1.A load of 800KW at 0.8 lagging pf is supplied by two 3Φ transformers rated 400

KVA and 600 KVA operating in parallel. The transformers have the same turn ratio

and their per unit resistance and reactance are:

Transformer Resistance (R) Reactance(X)

A 0.02 0.04

B 0.01 0.05

Calculate KVA output and pf of each Transformer.

Solution:

Load P in KVAfp

KWinLoadKVA 000,1

8.0

800

.

..

KVAP 09.36000,1

On the basis of 600 KVA:

% impedance of 400 KVA transformer

04.637.66342400

600 jjZA

% impedance of 600 KVA transformer

07.781.551 jZB

0708.111145163 jjjZZ BA

Load shared by 400 KVA Transformer;

0 00

0

1000 36.9 5.1 78.7432 28.2

11.8 70

BA

A B

Z xP P

Z Z

=432 KVA at lagging p.f 0.882 (lagging) Ans

Load shared by 600 KVA Transformer;

2.A 500 KVA single phase Transformer has 0.01 p.u resistance, 0.05 p.u reactance

and open circuit secondary voltage of 405V. It is connected in parallel with a 250

KVA single phase Transformer having 0.015 p.u resistance, 0.04 p.u reactance and

open circuit secondary voltage of 415V.

Find:

a) Circulating current at no-load.

b) Currents delivered by the secondary of each Transformer to a load of 750 KVA at 0.8 lagging p.f.

c) KVA and p.f of each transformer.

Take voltage across load to be 400V.

Solution:

Let 500 KVA Transformer be A

And 250 KVA Transformer be B

Rated current of Transformer A A1250400

500000

Rated current of Transformer B A625400

250000

04000.01 0.05 . 0.01 0.05 0.0032 0.016 0.0163 78.7

1250AZ j p u j j

04.690274.00256.00096.0625

40004.0015.0.04.0015.0 jjupjZB

Load KVA = 750 KVA at 0.8 p.f lagging

Therefore 7500002

Z

V

Or 08.362133.0750000

400400

xZ

a. Circulating current at no load

Since there is no load =>

0256.00096.0016.00032.0

415405

jjZZ

EEII

BA

BABA

A2309.720436.0

100

at a pf of Cos 72.90 i.e. 0.294 lagging.

b. From equation

)9.........(..........BALBA

LBABAA

ZZZZZ

ZEEZEI

and

)8.........(..........BALBA

LBAABB

ZZZZZ

ZEEZEI

, find the values of IA and IB.

3.)

A 33 KVA, 2200/220 V, 50 Hz single phase transformer has the following

parameters:

Primary winding (h.v side) – resistance r1 =2.4 Ω;

leakage reactance x1 = 6 Ω

Secondary winding (l.v side) – resistance r2 =0.03 Ω;

leakage reactance x2 = 0.07 Ω

(a). Find primary resistance and leakage reactance referred to

secondary.

(b). Find secondary resistance and leakage reactance referred to

primary.

(c). Find the total equivalent resistance and leakage reactance

referred to:

(i). Primary

(ii). Secondary

(d). Calculate the total ohmic loss at full load.

(e). Calculate the voltage to be applied to the h.v side, in order to

obtain a short circuit current of 160 A in the l.v. winding. Under

these conditions, find the power input.

Solution:

(a) Primary resistance referred to secondary

024.0

2200

2204.2

22

1

21

'

1N

Nrr

Primary leakage reactance referred to secondary

06.0

2200

22000.6

22

1

21

'

1N

Nxx

(b) Secondary resistance referred to primary

00.3

220

220003.0

22

2

12

'

2N

Nrr

Secondary leakage reactance referred to primary

00.71007.0

2

2

2

12

'

2N

Nxx

(c) (i) Equivalent resistance referred to primary

4.500.34.2'

211 rrre

Equivalent leakage reactance referred to primary

1300.700.6'

211 xxxe

(ii) Equivalent resistance referred to secondary

054.0024.003.0'

122 rrre

Equivalent leakage reactance referred to secondary

13.006.007.0'

122 xxxe

(d) Primary full load current AI 152200

330001

Secondary full load current AI 150220

330002

Therefore, ohmic loss at full load wattsrI e 12164.5152

1

2

1

wattsrI e 1216054.01502

2

2

2

(e) A current of 160A in the l.v. winding is equivalent to 16A in the

h.v. winding. The equivalent circuit of the transformer, referred to h.v.

side is illustrated in fig. below, from which equivalent leakage impedance

referred to h.v. side is

08.14134.5..

134.5

22

1

1

e

e

Zor

jZ

Fig. Equivalent circuit referred to h.v. side

Therefore, the voltage to be applied to the h.v. side is :

V= (16) (ze1) = (16) (14.08)=225 volts

Power input wattsrI e 13824.5162

1

2

1

Or power input wattsVICos 138208.14

4.516225

4.) 3. A 10 KVA, 2500/250 V, single phase transformer has resistance and

leakage reactance as follows:

r1 = 4.8Ω r2 = 0.48Ω

x1 = 11.2Ω x2 = 0.112 Ω

Subscripts 1 and 2 denotes high voltage and low voltage windings

respectively.

With primary supply voltage held constant at 2500V, calculate the

secondary terminal voltage, when

(a). The l.v. winding is connected to load impedance of 5 +j3.5Ω.

(b). The transformer delivers its rated current at 0.8 p.f lagging, on the

l.v. side.

Solution:

(a) All the quantities may be referred to either the h.v. side or the l.v.

side. In this example, l.v. winding is the secondary winding since load

connected across it. With all the quantities referred to the l.v. side, the

equivalent circuit is obtain, where

2

1

2122

N

Nrrre

096.0

10

18.4048.0

2

And

224.0

10

12.11112.0

22

1

2122

N

Nxxxe

The transformer leakage impedance re2 +jxe2 and the load impedance are in

series. Therefore, the total impedance in the secondary winding circuit is

02.3631.6724.3096.5 jz

Fig (a) and (b): Equivalent circuit for (a) and (b)

Hence the load current AI 65.3931.6

250

The secondary terminal voltage is

LIZV 2

V2421.665.395.3565.3922

(b). The phasor diagram for Fig. (b) is illustrated in Fig. (c) below.

Fig. (c) Phasor diagram for Fig. (b)

From the geometry of this diagram,

222 ABOAOB

Or 22222

2

2222250 ee xISinVrICosV

Or 22

2

2 96.86.084.38.0250 VV

Or 55.9590.16250 2

2

2

2 VV

Or 0404,6290.16 2

2

2 VV

Therefore VV 55.2412

616,2492869.162

Alternatively, the secondary terminal voltage V2 can be obtained as follows:

From the phasor diagram, it may be seen that OD is approximately equal to OB=250V

Therefore CDCDODV 2502

Now EDCECD

222222 SinxICosrI ee

V45.8448.86.096.88.084.3

Therefore secondary terminal voltage VV 55.24145.82502

The magnitude of secondary voltage V2 turns out to be same in both methods. However, the

computational labour in the second method is less than in the first method, therefore, the

second method should be preferred.