0195136047.pdf

484 MAGNETIC CIRCUITS AND TRANSFORMERS

R 2 ′=a^2 R 2 =

(

2400

240

) 2

( 0. 0075 )= 0. 75 

X′ 2 =a^2 X 2 =

(

2400

240

) 2

( 0. 01 )= 1. 0 

Note that the exciting admittance on the 240-V side is given. The exciting branch

conductance and susceptance referred to the high-voltage side are given by

1

a^2

( 0. 003 ) or

1

100

× 0. 003 = 0. 03 × 10 −^3 S

and

1

a^2

( 0. 02 ) or

1

100

× 0. 02 = 0. 2 × 10 −^3 S

(b) The equivalent circuit referred to the low-voltage side is given in Figure E11.3.1 (b).

Note the following points.

(i) The voltages specified on the nameplate of a transformer yield the turns ratio directly.

The turns ratio in this problem is 2400:240, or 10:1.

(ii) Since admittance is the reciprocal of impedance, the reciprocal of the referring factor

for impedance must be used when referring admittance from one side to the other.

−jBm = −j0.2 × 10−3 S

RC =^ jXm = j

R 1 = 0.75 Ω jX 1 =j^1 Ω R 2 = a^2 R 2 = 0.75 Ω jX 2 = ja^2 X 2 = j^1 Ω

(a)

GC = 0.03 × 10 −^3 S

GC

1

3

=^105 Ω

-jBm

1

2

=^104 Ω

or or

2400 : 240

Ideal transformer

(can be omitted)

′ ′

−j 0.02 S

RC /a^2 =^ j

R 1 = R 1 /a^2 = 0.0075 Ω

jX 1 = jX 1 /a^2 = j 0.01 Ω

R 2 = 0.0075 Ω

′

′ jX 2 = j 0.01 Ω

(b)

a^2 GC = 0.003 S

3

1000

2

(^100) = jX
Ω Ω m /a^2
or or
2400 : 240
Ideal transformer
(can be omitted)
= −ja^2 Bm
Figure E11.3.1Equivalent circuit.(a)Referred to high-voltage side.(b)Referred to low-voltage side.