2-3. A 1000-VA 230/115-V transformer has been tested to determine its equivalent circuit. The results of the tests are shown below.

Open-circuit test Short-circuit test VOC = 230 V VSC = 19.1 V IOC = 0.45 A ISC = 8.7 A POC = 30 W PSC = 42.3 W

All data given were taken from the primary side of the transformer. (a) Find the equivalent circuit of this transformer referred to the low-voltage side of the transformer. (b) Find the transformer’s voltage regulation at rated conditions and (1) 0.8 PF lagging, (2) 1.0 PF, (3) 0.8 PF leading. (c) Determine the transformer’s efficiency at rated conditions and 0.8 PF lagging.

SOLUTION

(a) OPEN CIRCUIT TEST:

1 1

0.45 0.001957230

30cos cos 73.15(230)(0.45)

0.001975 73.15 0.000567 0.0018731/ 17631/ 543

EX C M

oOC

OC OC

oEX C M

C C

M M

Y G jB S

PV I

Y G jB jR GX B j

θ − −

= − = =

= = =

= − = ∠− = − Ω

= = Ω= = Ω

SHORT CIRCUIT TEST:

1 1

19.1 2.28.7

42.3cos cos 75.3(19.1)(8.7)

2.20 75.3 0.558 2.128

0.558

2.128

EQ EQ EQ

oSC

SC SC

oEQ EQ EQ

EQ

EQ

Z R jX

PV I

Z R jX j

R

X j

θ − −

= − = = Ω

= = =

= + = ∠ = + Ω

= Ω

= Ω

To convert the equivalent circuit to the secondary side, divide each impedance by the square of the turns ratio (a = 230/115 = 2). The resulting equivalent circuit is shown below:

REQ,s = 0.532 Ω X EQ,s = j441 Ω RC,s = 134 Ω XM,s = 134 Ω

(b) To find the required voltage regulation, we will use the equivalent circuit of the transformer referred to the secondary side. The rated secondary current is

1000 /115 8.70SI A= = We will now calculate the primary voltage referred to the secondary side and use the voltage regulation equation for each power factor. (1) 0.8 PF Lagging:

' 115 0 (0.140 0.532 )(8.7 36.87 )

' 118.8 1.4

o oP S EQ S

oP

V V Z I j

V

= + = ∠ + + Ω ∠−

= ∠

118.8 115 100% 3.3%115

VR −= × =

(2) 1.0 PF:

' 115 0 (0.140 0.532 )(8.7 0 )

' 116.3 2.28116.3 115 100% 1.1%

115

o oP S EQ S

oP

V V Z I j

V

VR

= + = ∠ + + Ω ∠

= ∠−

= × =

(3) 0.8 PF Leading:

' 115 0 (0.140 0.532 )(8.7 36.87 )

' 113.3 2.24113.3 115 100% 1.5%

115

o oP S EQ S

oP

V V Z I j

V

VR

= + = ∠ + + Ω ∠

= ∠−

= × = −

(c) At rated conditions and 0.8 PF lagging, the output power of this transformer is

cos (115)(8.7)(0.8) 800OUT S SP V I Wθ= = =

The copper and core losses of this transformer are 2 2

,

2 2

(8.7) (0.140) 10.6

( ') (118.8) 32.0441

CU S EQ S

pcore

C

P I R W

VP W

R

= = =

= = =

Therefore the efficiency of this transformer at these conditions is

800100% 100% 94.9%800 10.6 32.0

OUT

OUT CU core

PP P P

η = × = × =+ + + +