Electrical Technology (EET 103/ERT105) Tutorial Module

UNIVERSITI MALAYSIA PERLIS 1

TUTORIAL 5 – TRANSFORMER Question 1 The secondary winding of a transformer has a terminal voltage of vs (t) = 282.8 sin 377t V. The turns ratio of the transformer is 100:200. If the secondary current of the transformer is (t) =7.07 sin (377t - 36.87o) A, the impedances of this transformer referred to the primary side are:

Req = 0.20Ω Rc = 300Ω Xeq = 0.750Ω XM = 80Ω

Determine:

i. the primary current of this transformer. ii. the voltage regulation and efficiency.

(Ans: A.I o

p 41111 −∠= , VR = 6.2%, η = 93.7%)

Question 2 A 20-kVA 8000/480-V distribution transformer has the following resistances and reactances:

Rp = 32Ω Rs = 0.05Ω Xp = 45Ω Xs = 0.06Ω RC = 250kΩ XM = 30kΩ

The excitation branch impedances are given referred to the primary side of the transformer.

a) Find the equivalent circuit of this transformer referred to the primary side. b) Find the per-unit equivalent circuit of this transformer. c) Assume that this transformer is supplying rated load at 480V and 0.8 PF

lagging. What is this transformer’s input voltage? What is its voltage regulation?

d) What is the transformer’s efficiency under the conditions of part (c)? (Ans: a) R s’ = 13.9Ω, Xs’ = 16.7Ω b) Rp = 0.01, Xp = j0.0141, Rs’ = 0.0043, Xs’ = j0.0052, RC = 78.125, XC = j9.375 c) V.V o

p 3808185∠= , VR = 2.31% d) η =

96.6%)

Electrical Technology (EET 103/ERT105) Tutorial Module

UNIVERSITI MALAYSIA PERLIS 2

Question 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 = 230V IOC = 0.45A POC = 30W

VSC = 19.1V ISC = 8.7A

PSC = 42.3W 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.

(Ans: a) R eq,s =0.140Ω, Xeq,s = j0.532Ω, RC,s =441Ω, XM,s = 134Ω b) VR = 3.3%, 1.1%, -1.5%, η = 94.9%) Question 4 A 15-kVA 8000/230V distribution transformer has an impedance referred to the primary of 80 + j300 Ω. The components of the excitation branch referred to the primary side are RC = 350kΩ and XM = 70kΩ.

a) If the primary voltage is 7967 V and the load impedance is ZL = 3.0 + j1.5Ω, what is the secondary voltage of the transformer? What is the voltage regulation of the transformer?

b) If the load is disconnected and a capacitor of –j4.0Ω is connected in its place, what is the secondary voltage of the transformer? What is its voltage regulation under these conditions?

(Ans: V..V o

s 138218 −∠= , VR = 4.7% b) V..V os 215246 −∠= , VR = -7.07%)

Electrical Technology (EET 103/ERT105) Tutorial Module

UNIVERSITI MALAYSIA PERLIS 3

Question 5 A 5000-kVA 230/13.8-kV single-phase power transformer has a per-unit resistance of 1 percent and a per unit reactance of 5 percent (data taken from the transformer’s nameplate). The open-circuit test performed on the low-voltage side of the transformer yielded the following data:

VOC = 138kV IOC = 15.1A POC = 44.9kW

a) Find the equivalent circuit referred to the low-voltage side of this transformer.

b) If the voltage on the secondary side is 13.8 kV and the power supplied is 4000 kW at 0.8 PF lagging, find the voltage regulation of the transformer. Find its efficiency.

(Ans: a) Req = 0.39 Ω, Xeq = 1.9Ω b) VR = 3.84%, η = 97.6%)