7/26/2019 ECE320 HW3 Solution

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HOME WORK # 3 SOLUTIONS

1. An ideal transformer has an input voltage of 480 V. The output current and

voltage are 10 A and 120 V. Determine the value of input current.

Solution:

( )( )2 21

1

120 102.5 A

480

V II

V= = =

2. A 15-kVA, 2400:240-V, 60 Hz transformer has the following equivalent circuitparameters:

1 2.5R = 2 0.025R = 1 7.0X = 2 0.070X =

32cR k= 11.5mX k=

If the transformer is supplying a 10-kW, 0.8 PF lagging load at rated voltage, calculate:

(a) Input voltage

(b) Input current, and

(c) Input power factor

Solution:

Reflect 2R and 2X to the primary side:.

( )2 2

12 2

2

24000.025 2.5

240

NR R

N

= = =

( )2 2

12 2

2

24000.070 7.0

240

NX X

N

= = =

(a,b) Assume 2V on reference; then,

( )( )2 10 240 0 2400 0 VaV = =

The load current is found by

( ) ( )2

22

10,00052.08 A

240 0.8

PI

V PF= = =

( )12 2 cos 0.8 52.08 36.87 AI I = =

The reflected load current is

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2 2

15.208 36.87 A

10I I = =

The excitation branch voltage is found by use of KVL.

( ) ( )( )2 2 2 2 5.208 36.87 7.433 70.35 2400 0E I R jX aV = + + = +

2432.381 0.50 VE=

The excitation branch current follows as

2432.381 0.50 2432.381 0.50

32,000 11,500 90o c m

c m

E EI I I

R jX

= + = + = +

0.224 69.70 AoI =

The input current can now be determined by KCL.

1 2 5.208 36.87 0.224 69.70oI I I= + = +

15.398 38.16 AI =

Application of KVL yields the input voltage.

( ) ( )1 1 1 1 5.398 38.16 7.433 70.35 2432.381 0.50V I R jX E = + + = +

1 2466.61 0.99 VV =

1 2466.61 VV =

(c) With 1V and 1I known, the input power factor is given by

( ) ( )1 1cos cos 0.99 38.16 0.775 lagginginPF V I = = + =

3. A 20 kVA, 2500/250 V, 50Hz, single-phase transformer gave the following test

result:Open-circuit test: 250V, 1.5A, and 100 watts measured on the low voltage side

Short-circuit test: 100V, 8A, and 320 watts measured on the on high voltage side

a) Compute the parameters of the equivalent circuits and draw the equivalent circuit

of the transformer showing all the values.

b) Calculate the equivalent circuit of the transformer in per unit.

Solution:

From the given test results data, we have

25scwdg

sc

PR

I= =

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2

2 12scl wdgsc

VX R

I

= =

and

( )2

625oc

core

oc

V

R P= =

2 2

oc ococ

core m

V VI

R X

= +

Therefore, .RcoreandXmare low-voltage side values, which can be

converted to the high voltage side: R_core =Rcore*(N

173mX =

1/N2)2= 62.5 k,and X_m =Xm

*(N1/N2)2= 17.3 k.

We can find the inductances:

38.2100

ll

XL mH

= =

550.68100

mm

XL mH

= =

(b) Equivalent circuit inper unit:

2

1

2500312.5

20base

Zk

= =

2

2

250

3.12520baseZ k= =

,

2

625200,core pu

base

R puZ

= =

,

2

17355.36,m pu

base

X puZ

= =

,

1

120.0384,l pu

base

X puZ

= =

,

1

50.016,

wdg pu

base

R puZ

= =