1 = 400 − (80 × 0.15 ) = 388 V
E b2 = 400 − 60 (0.15 + 0.7) = 349 V
@ cons tan t − field − exictation
E b1 N1
=
E b2 N2
388 350
=
349 N2
N 2 = 314 .82 rev / min

b)
Starter are use with a large d.c motor to protect the armature winding from high current during
loading.

c)

power − output − of − the − motor = torque × 2π × speedn

N
= T × 2π
60
6.4 ×10 3 × 60
Torque = = 48 .5 Nm
2π ×1260
power − output
operating − efficiency = ×100
power − input
6.4 ×10 3
= ×100 = 81 .84 %
230 × 34

June 2008

1a

KW Active − power 600

KVA = = = = 1000 KVA
COS φ Power − factor 0.6

With the KVA Value stays the same, the power in KW @ .095 lagging

kw = kva × cos φ = 1000 × 0.95 = 950 kw

The additional power that the transformer can generate @ 0.95 lagging = 950- 600= 350kw
b)

Cos φ = 0.7 φ = 45 .6 0 ; cos φ / = 0.95 φ / =18 .2 0

The horizontal component of load current (I) @ 0.7 pf lagging = 72 × 0.7 = 50 .4 A

The vertical component of load current (I) @ 0.7 pf lagging = 72 × sin 45 .6 = 55 .64 A

The horizontal component of load current ( I ′ ) @improved pf of 0.95lagging = I ′ × 0.95

Since the horizontal component of load current before improvement is equal to the horizontal
component of load current after improvement. It implies that;

I cos φ = I ′ cos φ′
50 .4
50.4 A = I ′ × 0.95 I′ = = 53 .1A
0.95

The new supply current= 53.1A

(iii) KVAR rating of the capacitor required fro this improvement is given by

230 × 50 .4
KVAR = KW (Tan φ − Tan φ′) = (Tan 45 .6 0 − Tan 18 .2 0 ) = 8.026 KVAR ≅ 8.0 KVAR
1000

c)

The advantage of connecting a static capacitor used for power factor improvement of an individual
load as near as possible to the load terminal is that it prevents over correction and voltage surge since
power factor vary with load. Being lower at low load.
2a

power − output
Efficiency = ×100
power − input
power − output 8 ×10 3
power − input = ×100 = ×100 = 9411 .8w
efficiency 85
power − input − of − the − motor = IV = 9411 .8w
9411 .8
I = = 40 .9 A
230
armature − current ( I a ) = I − I f
Vt 230
If = = = 2A
Rsh 115
I a = 40 .9 − 2 = 38 .9 A
Backemf ( E b ) = Vt − I a Ra = 230 − (38 .9 × 0.4) = 214 .44V
Totalcoppe r − loss = copperloss @ field − winding + copperloss @ armature − winding
2 2
= I f Rsh + I a Ra = 2 2 ×115 + 38 .9 2 × 0.4 = 1065 .3w
Rotational losses = Motorlosse s − Totalcoppe r − loss
= ( powerinput − poweroutpu t ) − totalcoppe rloss
= 1411 .8 −1065 .3 = 346 .5w

3a

Advantage of alternating current system over direct current system

i It can be step-up or step-down

Disadvantage
i It can not be transmitted over the same size line using the same size tower

b)

Separate protective earth (PE) and neutral (N) conductors from supply to consumer, which are not
connected together at any point after the building distribution point. That is PE and N are separate
conductors that are connected together only near the power source.

C)

The function of an isolator in distribution system is to cut-off supply from all or a discrete section of
the installation by separating the installation or section from every source of electrical energy for
reasons of safety.
d)
The main advantage of IDMT relay is that it has time-current characteristics in which the time
varies inversely with current at lower fault currents, but attains a constant minimum value at high
currents

4a

Resistors are connected in series with each of the phase winding as shown above when starting
three-phase wound-rotor induction motor.

b)

rotor − frequency ( f r ) = slip × synchfrequ ency

rotor − frequency 2
slip = = = 0.04
synchfrequ ency 50
@ − rotorspeed − of − 800 rev / min
synchspeed − rotorspeed
slip = ×100
synchspeed
120 × frequency 120 × 50
synchspeed = = = 1000 rev / min
numbers − of − poles 6
1000 − 800
% slip = ×100 = 20 % = 0.2
1000
rotor − frequency = slip × synchfrequ ency = 0.2 × 50 = 10 Hz
slip − speed − @ slip − of − 5%
rotorspeed = synchspeed (1 − slip )
= 1000 (1 − 0.05 ) = 950 rev / min
slipspeed = synchspeed − rotorspeed = 1000 − 950 = 50 rev / min

The three features of a three-phase synchronous machine when operating in the motor mode are;
i High torque
ii constant speed
iii High efficiency

5a

Four typical metering requirements

i Energy
ii Maximum demand
iii KVA
iv Power factor

b)

c)

The multiplying factor = 5 × 4 = 20

The actual power taken by the load=
scaling − reading − of − the − wattmeter × multipling − factor
= 850 × 20 = 17000 = 17 kw

6a
i The eddy-current loss in a transformer can be reduced by laminating the iron –core
ii Two reasons for copper loss in a transformer are due to primary and secondary windings

b)
no − loadpower = VI cos φ = 175 w
VI cos φ = 175
175 175
cos φ = = = 0.28
VI 250 × 2.5
no − load − powerfacto r = 0.28
ii − Efficiency @ full − load − unity − powerfacto r
poweroutpu t − in − kw @ unity − powerfacto r = 20 ×10 3 ×1 = 20 kw
totalloss = ironloss + copperloss = 175 + 400 = 575 w = 0.575 kw
20
efficiency = ×100 = 97 .2%
20 + 0.575
Efficiency @ 25 % full − load @ pf − of − 0.65 lagging
sin ce − (kva ) 2 α − copperloss
power − output − in − kw @ 25 % full − load @ pf − of − 0.65 = 20 ×10 3 × 0.25 × 0.65 = 3.250 kw
1
copperloss @ 25 % full − load = ( ) 2 × 400 = 25 w
4
Totalloss @ 25 % full − load = 175 + 25 = 200 w = 0.2kw
3.250
Efficiency = ×100 = 94 .2%
3.250 + 0.2
7a

b)

The field current of a shunt generator is a function of the terminal voltage. Increased loading cause a
drop in excitation current. The induced voltage of shunt generator is easily controlled by varying the
excitation current by means of a reheostat connected in series with the shunt field winding.

c)

Critical field resistance when referring to a d.c generator is a certain value of field resistance at
which generation does not possible.

d)
Two forms of loss occurring in a d.c machine are;
i constant losses
ii variable losses

8a

i
rotor − grossoutpu t × slip
rotor − copperloss =
1 − slip
rotor − grossoutpu t = motor − output + rotational losses
= 28 kw +1.4kw = 29 .4kw
synchspeed − rotorspeed synchspeed − 96 % synchspeed
% slip = ×100 = ×100
synchspeed synchspeed
synchspeed (1 − 0.96 )
= ×100 = 4% = 0.04
synchspeed
29 .4 × 0.04
rotor − copperloss = = 1.225 kw
1 − 0.04

power − output 28
Operating − efficiency = ×100 = ×100 = 83 .02 %
power − output + losses 28 + (1.225 + 3.1 +1.4)
b)

The starting current is 5 to 7 times of the full-load current of a typical three-phase cage rotor
induction motor.

c)

For maximum torque to occur, the rotor resistance must be equal to the rotor reactance
@ Standstill

Rr = X r

@ Running

Rr = SX r

d)
Rr = Rotor − resis tan ce
X r = Rotor − reac tan ce
r = External − resis tan ce
For − max imum − torque − condition
( Rr + r ) = X r
r = X r − Rr = 4 − 0.2 = 3.8Ω

9a

b)
Advantages of using relay for system protection

i It detects system failures when they occur and isolate the faulted section from the remaining of the
system
ii It mitigates the effects of failures after they occur. Minimising risk of fire, danger to personal and
other high voltage system.

c)

Fusing Factor- Is the ratio greater than unity of the minimum fusing current to the current rating.

d)

The function of arc chute in circuit breaker is to divide the arc and cool it.

10 a

b)

Advantage of autotransformer over double wound

i Cost saving since less copper is needed
 Disadvantage
i Since the primary and the secondary winding are not electrically separated, an open circuit in the
secondary causes a full primary voltage appearing across the secondary.

c)

kw 20000
Vp I p = = = 25000 VA V p = primary − voltage
cos φ 0.8
25000 Vs = sec ondary − voltage
Ip = =108 .7 A I p = primary − current
230
kw 20000 I s = sec ondary − current
Vs I s = = = 25000 VA
cos φ 0.8 N p = primary − turns
25000 N s = sec ondary − turns
Is = =166 .7 A
150 I s − I p = current − through − the − common − winding
( I s − I p ) =166 .7 −108 .7 = 58 A

d)

The reason for an autotransformer having a primary voltage of 230V not suitable for use in circuit
requiring a secondary voltage of 12V is because of the possibility of open circuiting of secondary
winding which will cause a 230V to be applied across 12V circuit resulting in over voltage.

Dec 2008

1a

b)
i
Advantage of autotransformer over double wound
 Cost saving since less copper is needed

ii The danger which may arise if an autotransformer is operated with a high voltage transformer
ratio is that if the circuit of primary winding is open circuited, it will results to high voltage across
the secondary circuit(side ) of the transformer..

c)

copperloss − @ 25 % full − load = (0.25 ) 2 ×800 = 50 w

losses = copperloss + ironloss = 50 + 480 = 530 w

d)

Voltage regulation of a transformer is the difference in the terminal voltage of the transformer at no-
load and at full-load

120 × frequency 120 × 50

synchspeed = = = 1000 rev / min
number − of − poles 6
Nr
Rotational − powerloss = Torqueabso rb × 2π
60
where − N r = rotorspeed
N r = synchspeed − slip × synchspeed = 1000 − 0.04 ×1000 = 960 rev / min
 960
Rotational − powerloss = 12 × 2 ×π = 1206 .4 w
60
grosstorqu e = torqueabso rb + torqueshaf t
power − output 40 ×10 3
torqueshaf t = = = 397 .9 Nm
N 960
2π r 2π
60 60
grosstorqu e = 12 + 397 .9 = 409 .9 Nm
960
Rotorgross − outputpowe r = grosstorqu e × 2π = 409 .9 × 2π ×16 = 41207 .6 w
60
slip 0.04
Rotor − copperloss = rotorgross − outputpowe r × = 41207 .6 × = 1716 .98 w ≅ 1.72 kw
1 − slip 0.96
inputpower = outputpowe r + losses
= 40000 + (1716 .98 +1716 .98 +1206 .4) = 44640 .36 ≅ 4.5kw
output − power 40000
Full − load − efficiency = ×100 = ×100 = 88 .9%
input − power 45000
input − power = 3VI cos φ
input − power 45000
I = = = 81 .2 A
3V cos φ 3 × 400 × 0.8

3a

b)
c)
The reason for one wattmeter giving a reverse reading is that the connection of the current( fixed )
coil of the wattmeter has been reserved.

4a

Armature reaction occurs in a d.c motor because of the magnetic field set-up by the armature current
which distorts the flux distribution of the main flux as show in the diagrams below.

b)

The two methods of overcoming the effect of armature reaction on commutation in a d.c motor are:

i Compensating winding ( interpoles ) ; A set of compensating windings is introduced or slotted

in the field winding arrangement to compensate for the loss due the distortion caused by the
armature reaction on the flux distribution.

ii Moving Brush Gear ; The movable or adjustable brush gear is used for adjusting the brush
contacts for proper and firm connection with the commutator to avoid sparking.

c)
The speed of a d.c shunt motor increases if resistance is connected in series with the field winding
because, the field current decreases with increase in the field resistance results in reduction of flux
produced. The relationship between the speed and flux is inversely proportional. i.e as flux
increases, speed reduces .

1
Speed α
Φ

5a
Uses of Transformer in transmission and distribution systems.

i For step-up and step-down of electrical power, voltage and current

ii For coupling circuit of different voltage levels

b)

Three wire three-phase distribution system is use where neutral connection is not needed on the load.
Such as delta connected load. The disadvantage here is that single- phase load can not be obtained on
this distribution system.

Four wire three-phase distribution system is use where neutal connection is needed on the load. Such
as star connected load. The advantage here is that single-phase load can be obtained on this
distribution system.

c)

Advantages of copper conductors

i lower resistivity and higher conductivity

ii copper conductors may be annealed or hard-drawn

Advantages of mineral ( magnesium oxide ) insulation

i It is water proof and resistance to ultraviolet light and many corrosive elements
ii It does not initiate an explosion even during circuit fault conditions

6 a i)
T
he −lo
ad −p
owe
r =IV c
os φ
T
he − lo
ad −p
ower 18 × 10 3
c
urre
n t −ta
ken −b
y −lo
ad = = = 113 .42 A
V cos φ 2
30 × 0.69
T
he −im
pro
ved −p
owe
rfa
cto r − r −
fo the −re
d u
ced − c
urre
nt −is − giv
ing −b
y
×
c
os φ′=
1
8
2
8
×
0 3
1
230
≅0.9
5

φ ′=1
8 .2 0

c
os φ = 0 .6
9 =p
owe
rfa
cto r −b
efo
re −c
urre
nt −re
duc
tio
n
φ= 46 .4 0
ii)
18 ×10 3
KVAR − rating = KW (Tan φ − Tan φ′) = (Tan 46 .4 0 − Tan 18 .2 0 ) = 12 .984 KVAR ≅ 13 .0 KVAR
1000

2
V
VAR = c ; 1 1
Xc Xc = C= = 776 µF
2π × 50 × C 2π × 50 × 4.1
2
Vc 230 2
Xc = = = 4.1Ω
VAR 13000

7 a)
ii )

magneti sin g − component − of − the − no − load − current = no − load − current × sin φ0

I m = I 0 sin φ0 = 3.8 × sin 71 .94 0 = 3.61 A
core − loss − component − of − the − no − load − current = no − load − curret × cos φ0
I c = I 0 cos φ0 = 3.8 × 0.31 = 1.178 A
b)

I p = primary − current
I s = sec ondary − current
′
I p = primary − current − due − to − load
V p = primary − voltage
Vs = sec ondary − voltage
φ p = phase − angle − of − primary − current − due − to − load
I 0 = no − load − current
R p = primary − resis tan ce
Rs = sec ondary − resis tan ce
R L = Load

8a

b)
The salient-pole type of rotors are not suitable for high speed machines because they can not
withstand the centrifugal forces developed in the large sizes at high speed.

c)
Essential differences in the construction between cage rotors and wound rotors for induction motors
are:
The cage rotor consists of a cylindrical laminated core with parallel slots for carrying the rotor
conductors which are not wires but heavy bars of copper , aluminum or alloys. The rotor bars are
brazed or electrically welded or bolted to two heavy and stout short-circuiting end-rings , hence , it is
not possible to add any external resistance in series with the rotor circuit for starting purposes. The
rotor slots are usually not quite parallel to the shaft but are purposely given a slight skew. While, the
wound rotor is provided with three-phase , double- layer distributed winding consisting of coils as
used in alternators. The rotor is wound for as many poles as the number of stator poles and is always
wound three-phase even when the stator is wound two- phase . The three- phases are starred
internally. The other three winding terminals are brought out and connected to three insulated slip-
rings mounted on the shaft with brushes resting on them. These three brushes are further externally
connected to three-phase star-connected rheostat. This makes possible the introduction of additional
resistance in the rotor circuit during the starting period for increasing the starting torque of the
motor.

d) The percentage slip when the rotor of a three-phase machine is at standstill is 100%

9a

15000
loadcurren t ( I ) = = 65 .22 A
230
230
fieldcurre nt ( I f ) = = 2A
115
2 2
inputpower − to − the − generator = VI + I a Ra + I f Rhs + rotational losses
= 15000 + 67 .22 2 × 0.4 + 2 2 ×115 + 700 = 17 .96 KW
VI 15000
Operationa l − efficiency = 2 2
×100 = ×100 = 83 .5%
VI + I a Ra + I f Rsh + rotational losses 17967 .4
b)

The field current of a shunt generator is a function of the terminal voltage. Increased loading cause a
drop in excitation current. The induced voltage of shunt generator is easily controlled by varying the
excitation current by means of a reheostat connected in series with the shunt field winding.
ii The method of restoring the terminal voltage of the above generator to its previous value is by
connecting a rheostat or variable resistor in series with the field winding as shown below.

10 a

TNC-S Supply system

In this system of earthing, the neutral and the earth ( protective earth conductor ) are combined at
the source but separated at the consumer end.

b)

The purpose of earthing system are :

i To protect the personnel from severe electric shock
ii To protect electrical appliances from damage

c)

The advantage of a closed ring main system of distribution compared with a radial system is that it
is more reliable in terms of provision of constant supply of electricity to the final circuit.

d)
The function of circuit breaker is to automatically and quickly break electrical circuit during fault
condition.
June 2007

1a

b)
P1 = Re ading − of − wattmeter − one
P2 = Re ading − of − wattmeter − two
φ = phase − angle
 − 1
P2 − P1 P2 = 3.8 KW
φ = tan 3( )
P2 + P1 P1 =1.2 KW

3.8 −1.2
φ = tan −1 3 ( ) = 42 0
3.8 + 1.2

Power factor is the cosine of the phase angle= cos42 = 0.74

When the connection to the current coil of wattmeter is reversed, it gives a negative reading. Thus,
P1 = −1.2 KW
P2 − (−P1 ) 5
φ = tan −1 3 ( ) = tan −1 3 ( ) = 73 .3 0
P2 − P1 6
Power − factor = cos 73 .3 = 0.29 ≅ 0.3

2a

Vt 408
If = = = 2.55 A
RSH 160
E g = Vt + I a Ra
P 55000
Ia = I f +I I = = = 134 .8 A
Vt 408
I a = 2.55 +134 .8 = 137 .35 A

E g − Vt 420 − 408
Ra = = = 0.0874 Ω
Ia 137 .35

ii
 Vt 402
If = = = 2.513 A
RSH 160
P 70000
I = = = 174 .13 A
Vt 402
I a = 2.513 +174 .13 = 176 .64 A

E g =Vt + I a Ra = 402 +176 .64 ×0.0874 = 417 .44 V

b)

ii

3a
 b) The use of direct-on-line starting is sometimes limited in terms of the power rating of the motor
because of the high starting current associated with induction motor in which the supply voltage to
the motor need to be reduced at starting in order not to affect the operation of other appliances in the
same circuit. The direct-on-line starting lack the provision for voltage reduction at starting thus,
limited to small induction motor of fractional horse- power.

4a

The e.m.f induced in the primary winding of a transformer E p = 4.44 ϕfN p

The e.m.f induce in the secondary winding of the transformer E s = 4.44 ϕfN s
For an ideal transformer, the power input is equal to the power output
powerinput = poweroutpu t , E p 4.44ϕfN p Np
= =
V p I p = Vs I s E 4.44ϕfN N
s s s

Vp Np Is Ep V p = primary − voltage
= = =
Vs Ns Ip Es Vs = sec ondary − voltage
I p = pimary − current
I s = sec ondary − current
N p = primary − turns
N s = sec ondary − turns
b)
The core construction of a shell-type transformer in which the primary and secondary windings are
wound on the centre limb of the transformer core.

E p = primary − induced − e.m. f

E s = sec ondary − induced − e.m. f
V p = primary − voltage ϕm = main − flux
Vs = sec ondary − voltage
I p = pimary − current
I s = sec ondary − current
N p = primary − turns
N s = sec ondary − turns

c)

The main advantage of the shell-type of construction compared with the core-type is that the leakage
flux is reduced.

5a
The power factor of a single-phase circuit may be determined by connecting voltmeter across the
circuit, ammeter in series and the wattmeter as shown in the circuit above.

b)

Cos φ = 0.72 φ = 44 0 ; cos φ / = 0.95 φ / =18 .2 0

The horizontal component of load current (I) @ 0.72 pf lagging = 75 × 0.72 = 54 A

The vertical component of load current (I) @ 0.72 pf lagging = 75 × sin 44 = 52 .1A

The horizontal component of load current ( I ′ ) @improved pf of 0.95lagging = I ′ × 0.95

Since the horizontal component of load current before improvement is equal to the horizontal
component of load current after improvement. It implies that;

I cos φ = I ′ cos φ′
54
54 A = I ′ × 0.95 I′ = = 56 .8 A
0.95

The new supply current= 56.8A

KVAR rating of the capacitor required fro this improvement is given by

230 × 54
KVAR = KW (Tan φ − Tan φ′) = (Tan 44 0 − Tan 18 .2 0 ) = 7.910 KVAR ≅ 8.0 KVAR
1000

OR

Vertical component of the load current (I ) after improvement

= I ′ sin φ′ = 56 .8 ×sin 18 .2 = 17 .74 A

The capacitor current required to be injected to achieve the improvement is given by

I C = I sin φ − I ′ sin φ ′ = 52 .1 −17 .74 = 34 .40 A

For single- phase,

 VP I C 230 × 34 .4
The KVAR rating = = = 7.912 KVAR ≅ 8.0 KVAR
1000 1000

c)

i Suitable method of power factor improvement for individual loads is the use of shunt capacitor
connected across the load
ii Suitable method of power factor improvement for overall systems is by the use of synchronous
motor.

6a
In an IT supply system of earthing , the distribution system has no connection to earth at all, or it
has only high impedance connection . In such system , an insulation monitoring device is used to
monitor the impedance.

b)

Exposed-conductive-part is the metalwork of an electrical appliance or the trunking and conduit of

an electrical system which can be touched because they are not normally live , but which may
become live under fault conditions.

c)
Four advantages of mineral-insulated, metal sheathed

i It is mechanically robust and resistance to impact

ii It is water proof and resistance to ultraviolet light and many corrosive elements
iii It does not initiate an explosion even during circuit fault conditions
iv It does not contribute fuel or hazardous combustion products to a fire and cannot propagate a fire
within a building

7
a
Inputpower = 3VI cos φ = 3 ×95 × 400 ×0.8 = 52 .66 kw
Inputpower −to − rotor = (inputpower − of −the − motor ) − statorloss es
= 52 .66 −1.6 = 51 .1kw
rotor − copperloss = slip ×inputpower −to − rotor = 0.04 ×51 .1 = 2.04 kw

1
Total − mechanical power − developed ( Pm ) = rotorcoppe rloss ×(1 − slip ) ×
slip
1
= 2.04 kw ×(1 − 0.04 ) × = 48 .96 kw
0.04
Outputpowe r = inputpower − losses = 52 .66 − (2.04 + 0.84 +1.6) = 48 .2kw

Rotorcurre nt − frequency = Slip × Synchronou s − frequency = 0.04 ×50 = 2 Hz

8
ii − Efficiency @ full − load − unity − powerfacto r
poweroutpu t − in − kw @ unity − powerfacto r = 50 ×10 3 ×1 = 50 kw
ironloss = VI 0 cos φ0 = 230 × 4.5 × 0.28 = 289 .8w
totalloss = ironloss + copperloss = 289 .8 + 600 = 889 .8w = 0.89 kw
50
efficiency = ×100 = 98 .3%
50 + 0.89
Efficiency @ 50 % full − load @ pf − of − 0.8lagging
sin ce − (kva ) 2 α − copperloss
power − output − in − kw @ 50 % full − load @ pf − of − 0.8 = 50 ×10 3 × 0.8 × 0.5 = 20 kw
1
copperloss @ 50 % full − load = ( ) 2 × 600 = 150 w
2
Totalloss @ 50 % full − load = 289 .8 +150 = 439 .8w = 0.4398 kw
20
Efficiency = ×100 = 97 .9%
20 + 0.4398

i
Efficiency @ 25 % full − load @ pf − of − 0.6lagging
sin ce − (kva ) 2 α − copperloss
power − output − in − kw @ 25 % full − load @ pf − of − 0.6 = 50 ×10 3 × 0.6 × 0.25 = 7.5kw
1
copperloss @ 25 % full − load = ( ) 2 × 600 = 37 .5w
4
Totalloss @ 25 % full − load = 289 .8 + 37 .5 = 327 .3w = 0.3273 kw
7.5
Efficiency = ×100 = 95 .8%
7.5 + 0.3273

9a

b)

Earth fault loop impedance is the impedance of the line-earth loop path and neutral –earth loop path
during fault or under fault condition.

c)

Resistance area is the area where voltage gradients exist.

ii The reason for considering resistance area when measuring earth electrode resistance is to
prevent overlapping of the resistance area of the actual electrode and the auxiliary earth electrode .
10 a

E b = Vt − I a Ra
Ia = I −I f
I a = 14 −1.25 = 12 .75 A
E b = 200 − (12 .75 × 0.4) = 194 .9V
Poweroutpu t
Operating − efficiency = ×100
Powerinput
2500
= ×100 = 89 .3%
200 ×14
Outputpowe r = Torque × angularvel ocity
2500
T = = 24 .9 Nm =
960
2π ×
60

Vt 200
If = = = 1.25 A
Rsh 160
I = 14 A

b) 8% of 12.75=1.02A
The new armature current= 12.75-1.02 =11.73 A
E b1 = Vt − I a1 Ra = 200 − (11 .73 × 0.4) = 195 .31V
N × E b1 960 × 195 .3
new − speed ( N 1 ) = = = 962 rev / min
Eb 194 .9
Note: it is assumed that the field excitation is kept constant
Dec 2007

1a

Eb1 = V − I a Ra = 230 − (8 × 0.25 ) = 228 V ii)

Eb2 = V − IR a = 230 − (60 × 0.25 ) = 215 V
@ no − load
Eb1 = N1Φ1
@ full − load
Eb2 = N 2 Φ2
if − the − flux − is − maitained − cons tan t ,
Φ1 = Φ2
N1 × Eb2 750 × 215
N2 = = = 707 .24 rev / min
Eb1 228

flux − is − reduced − by − 50 %
Φ1 = Φ
Φ2 = 0.95 Φ
Eb1 Φ1 N 1
=
Eb2 Φ2 N 2
Eb2 × N 1 215 × 750
N2 = = = 744 .5rev / min
Eb1 × 0.95 228 × 0.95

b) When resistance is connected in series with fielding winding, the field current is reduced,
reducing the flux produced. Since the flux is inversely proportional to the speed. The speed
1
of the motor increases. i. e Nα
Φ
c) The two signs of poor commutation , resulting from the effect of armature reaction that may
be visible in d.c machine are;
i) Sparking at the commutator contact
ii) Wearing of carbon brush
Resulting in increase in motor speed and reduction in generator terminal voltage

2a
Three disadvantages of low power factor
CONSUMER
i) Extra-cost of power consumption
ii) Poor power efficiency
iii) High installation cost
SUPPLIER
i) Excessive power loss due to reactive power
ii) Poor voltage regulation
iii) High cost of installation
b)

Cos φ = 0.71 φ = 44 .8 0 ; cos φ / = 0.95 φ / =18 .2 0

The horizontal component of load current (I) @ 0.71 pf lagging = 77 × 0.71 = 54 .7 A
The vertical component of load current (I) @ 0.71 pf lagging = 77 × sin 44 .8 = 54 .3 A

The horizontal component of load current ( I ′ ) @improved pf of 0.95lagging = I ′ × 0.95

Since the horizontal component of load current before improvement is equal to the horizontal
component of load current after improvement. It implies that;

I cos φ = I ′ cos φ′
54 .7
54.7 A = I ′ × 0.95 I′ = = 57 .6 A
0.95
The new supply current= 57.6A

KVAR rating of the capacitor required fro this improvement is given by

=
3 × 400 ×54 .7
KVAR = KW (Tan φ −Tan φ′) = (Tan 44 .8 0 −Tan 18 .2 0 ) = 25 .174 KVAR ≅ 25 .0 KVAR
1000
3a
i) Dielectric stress is the stress placed upon a material when voltage is applied across it.
ii) Dielectric strength is the material’s ability to withstand voltage breakdown. It is the
insulation ability to contain or
withstand voltage without breaking down.

b) One method by which a more uniform distribution of dielectric stress may be achieved in
high voltage cable is by
inter-sheathing.
Four environmental conditions which may influence the choice of cable for particular
applications are :
i) Tempature
ii) Chemical
iii) Underground
iv) Water or oil
 d) Two advantages of the materials used in PVC-Insulated , PVC Sheath are :
i) For insulation
ii) For mechanical protection

4a
i) Purpose of measuring the insulation resistance of a wiring system is to help ensure
specifications are met and to verify proper hook-up and prevent electric shock.

iii)
Two dangers which could develop if a wiring system has an unacceptable low
value of insulation resistance are; Electric shock and Short circuit
b) The combined insulation conductance
1 1 1 6 1
= + + = =
4 12 6 12 2
The − combined − insulation − resis tan ce = 2 MΩ

c) Testing of continuity of circuit protective conductors and bonding conductors of an

electrical installation is necessary before it is put into service to ensure the safety of the
personnel and prevent the electrical appliances from damage incase of faulty condition.
5 a) A synchronous motor is not self- starting because the magnetic flux produced by the stator
ran
Ahead ( synchronous speed) of the magnetic flux produced by the rotor. For torque to be
developed , the magnetic flux of the rotor must be brought to the speed of the magnetic flux
produced by the stator.

b)
i) Using a prime-mover ( induction motor )
ii) Using damper winding

c)

i) Low speed------------------ Salient pole rotor

ii) High speed------------------- Cylindrical rotor

120 f 120 × 50 .4
d) Speed − of − a − turbine = = = 1512 rev / min
p 4

6 a)
 E p = primary − induced − e.m. f
E s = sec ondary − induced − e.m. f
V p = primary − voltage ϕm = main − flux
Vs = sec ondary − voltage
I p = pimary − current
I s = sec ondary − current
N p = primary − turns
N s = sec ondary − turns

Ironloss = 620 w(cons tan t )

2
1 
b) Copperloss − @ − half − full − load =   × 960 = 240 w
2
Transforme r − loss @ half − full − load = 620 + 240 = 860 w

c) Advantage of autotransformer over double-wound transformer

i) Cost saving since less copper is needed

Disadvantage
i) Since the primary and secondary windings are not electrically separated, an
open-circuit in the secondary winding causes a full primary voltage to
appear across the secondary winding
ii) For starting induction motor and for inter-connecting systems that are
operating at approximately the same voltage.
7 a)
A combined protective earth – neutral (PEN ) conductor fulfills the functions of both a protective
earth ( PE ) and a neutral ( N) conductor. That the neutral and earth are connected together both at
the supply and at the consumer.

b)
Exposed-conductive-part is the metalwork of an electrical appliance or the trunking and conduit of
an electrical system which can be touched because they are not normally live , but which may
become live under fault conditions.

c)
Disadvantages of rewirable fuse

i) The fuse element may be replaced with wire of the wrong size either deliberately or
by accident.
ii) The circuit cannot be restored quickly since the fuse element requires screw fixing

The main contactor K1 will energize only when the control circuit fuse (F3), backup fuse
(F1), and the overload relay (F2) are healthy and the start pushbutton (S1) is pressed.
Reduced-voltage configuration (star configuration)
Star–delta timer coil (K4) gets power through fuses F3, F1, NC contact of stop
pushbutton (S0), and NO contact of start push button. As start PB (S1) is pressed, the
timer coil K4 will pickup and in turn energize the star contactor coil K2. The main line
contactor (K1) coil gets power via the NC contact of S0, NO contact of S1, NO contact of
K2 and remains latched unless the stop pushbutton (S0) is pressed.
Now, the main line contactor (K1) and the star contactor (K2) are in a pickup condition,
which will drive the motor in the star configuration.
Full voltage (delta configuration)
As the time duration set on a K4 timer (star to delta timer) elapses, the contactor coil (K3)
is picked up and at the same time, the star contactor (K2) is de-energized.
Now, the main line contactor (K1) and the delta contactor (K3) are in a pickup
condition, which will drive the motor in a delta configuration. When the motor trips in an
overload condition either in a star or delta configuration, the control circuit always
ensures that the motor restarts in a star configuration, rather than the delta configuration.

b) For direct-on-line starting (DOL )

The starting current = 5 x full-load

For star- delta starting

1
The starting current = 5 × line − voltage ×
3
For autotransformer starting on 70% tapping

The starting current = 5 x 0.7 x full - current

(R +r) = X
where ( r ) = external − resis tan ce
R = rotor − resis tan ce − per − phase = 0.4
X = rotor − reac tan ce − per − phase = 5Ω
c) For − max imum − torque − to − occur
@ starting
(R +r) = X
( 0.4 + r ) = 5
r = 5 − 0.4 = 4.6Ω

9a

Iron-loss is due to eddy current and hysteresis loss. The eddy current loss is due to e.m.f
being induced in the winding and core of the transformer. While hysteresis loss is due to
molecular structure of the material from which the core is made from. The iron- loss is not
affected by transformer load. It remains constant for a specific transformer. Copper loss is a
loss that occurs due o the resistance of the winding or conductor. Copper loss varies with the
transformer load by the relationship stated below

copperloss α( KVA )
2

b)
 poweroutpu t
Efficiency = ×100
powerinput
poweroutpu t = 200 ×10 3 × 0.8 = 160 kw
loss = iron − loss + copperloss = 3 + 4.15 = 7.15 kw
160
Efficiency = ×100 = 95 .7%
160 + 7.15

c) The main advantage of using a sandwich type of winding in transformer

construction is to reduce the leakage fluxes.

10 a

Vt 230
If = = = 2A
RSH 115
E g = Vt + I a Ra
P 15000
Ia = I f +I I = = = 65 .2 A
Vt 230
I a = 2 + 65 .2 = 67 .2 A

E g = Vt + I a Ra = 230 + (67 .2 × 0.22 ) = 244 .8V

2 2
Total − copperloss = I f Rsh + I a Ra = 2 2 ×115 + 67 .2 2 × 0.22 = 460 + 993 .5 = 1453 .5w ≅ 1.5kw

ii
Vt 226
If = = = 1.97 A
RSH 115
P 18000
I = = = 79 .7 A
Vt 226
I a = 1.97 + 79 .7 = 81 .7 A

E g =Vt + I a Ra = 226 +81 .7 ×0.22 = 243 .974 V ≅ 244 V

 SUPPLEMENTARY QUESTIONS & ANSWERS

DC MACHINES

1 a)

State how the torque developed by a dc motor varies with EACH of the following
i) field excitation
ii) armature current

b)
List FOUR losses occurring in a dc machine.
c)
Sketch a circuit diagram showing the field connection for a separately excited dc machine

2 a) Show , by means of connection diagrams , two different methods by which a

variable resistance may be used to control the speed of a dc shunt –wound motor. In each
case state the effect on the speed when the resistance is increased.

SOLUTION

1 a)
i) TαI f ( Torque is directly proportional to the field current )
TαI a ( Torque is directly proportional to the square of the armature current )
2
ii)
 b)
i) copper loss
ii) friction and windage loss
iii) iron loss
iv) brush contact loss

c)

In flux or field control method, the flux can be varied with the help of variable resistance
connected in series with field winding. Since the speed varies inversely with the flux, hence, by
increase the resistance , the flux reduces and the speed increases.

In armature control method, the voltage across the motor armature is changed with the help
of controller resistance connected in series with the armature, an increase in the resistance causes
the potential difference across the armature to decrease, thereby decreases the speed. Speed is
directly proportional to armature current.

AC MACHINE

1 a)

A 90kw three-phase , six pole induction motor supplied at 415V,50Hz operates on full-load
at 0.86 power factor lagging with a slip of 4%. The rotational losses absorb a torque of
16Nm and the stator and rotor copper losses are of equal value. Calculate the
i) rotational power losses
ii) rotor copper loss
v) input power
vi) full-load efficiency

SOLUTION

1 a)

torque × 2π × rotorspeed
Rotational powerloss =
60
120 × frequency 120 ×50
synchspeed = = =1000 rev / min
number − of − poles 6
rotorspeed = synchspeed (1 − slip ) =1000 (1 − 0.04 ) = 960 rev / min
 torque × 2π × rotorspeed 16 × 2π × 960
Rotational powerloss = = = 1608 .5w
60 60
slip × rotorgross output − power slip × (motoroutpu t + rotational power )
Rotorcoppe rloss = =
1 − slip 1 − slip
0.04 × (90000 +1608 .5)
= = 3817 .02 w
1 − 0.04
inputpower − of − the − motor = motoroutpu t + totallosse s
= 90000 + (1608 .5 + (2 × 3817 .02 )) = 99242 .54 w
motoroutpu t 90000
full − load − efficiency = ×100 = ×100 = 90 .7%
motorinput 99242 .54

2 a)

A three- phase, 550V, 50Hz, 18.65kw, 4 pole induction motor is supplied at the rated
voltage and frequency. The starting torque is equal to the full-load torque and at full-load
the slip is 4%. Calculate the starting torque in Nm. What will be the approximate starting
torque if the supply voltage falls to 520V with the frequency and slip remaining constant.

SOLUTION

motoroutpu t 18650
Full − torque = =
rotorspeed rotorspeed
2π × 2π ×
60 60
120 × frequency 120 ×50
synchspeed = = = 1500 rev / min
number − of − pole 4
rotorspeed = synchspeed (1 − slip ) = 1500 (1 − 0.04 ) = 1440 rev / min

motorput 18650
Full − torque = = = 123.7 Nm
rotorspeed 1440
2π × 2π ×
60 60
TαsV 2
2
TST sVST
= 2
TF sVF
2
VST × TF 520 2 × 123.7
TST = 2
= = 110.6 Nm
VF 550 2

b)
 A 6 pole, three-phase induction motor runs at a speed of 960rev/min when the shaft torque
is 136Nm and the frequency 50Hz. Calculate the rotor copper loss if the friction and
windage losses are 150w

Shafttorqu e × 2π × rotorspeed 136 × 2π × 960

motoroutpu t = = = 13672 .2 w
60 60
slip × rotorgross output
rotorcoppe rloss =
1 − slip
synchspeed − rotorspeed
slip =
synchspeed
120 × frequency 120 × 50
synchspeed = = = 1000 rev / min
number − of − poles 6
synchspeed − rotorspeed 1000 − 960
slip = = = 0.04
synchspeed 1000
slip × rotorgross output 0.04 ×(13672 .2 +150 )
rotorcoppe rloss = = = 575 .93 w
1 − slip 0.96

c)
A 50Hz , four-pole , three-phase induction motor operates under a variety of
load conditions. Calculate the
i) slip when the rotor current frequency is 3Hz
ii) rotor current frequency when the rotor speed is 750 rev/min
vii) slip speed when slip is 4%

SOLUTION
120 ×50
synchspeed = =1500 rev / min
4
rotorfrequ ency 3
slip = = = 0 .6
sup plyfrequen cy 5
synchspeed − rotorspeed 1500 − 750
slip = = = 0 .5
synchspeed 1500
rotorfrequ ency ( f r ) @ 750 rev / min@ 0.5slip = slip ×sup plyfrequen cy = 0.5 ×50 = 25 Hz
slipspeed = synchspeed − rotorspeed
rotorspeed @ 0.04 slip = synchspeed (1 − slip ) = 1500 (1 − 0.04 ) =1500 × 0.96 = 1440 rev / min
slipspeed =1500 −1440 = 60 rev / min

d) Define the following terms when referring to ac induction motors

i) synchronous speed
ii) slip
SOLUTION

Synchronous speed is the speed of rotating magnetic field produced by the

stator winding of a three-phase induction motor.
 Slip is the fractional difference in speed of synchronous speed and the rotor
speed

e) state the type of rotor construction suitable for each of the following
synchronous machines
i) low speed
ii) high speed

SOLUTION
Low speed……………. Salient pole rotor

High speed………………Cylindrical rotor

TRANSFORMER

1 a) A 500 kVA transformer has a full load copper loss of 4kW and an iron loss
of 2.5kW. Determine (a) the output kVA at which the efficiency of the
transformer is a maximum, and (b) the maximum efficiency, assuming
the power factor of the load is 0.75
b) Define the term’ voltage regulation’ of a transformer
c) A 500/250V , 10KVA single-phase transformer has primary and secondary winding
resistances of 0.3 Ω and 0.02 Ω respectively. The primary and secondary leakage
reactances are 1.2 Ω and 0.05 Ωrespectively. Calculate the
i) equivalent impedance referred to the primary
ii) full-load voltage regulation at a power factor of 0.8 lagging

SOLUTION
1 a)

Let x = the fraction of the output at which maximum efficiency occur

→ copperloss ( x 2 ) = 2500
2500
x= = 0.8
4000
the − output − at − which − efficiency − is − max imum = 0.8 × 500 = 400 KVA
Rated − outputpowe r − in − kw = KVA cos θ = 500 × 0.75 = 375 kw
outputpowe r 375000
max imum − efficiency = ×100 = ×100 = 98 .7%
outputpowe r + losses 375000 + ( 2 × 2500 )
b)

Voltage regulation is the difference in transformer terminal voltage at no-load and full-load.

c)
 2 2
Ze = Re + X e
2 2
V p 
 = 0.3 + 0.02 
500 
Re = R p + Rs   = 0.38 Ω
 Vs   250 
2 2
V p 
 = 1.2 + 0.05 
500 
X e = X p + X s   = 1.4Ω
 Vs   250 
Z e = 0.38 2 + 1.4 2 = 1.451 Ω
10000
primary − current = = 20 A
500
I p ( Re cos θ + X e sin θ ) 20 (0.38 × 0.8 + 1.4 × sin 36 .9)
voltage − regulation = = = 0.05
Vp 500

2 a) A 15KVA , 1000/ 250V single- phase transformer has primary and secondary winding
resistances of 0.8 Ω and 0.1 Ω respectively. The primary and secondary leakage reactances are
1.6 Ω and 0.06 Ω
Calculate the equivalent
i) resistance referred to the secondary
ii) leakage reactance referred to the secondary.
b) Calculate for full-load conditions, the
i) voltage drop at a power factor of 0.6 lagging
ii) per unit regulation at a power factor of 0.95 lagging

2
V  2

Re = R s + R p  s  = 0.1 + 0.8

250 
 = 0.15 Ω
V  1000 
 p 
2
V  2

Xe = Xs + X p s  = 0.06 +1.6 
250 
 = 0.16 Ω
V   1000 
 p 
15000
sec ondary − current = = 60 A
250
I ( R cos θ + X e sin θ ) 60 (0.15 × 0.6 + 0.16 × sin 53 .1)
voltage − drop = s e = = 13 .1V

60 (0.15 × 0.95 + 0.16 sin 18 .2)

per − unit − regulation @ 0.95 lagging = = 0.05
250

TRANSMISSION & DISTRIBUTION

1 a)
State two advantages of the insulation and sheathing materials used in each of the following high
voltage cables
i) p.v.c insulated, p.v.c sheathed
ii) paper insulated, metal sheathed

b) Define the meaning of the term ‘dielectric stress’.

c) Define the term ‘ fusing factor’

d) state

i) the main effect of excessive dielectric stress in high voltage cables and cable terminations
ii) two methods by which the stress may be controlled.

SOLUTION

Advantages of pvc insulated , pvc sheathed

i) pvc insulation protection against short-circuit and electric shock
ii) pvc sheathed provides mechnical protection
Advantages of paper insulated , metal sheathed
i) paper insulation provides insulation against short-circuit and also reduces/ control
the
dielectric stress on the cable
ii) metal sheathed provides mechanical protection
b) Dielectric stress is the stress place on a high voltage cable when a voltage is applied
across it.
c) Fusing factor is the ratio of the fusing current to the rating capacity of fuse.
d) The effect of excessive dielectric stress in high voltage cables and cable terminations
are insulation breakdown and short-circuit.
Methods of controlling stress is by grading and intersheathing

2 a) i) Describe , with the aid of a diagram , the principle and method of earthing of a TT supply
system.
ii) Describe the main function of an inverse definite minimum time (IDMT) protection relay.
b) Define the meaning of the following terms
i) earth
ii) earth electrode
c) state two services where the metalwork shall not be used as a protective earth electrode.

SOLUTION

a)
 ii) The main function of an inverse definite minimum time(IDMT) protection relay is for
discrimination of fault current.

b)
Earth is the general mass of the earth
Earth electrode is the electrode that connect the protective conductor to the general mass of
the earth

c)

3 a) Show , by a labeled diagram , how a power factor meter is connected into a single- phase
installation.
b) Explain why the used of a residual current device (RCD) is required for installations having
an increased shock risk.

SOLUTION
a)

b)
Unlike earthing system, the RCD trips immediately it senses imbalanced in the current
flowing in the neutral and live conductors. This action prevents the personnel from having
contact with the fault current.