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:
0.45
= 0.001957 S
230
30
= cos −1
= 73.15o
(230)(0.45)
YEX = GC − jBM =
θ = cos −1
POC
VOC I OC
YEX = GC − jBM = 0.001975∠ − 73.15o = 0.000567 − j 0.001873 Ω
RC = 1/ GC = 1763 Ω
X M = 1/ BM = j 543 Ω
SHORT CIRCUIT TEST:
19.1
= 2.2Ω
8.7
42.3
= cos −1
= 75.3o
(19.1)(8.7)
Z EQ = REQ − jX EQ =
θ = cos −1
PSC
VSC I SC
Z EQ = REQ + jX EQ = 2.20∠75.3o = 0.558 + j 2.128 Ω
REQ = 0.558 Ω
X EQ = j 2.128 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
I S = 1000 /115 = 8.70 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:
VP ' = VS + Z EQ I S = 115∠0o + (0.140 + j 0.532Ω)(8.7∠ − 36.87 o )
VP ' = 118.8∠1.4o
VR =
118.8 − 115
× 100% = 3.3%
115
(2) 1.0 PF:
VP ' = VS + Z EQ I S = 115∠0o + (0.140 + j 0.532Ω)(8.7∠0o )
VP ' = 116.3∠2.28o
VR =
116.3 − 115
× 100% = 1.1%
115
(3) 0.8 PF Leading:
VP ' = VS + Z EQ I S = 115∠0o + (0.140 + j 0.532Ω)(8.7∠36.87 o )
VP ' = 113.3∠2.24o
VR =
113.3 − 115
× 100% = −1.5%
115
(c) At rated conditions and 0.8 PF lagging, the output power of this transformer is
POUT = VS I S cos θ = (115)(8.7)(0.8) = 800 W
The copper and core losses of this transformer are
PCU = I S 2 REQ , S = (8.7) 2 (0.140) = 10.6 W
Pcore =
(V p ') 2
RC
=
(118.8) 2
= 32.0W
441
Therefore the efficiency of this transformer at these conditions is
POUT
800
η=
× 100% =
× 100% = 94.9%
800 + 10.6 + 32.0
POUT + PCU + Pcore