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Practice (3.7): The primary current to an ideal transformer rated at 2200/110 V is 25A. Calculate: (a) the turns ratio, (b) the kVA rating, (c) the secondary current. Ans: (a) 1/20, (b) 55 kVA, (c) 500 A.

Practice (3.7): The primary current to an ideal transformer rated at 2200/110 V is 25A. Calculate: (a) the turns ratio, (b) the kVA rating, (c) the secondary current. Ans: (a) 1/20, (b) 55 kVA, (c) 500 A.

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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1:1 VI O & 4G 4G K/s Z Electric Circui... -> Dr. Jamal A.-K. Mohammed Electric Circuits - CH. 3 9600 = 4A and 12 = V2 9600 (c) S, = V,1, = V,1, = 9.6 kVA = 1, = 2600 = 9600 = 4A and = 80A 120 !! %3D %3D 2400 Or 2 =4 =80A 0.05 %3D Practice (3.7): The primary current to an ideal transformer rated at 2200/110 V is 25A. Calculate: (a) the turns ratio, (b) the kVA rating, (c) the secondary current. Ans: (a) 1/20, (b) 55 kVA, (c) 500 A. Ex. (3.8): For the ideal transformer 1:2 circuit of Fig. 3.30, find: (a) the ww- + source current I. (b) the output 120/0° v rms + , 3||E v, 20 2 voltage Vo, and (c) the complex power supplied by the source. Fig. 3.30 Sol: (a) The 20-2 impedance can be reflected to the primary side and we get: Zg = 4 = = 5n 22 Thus, Zn = 4 - j6 + ZR = 4 - j6 + 5 = 9 - j6 = 10.822 – 33.69° N 12020° Vin Zin 11.09433.69° A 10.822-33.69 (b) Because both I and I; leave the dotted terminals, therefore n has negative sign: 11.09233.69 12 = = - - 12 = -5.545433.69° A 2 V, = Z,1, = 201, = 20(-5.545433.69°) = 110.94213.69° V (c) The complex power supplied is: S = V, 1; = (12020°)(11.092 – 33.69°) = 1,330.84 - 33.69° VA Dr. Jamal A.-K. Mohammed Electric Circuits - 3 Practice (3.8): In the ideal 2 16 2 1:4 ww transformer circuit of Fig. 3.31, find V. and the complex power 240/0° V rms E v, E V2 V. -1242 supplied by the source. Fig, 3.31 Ans: 429.4 2116.57° V, 17.1742 - 26.57° kVA
Transcribed Image Text:1:1 VI O & 4G 4G K/s Z Electric Circui... -> Dr. Jamal A.-K. Mohammed Electric Circuits - CH. 3 9600 = 4A and 12 = V2 9600 (c) S, = V,1, = V,1, = 9.6 kVA = 1, = 2600 = 9600 = 4A and = 80A 120 !! %3D %3D 2400 Or 2 =4 =80A 0.05 %3D Practice (3.7): The primary current to an ideal transformer rated at 2200/110 V is 25A. Calculate: (a) the turns ratio, (b) the kVA rating, (c) the secondary current. Ans: (a) 1/20, (b) 55 kVA, (c) 500 A. Ex. (3.8): For the ideal transformer 1:2 circuit of Fig. 3.30, find: (a) the ww- + source current I. (b) the output 120/0° v rms + , 3||E v, 20 2 voltage Vo, and (c) the complex power supplied by the source. Fig. 3.30 Sol: (a) The 20-2 impedance can be reflected to the primary side and we get: Zg = 4 = = 5n 22 Thus, Zn = 4 - j6 + ZR = 4 - j6 + 5 = 9 - j6 = 10.822 – 33.69° N 12020° Vin Zin 11.09433.69° A 10.822-33.69 (b) Because both I and I; leave the dotted terminals, therefore n has negative sign: 11.09233.69 12 = = - - 12 = -5.545433.69° A 2 V, = Z,1, = 201, = 20(-5.545433.69°) = 110.94213.69° V (c) The complex power supplied is: S = V, 1; = (12020°)(11.092 – 33.69°) = 1,330.84 - 33.69° VA Dr. Jamal A.-K. Mohammed Electric Circuits - 3 Practice (3.8): In the ideal 2 16 2 1:4 ww transformer circuit of Fig. 3.31, find V. and the complex power 240/0° V rms E v, E V2 V. -1242 supplied by the source. Fig, 3.31 Ans: 429.4 2116.57° V, 17.1742 - 26.57° kVA
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