The secondary winding of a transformer has a terminal voltage of v (1) = 282.8 sin 377t V. The turns If the secondary current of the transformer is i,(1) = 7.07 sin (377t- 36.87°) A, what is the primary current of this transformer? What are its voltage ratio of the transformer is 100:200 (a = 0.50). regulation and efficiency? The impedances of this transformer referred to the primary side are R = 0.20 2 R = 300 2 %3D X =0.750 2 Xy = 80 2 eq Is Ip Reap jXeap + O RC jXM aVs Answer: Ip = 11.1 < -41.0°A VR = 6.2 % Reap = R, + a'Rs (a) %3D n = 93.7 % XeaP = X, + a°Xs %3D

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The secondary winding of a transformer has a terminal voltage of v,(t) 282.8 sin 377t V. The turns
ratio of the transformer is 100:200 (a = 0.50).
If the secondary current of the transformer is
i,(1) = 7.07 sin (377t- 36.87°) A, what is the primary current of this transformer? What are its voltage
regulation and efficiency? The impedances of this transformer referred to the primary side are
R = 0.20 2
R = 300 Q
= 0.750 2
X = 80 2
Ip
Reap
jXeap
RC
jXM
aVs
Answer:
= 11.1 < -41.0°A
p =
VR = 6.2 %
Reap = R, + aRS
Xegp = X, + a²Xs
(a)
n = 93.7 %
Transcribed Image Text:The secondary winding of a transformer has a terminal voltage of v,(t) 282.8 sin 377t V. The turns ratio of the transformer is 100:200 (a = 0.50). If the secondary current of the transformer is i,(1) = 7.07 sin (377t- 36.87°) A, what is the primary current of this transformer? What are its voltage regulation and efficiency? The impedances of this transformer referred to the primary side are R = 0.20 2 R = 300 Q = 0.750 2 X = 80 2 Ip Reap jXeap RC jXM aVs Answer: = 11.1 < -41.0°A p = VR = 6.2 % Reap = R, + aRS Xegp = X, + a²Xs (a) n = 93.7 %
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