1. The high-voltage coil of a transformer is wound with 700 turns of wire, and the lowvoltage coil is wound with 292 turns. When used as a step-up transformer (the lowvoltage coil is used as the primary), the load current is 10.5 A. Find the load component of the primary current. A. 43.5 A B. 4.38 A C. 25.18 A D. 2.518 A Solution: 𝑁𝑝 𝑁𝑠

=

𝐼𝑝 =

𝐼𝑠 𝐼𝑝 𝐼𝑠 𝑁𝑠 𝑁𝑝

=

(700)(10.5) 292

𝑰𝒑 = 𝟐𝟓. 𝟏𝟕 𝑨𝒎𝒑𝒆𝒓𝒆𝒔 REE – May 2008 2. A transformer has a primary winding of 2, 000 turns and of 2, 400 Volts and current of 8.66 − 𝑗5 Ampere with an impedance 𝑍2 connected across the secondary winding. If the secondary winding has 500 turns, what is the value of the secondary current? A. 20 − 𝑗34.64 𝐴 B. 𝟑𝟒. 𝟔𝟒 − 𝒋𝟐𝟎 𝑨 C. 34.64 + 𝑗20 𝐴 D. 20 + 𝑗34.64 𝐴 Solution: 𝑁𝑝 𝑁𝑠

=

𝐼𝑝 =

𝐼𝑠 𝐼𝑝 𝐼𝑠 𝑁𝑠 𝑁𝑝

=

(8.66−𝑗5)(2,000) 500

𝑰𝒑 = 𝟑𝟒. 𝟔𝟒 − 𝒋𝟐𝟎 𝑨𝒎𝒑𝒆𝒓𝒆𝒔

3. A 120 V to 27.5 V, 400 Hz step-down transformer is to be operated at 60 Hz. What is the highest safe input voltage? A. 200 V B. 400 V C. 120 V D. 18 V Solution: 𝐸1 𝐸2

=

𝐸2 =

𝑓1 𝑓2 𝐸1 𝑓2 𝑓1

=

(120)(60) 400

𝑬𝟐 = 𝟏𝟖 𝑽𝒐𝒍𝒕𝒔 REE – September 2011 4. When a welding transformer is used in a resistance welding, it will A. step up voltage B. step down voltage C. step up current D. step down current ∗ 𝑁𝑜𝑡𝑒: 𝐼𝑛 𝑎 𝑤𝑒𝑙𝑑𝑖𝑛𝑔 𝑡𝑟𝑎𝑛𝑠𝑓𝑜𝑟𝑚𝑒𝑟, 𝑎 ℎ𝑖𝑔ℎ 𝑣𝑎𝑙𝑢𝑒 𝑜𝑓 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑖𝑠 𝑛𝑒𝑒𝑑𝑒𝑑 𝑡𝑜 𝑚𝑒𝑙𝑡 𝑒𝑎𝑠𝑖𝑙𝑦 𝑡ℎ𝑒 𝑖𝑟𝑜𝑛.

5. A 4, 600/230 V, 60 Hz step-down transformer has core dimension of 76.2 mm by 111.8 mm. A maximum flux density of 0.93 𝑊𝑏/𝑚2 is to be used. Assuming 9 percent loss of area due to stacking factor of laminations, calculate the primary and secondary turns required. A. 2, 395 and 120 B. 120 and 2, 395 C. 2, 180 and 109 D. 109 and 2, 180 Solution:

𝐴𝑒𝑓𝑓 = (1 − 0.09)(76.2 𝑚𝑚 × 111.8 𝑚𝑚) 𝐴𝑒𝑓𝑓 = 7, 752.436𝑚𝑚2 (

1𝑚 1,000 𝑚𝑚 −3 2

𝐴𝑒𝑓𝑓 = 7.752 × 10 ∅𝑚 = 𝛽𝑚 𝐴𝑒𝑓𝑓 ∅𝑚 = (0.93

𝑊𝑏 𝑚2

)

2

𝑚

) (7.752 × 10−3 𝑚2 )

∅𝑚 = 7.209 𝑚𝑊𝑏 𝐸𝑟𝑚𝑠 = 4.44𝑓∅𝑁 𝑁𝑝 =

𝐸𝑟𝑚𝑠 4.44𝑓∅

=

4,600 (4.44)(60)(7.209×10−3 )

𝑵𝒑 = 𝟐, 𝟑𝟗𝟓 𝒕𝒖𝒓𝒏𝒔 𝑁𝑠 =

𝐸𝑟𝑚𝑠 4.44𝑓∅

=

230 (4.44)(60)(7.209×10−3 )

𝑵𝒔 = 𝟏𝟐𝟎 𝒕𝒖𝒓𝒏𝒔 REE – October 1997 6. A small single-phase transformer has 10.2 watts no-load loss. The core has a volume of 750 cubic cm. The maximum flux density is 10, 000 gauss and the hysteresis constant of the core is 5 × 10−4 , using the Steinmetz law to find the hysteresis, determine the eddy current loss. A. 4.55 Watts B. 5.55 Watts C. 3.55 Watts D. 2.55 Watts Solution: ∗ 𝑆𝑡𝑒𝑖𝑛𝑚𝑒𝑛𝑡𝑧 𝑙𝑎𝑤

𝑃ℎ 𝛼 𝛽𝑚 1.6 ∗ 𝑘ℎ = (5 × 10−4 )(750) 𝑘ℎ = 0.375 𝑃ℎ = (0.375)(60)(10, 000)1.6 𝑃ℎ = 5.652 𝑊𝑎𝑡𝑡𝑠 𝑃𝑒 = 𝑃𝑖𝑛 − 𝑃ℎ 𝑃𝑒 = 10.2 − 5.652 𝑷𝒆 = 𝟒. 𝟓𝟒𝟖 𝑾𝒂𝒕𝒕𝒔

𝑑𝑦𝑛𝑒−𝑐𝑚 𝑠𝑒𝑐

(

1𝑁 105 𝑑𝑦𝑛𝑒

)(

1𝑚 100 𝑐𝑚

)

REE – September 2006 7. The primary of transformer has 200 turns and is excited by a 240 V, 60 Hz source. What is the maximum value of the core flux? A. 4.04 mWb B. 4.40 mWb C. 4.13 mWb D. 4.32 mWb Solution:

∅𝑚 =

𝐸

240

4.44𝑓𝑁

= (4.44)(60)(200)

∅𝒎 = 𝟒. 𝟓𝟎 𝒎𝑾𝒃 REE – September 2008 8. A transformer is rated 1 kVA, 220/110 V, 60 Hz. Because of an emergency this transformer has to be used on a 50 Hz system. If the flux density in the transformer core is to be kept the same as at 60 Hz and 220 V, what is the kilovolt-ampere rating at 50 Hz. A. 0.890 kVA B. 0.833 kVA C. 0.909 kVA D. 0.871 kVA Solution:

𝑆𝛼𝑓 𝑆1 𝑆2

𝑓1

=

𝑆2 =

𝑓2 𝑆1 𝑓2 𝑓1

=

(1)(50) 60

𝑺𝟐 = 𝟎. 𝟖𝟑𝟑 𝒌𝑽𝑨 9. A single-phase transformer has a no-load power input of 250 Watts, when supplied at 250 V, 50 Hz has a p.f of 0.25. What is the magnetizing component of the no-load current? A. 4.00 A B. 3.87 A C. 1.00 A D. none of these Solution:

𝑆=

𝑃 𝑝.𝑓

=

250 0.25

𝑆 = 1, 000 𝑉𝐴 𝑝𝑓 = 𝑐𝑜𝑠 −1 0.25 𝑝𝑓 = 75.52° 𝑆 = 𝑉𝐼 𝐼=

1,000∠75.52 250

= 4∠75.52° 𝐴𝑚𝑝𝑒𝑟𝑒𝑠

𝐼 = 1 + 𝑗 3.873 𝐴𝑚𝑝𝑒𝑟𝑒𝑠 𝑰𝒎 = 𝟑. 𝟖𝟕𝟑 𝑨𝒎𝒑𝒆𝒓𝒆𝒔

REE – September 2011 10. A 4, 400 V, 60 Hz transformer has a core loss of 840 Watts, of which one-third is eddy current loss. What is the core loss when the x’former is connected to a 4, 600 V, 50 Hz source? A. 977 Watts B. 907 Watts C. 927 Watts D. 944 Watts Solution: 1 3

𝑃𝑐𝑜𝑟𝑒 = 𝑃𝑒𝑑𝑑𝑦 1

𝑃𝑒1 = (840) 3

𝑃𝑒1 = 280 𝑊𝑎𝑡𝑡𝑠 𝑃𝑐𝑜𝑟𝑒 = 𝑃ℎ + 𝑃𝑒 𝑃ℎ1 = 840 − 280 𝑃ℎ1 = 560 𝑊𝑎𝑡𝑡𝑠 𝑃𝑒2

=

𝑃𝑒1

𝑘2 (𝐸2 )2 𝑘1 (𝐸1 )2 4,6002

𝑃𝑒2 = 280 (

4,4402

)

𝑃𝑒2 = 306.033 𝑊𝑎𝑡𝑡𝑠 1.6

𝑃ℎ2 𝑃ℎ1

=

𝐸 𝑘2 ( 20.6 ) 𝑓 2 𝐸1 1.6 𝑘1 ( 0.6 ) 𝑓1

𝑃ℎ2 = 560 [

4,6001.6 ) 500.6 4,4001.6 ( ) 600.6

(

]

𝑃ℎ2 = 670.788 𝑊𝑎𝑡𝑡𝑠 𝑃𝑐𝑜𝑟𝑒2 = 𝑃ℎ2 + 𝑃𝑒2 = 670.788 + 306.033 𝑷𝒄𝒐𝒓𝒆𝟐 = 𝟗𝟕𝟔. 𝟖𝟐𝟏 𝑾𝒂𝒕𝒕𝒔 REE – September 2004 11. In an ideal transformer, what is the efficiency? A. 100% B. 90% C. 80%

D. 70%

12. A 100 kVA distribution transformer has a full-load copper loss of 1, 180 Watts. For what kilowatt load, at a power factor of 0.71, will the copper losses in the transformer be 1, 500 Watts? A. 90.25 B. 71 C. 112.75 D. 80.05 Solution: 𝑃𝑐𝑢−𝑎𝑛𝑦 2 2

𝑃𝑐𝑢−𝐹𝐿 𝑃

𝑆=

1,500 1,180

(

=(

𝑆𝑎𝑛𝑦 𝑆𝑟𝑎𝑡𝑒𝑑

)

2

𝑝.𝑓

=(

𝐾𝑊𝑙𝑜𝑎𝑑 /0.71 2 100

𝐾𝑊𝑙𝑜𝑎𝑑 2 0.71

)

) = 1002 (

1,500 1,180

)

𝐾𝑊𝑙𝑜𝑎𝑑 = √(0.71)2 (100)2 (

1,500 1,180

)

𝑲𝑾𝒍𝒐𝒂𝒅 = 𝟖𝟎. 𝟎𝟓𝟎 𝒌𝑾 13. Given a 10-kVA transformer with full-load losses amounting to 70 Watts in the iron and 140 Watts in the copper. Calculate the efficiency at half-load unity power factor. A. 98.62% B. 97.97% C. 97.28% D. 97.94% Solution: 𝑃𝐶𝑢 𝐻𝐿 𝑃𝐶𝑢 𝐹𝐿

=(

𝑃𝐶𝑢 𝐻𝐿 =

𝑆𝐻𝐿 2 𝑆𝐹𝐿

)

140(52 ) 102

𝑃𝐶𝑢 𝐻𝐿 = 35 𝑊𝑎𝑡𝑡𝑠 ɳ 𝐻𝐿 =

𝑃𝑜 𝐻𝐿 𝑃𝑜 𝐻𝐿+𝑃𝑐𝑜𝑟𝑒 +𝑃𝐶𝑢 𝐻𝐿

× 100% =

5,000 5,000+35

× 100%

ɳ 𝑯𝑳 = 𝟗𝟕. 𝟗𝟒% REE – April 2004 14. Instrument transformers are used in indicating and metering and with protective devices, they are used for . A. measuring B. detecting C. relaying D. sensing REE – September 2003 15. What type of transformer bank is used to convert 2-phase to 3-phase power? A. open-delta B. scott-T C. wye-delta D. delta-wye

16. A 100-kVA 2, 400/240-Volt 60 cycle transformer has the following constants: 𝑟𝑝 = 0.42 Ω, 𝑋𝑝 = 0.72 Ω; 𝑟𝑠 = 0.0038 Ω, 𝑋𝑠 = 0.0068 Ω. What is the equivalent impedance in primary terms? A. 0.016 Ω B. 1.612 Ω C. 0.161 Ω D. 16.12 Ω Solution:

𝑍𝑒 𝑝 = (𝑟𝑝 + 𝑎2 𝑟𝑠 ) + 𝑗(𝑋𝑝 + 𝑎2 𝑋𝑠 ) 𝑎=

𝑉𝑝 𝑉𝑠

=

2,400 240

𝑎 = 10 𝑍𝑒 𝑝 = (0.42) + (102 )(0.0038) + 𝑗(0.72) + (10)2 (0.0068) 𝑍𝑒 𝑝 = 0.8 + 𝑗1.4 = 1.61∠60.26° |𝒁𝒆 𝒑 | = 𝟏. 𝟔𝟏 𝒐𝒉𝒎𝒔 17. Calculate the all-day efficiency of a 100-kVA transformer operating under the following conditions: 6 hours on a load of 50 kW at 0.73 power factor; 3 hours on a load of 90 kW at 0.82 power factor; 15 hours with no load on secondary. The iron loss is 1, 000 Watts and the full-load copper loss is 1, 060 Watts. A. 96.31% B. 94.87% C. 95.33% D. 95.29% Solution: 𝑃𝑜𝑢𝑡 = (50, 000)(6) + (90, 000)(3) + (0)(15) 𝑃𝑜𝑢𝑡 = 570 𝑘𝑊ℎ𝑟 𝑃𝑐𝑜𝑟𝑒 = (1, 000)(24) 𝑃𝑐𝑜𝑟𝑒 = 24 𝑘𝑊ℎ𝑟 𝑃𝐶𝑢 1 𝑃𝐶𝑢 𝐹𝐿

𝑘𝑉𝐴

2

= (𝑘𝑉𝐴 1 ) 𝐹𝐿

𝑃𝐶𝑢 1 = 1, 060 (

50/0.73 2 100

)

𝑃𝐶𝑢 1 = 497.279 𝑊𝑎𝑡𝑡𝑠 𝑃𝐶𝑢 2 = 1, 060 (

90/0.82 2 100

)

𝑃𝐶𝑢 2 = 1, 276.919 𝑊𝑎𝑡𝑡𝑠 𝑃𝐶𝑢 3 = 0 𝑊𝑎𝑡𝑡𝑠 𝑃𝐶𝑢 𝑡𝑜𝑡𝑎𝑙 = 𝑃𝐶𝑢 1 (6) + 𝑃𝐶𝑢 2 (3) + 𝑃𝐶𝑢 3 (15) 𝑃𝐶𝑢 𝑡𝑜𝑡𝑎𝑙 = (487.279)(6) + (1, 276.919)(3) + (0)(15) 𝑃𝐶𝑢 𝑡𝑜𝑡𝑎𝑙 = 6.814 𝑘𝑊ℎ𝑟 ɳ𝑎𝑙𝑙 𝑑𝑎𝑦 =

𝑃𝑜𝑢𝑡 𝑃𝑜𝑢𝑡 +𝑃𝑐𝑜𝑟𝑒+𝑃𝐶𝑢

× 100% =

570 570+24+6.814

× 100%

ɳ𝒂𝒍𝒍 𝒅𝒂𝒚 = 𝟗𝟒. 𝟖𝟕% REE – September 2005 18. A 50-kVA, single-phase transformer has 96% efficiency when it operates at full-load unity power factor for 8 hours per day. What is the all-day efficiency of the transformer if the copper loss is 60% of full-load losses? A. 92% B. 90% C. 89.5% D. 93% Solution:

0.96 =

50

50+𝑃𝑙𝑜𝑠𝑠𝑒𝑠 𝐹𝐿 50 𝑃𝑙𝑜𝑠𝑠𝑒𝑠 𝐹𝐿 = − 50 0.96

𝑃𝑙𝑜𝑠𝑠𝑒𝑠 𝐹𝐿 = 2.083 𝑘𝑊 𝑃𝐶𝑢 𝐹𝐿 = (0.6)(2, 083) 𝑃𝐶𝑢 𝐹𝐿 = 1, 249.8 𝑊 𝑃𝑐𝑜𝑟𝑒 = 𝑃𝑙𝑜𝑠𝑠 + 𝑃𝐶𝑢 𝐹𝐿 = 2, 083 − 1, 249.8 𝑃𝑐𝑜𝑟𝑒 = 833.2 𝑊𝑎𝑡𝑡𝑠 𝑃𝑜𝑢𝑡 𝑡𝑜𝑡𝑎𝑙 = (50 𝑘𝑉𝐴)(1.0)(8) 𝑃𝑜𝑢𝑡 𝑡𝑜𝑡𝑎𝑙 = 400𝑘𝑊ℎ𝑟 𝑃𝑐𝑜𝑟𝑒 𝑡𝑜𝑡𝑎𝑙 = (833.2)(24) 𝑃𝑜𝑢𝑡 𝑡𝑜𝑡𝑎𝑙 = 19.997 𝑘𝑊ℎ𝑟 𝑃𝐶𝑢 𝑡𝑜𝑡𝑎𝑙 = (1, 249.8)(8) 𝑃𝐶𝑢 𝑡𝑜𝑡𝑎𝑙 = 9.998 𝑘𝑊ℎ𝑟 ɳ𝑎𝑙𝑙 𝑑𝑎𝑦 =

400 400+19.997+9.998

× 100%

ɳ𝒂𝒍𝒍 𝒅𝒂𝒚 = 𝟗𝟑. 𝟎𝟐% Asst. EE – October 1991 19. A 10 kVA, 2, 400/240 V, single-phase transformer has the following resistances and leakage reactances; 𝑟𝑝 = 3 Ω 𝑟𝑠 = 0.03 Ω 𝑋𝑝 = 15 Ω 𝑋𝑠 = 0.15 Ω Find the primary voltage required to produce 240 V at the secondary terminals at full-load, when the load power factor is 0.8 lagging. A. 2, 400 V B. 2, 496.5 V C. 2, 348 V D. 2, 445.5 V Solution:

𝑎=

2,400 240

𝑎 = 10

𝑎𝑉𝑠 = 10(240) 𝑎𝑉𝑠 = 2, 400 𝐼𝑡 =

10,000 ∠𝑐𝑜𝑠 −1 (0.8) 2,400

𝐼𝑡 = 4.17 ∠ − 36.87° 𝐴𝑚𝑝𝑒𝑟𝑒𝑠 𝑍𝑒 𝑝 = [3 + (10)2 (0.03)] + 𝑗[15 + (10)2 (0.15)] 𝑍𝑒 𝑝 = 6 + 𝑗30 𝑜ℎ𝑚𝑠 𝑉𝑝 = 𝐼𝑡 𝑍𝑒 𝑝 + 𝑎𝑉𝑠 𝑉𝑝 = (4.17 ∠ − 36.87)(6 + 𝑗30 ) + 2, 400 𝑉𝑝 = 2, 496.526 ∠1.95° 𝑉𝑜𝑙𝑡𝑠 |𝑽𝒑 | = 𝟐, 𝟒𝟗𝟔. 𝟓𝟐𝟔 𝑽𝒐𝒍𝒕𝒔 20. A 500 kVA, single-phase, 13, 200/2, 400 Volts transformer has 4% reactance and 1% resistance. The leakage reactance and resistance of the high voltage (primary) winding are 6.34 Ω and 1.83 Ω, respectively. The core loss under rated condition is 1, 800 Watts. Calculate the leakage reactance and resistance of the low voltage (secondary) winding. A. 7.56 Ω and 1.66 Ω B. 13.69 Ω and 3.42 Ω C. 0.25 Ω and 0.055 Ω D. 13.9 Ω and 3.48 Ω Solution:

𝑍𝑏 =

(𝑉𝑏 )2 𝑆𝑏

=

(13,200)2 500,000

𝑍𝑏 = 348.48 𝑜ℎ𝑚𝑠 𝑅𝑒 𝑝 = 𝑅𝑝𝑢 𝑍𝑏𝑎𝑠𝑒 = (0.01)(348.48) 𝑅𝑒 𝑝 = 3.4848 𝑜ℎ𝑚𝑠 𝑋𝑒 𝑝 = 𝑋𝑝𝑢 𝑍𝑏𝑎𝑠𝑒 = (0.04)(348.48) 𝑋𝑒 𝑝 = 13.9392 𝑜ℎ𝑚𝑠 𝑟𝑒𝑓𝑒𝑟𝑟𝑒𝑑 𝑡𝑜 𝑝𝑟𝑖𝑚𝑎𝑟𝑦 𝑅𝑒 𝑝 = 𝑟𝑝 + 𝑎2 𝑟𝑠 𝑎=

13,200 2,400

𝑎 = 5.5 𝑟𝑠 =

𝑅𝑒 𝑝 −𝑟𝑝 𝑎2

=

3.4848−1.83 5.52

𝒓𝒔 = 𝟎. 𝟎𝟓𝟓 𝑋𝑒 𝑝 = 𝑋𝑝 + 𝑎2 𝑋𝑠

𝑋𝑠 =

𝑋𝑒 𝑝−𝑋𝑝 𝑎2

=

13.9392−6.34 5.52

𝑿𝒔 = 𝟎. 𝟐𝟓𝟏 𝒐𝒉𝒎𝒔 21. In Problem No.20, calculate the %V.R and efficiency of the transformer at full-load, 0.85 p.f. lagging and 2, 400 Volts. A. 4% and 97.8% B. 6% and 95.4% C. 5% and 96.8% D. 3% and 98.4% Solution: 𝑟𝑝 𝑎2

1.83

=

5.52 𝑟𝑝

𝑋𝑝 𝑎2

=

𝑎2 6.34 5.52 𝑋𝑝

= 0.0605 𝑜ℎ𝑚𝑠

= 0.21 𝑜ℎ𝑚𝑠

𝑎2 −1

𝜃 = 𝑐𝑜𝑠 (0.85) 𝜃 = 31.79° 𝐼𝑇 =

500,000 ∠−31.79° 2,400

𝐼𝑇 = 208.33 ∠ − 31.79° 𝑍𝑒 𝑠 = (

𝑟𝑝 𝑎2

+ 𝑟𝑠 ) + 𝑗 (

𝑋𝑝 𝑎2

+ 𝑋𝑠 ) = (0.0605 + 0.055) + 𝑗(0.21 + 0.25)

𝑍𝑒 𝑠 = 0.1155 + 𝑗0.46 Ω 𝑉𝑝 = 𝐼𝑇 𝑍𝑒 𝑠 + 𝑉𝑠 𝑉𝑝 = (208.33 ∠ − 31.79°)(0.1155 + 𝑗0.46) + 2, 400 𝑉𝑝 = 2, 471.89 ∠1.6° 𝑉𝑜𝑙𝑡𝑠 %𝑉𝑅 =

2,471.89−2,400 2,400

× 100

%𝑽𝑹 = 𝟑% 𝑃𝑜𝑢𝑡 = (500 𝑘𝑉𝐴)(0.85) 𝑃𝑜𝑢𝑡 = 425 𝑘𝑊 𝑃𝑐𝑜𝑟𝑒 = 1, 800 𝑊 𝑃𝑐𝑢 = 𝐼𝑇 2 𝑅𝑇 = (208.33)2 (0.1155) 𝑃𝑐𝑢 = 5, 012.86 𝑊 ɳ=

425,000 425,000+1,800+5,012.86

ɳ = 𝟗𝟖 %

× 100

22. An 11, 000/230 V, 150 kVA, single-phase, 50 Hz transformer has a core loss of 1.4 kW and a full-load copper loss of 1.6 kW. What is the value of maximum efficiency at unity p.f? A. 98.17% B. 98.04% C. 97.22% D. 97.64% Solution:

𝑃𝑐𝑢 𝑚𝑎𝑥 = 𝑃𝑐𝑜𝑟𝑒 𝑃𝑐𝑢 𝑚𝑎𝑥 = 1, 400 𝑊 𝑃𝑐𝑢 𝑚𝑎𝑥 𝑃𝑐𝑢 𝐹𝐿

=(

𝑘𝑉𝐴𝑚𝑎𝑥 2 𝑘𝑉𝐴𝐹𝐿

)

𝑘𝑉𝐴𝑚𝑎𝑥 = √(1502 ) (

1,400 1,600

)

𝑘𝑉𝐴𝑚𝑎𝑥 = 140.31 𝑘𝑉𝐴 𝑘𝑊𝑚𝑎𝑥 = 𝑘𝑉𝐴𝑚𝑎𝑥 (𝑝. 𝑓) = (140.31)(1.0) 𝑘𝑊𝑚𝑎𝑥 = 140.13 𝑘𝑊 ɳ=

140.13 140.13+2(1.4)

× 100

ɳ = 𝟗𝟖. 𝟎𝟒 % 23. A 300-kVA, single-phase transformer is designed to have a resistance of 1.5% and maximum efficiency occurs at a load of 173.2 kVA. Find its efficiency when supplying full-load at 0.8 p.f. lagging at normal voltage and frequency. A. 97.56% B. 96.38% C. 98.76% D. 95.89% Solution:

𝑟𝑝.𝑢 =

𝑃𝑐𝑢 𝐹𝐿 𝑆𝐹𝐿

𝑃𝑐𝑢 𝐹𝐿 = 𝑟𝑝.𝑢 𝑆𝐹𝐿 = (0.015)(300, 000) 𝑃𝑐𝑢 𝐹𝐿 = 4, 500 𝑊𝑎𝑡𝑡𝑠 𝑃𝑐𝑢 𝑚𝑎𝑥 4,500

=(

173.2 2 300

)

𝑃𝑐𝑢 𝑚𝑎𝑥 = 1, 499.91 𝑊𝑎𝑡𝑡𝑠 𝑃𝑐𝑢 𝑚𝑎𝑥 = 𝑃𝑐𝑜𝑟𝑒 ɳ𝐹𝐿 =

𝑃𝑜𝑢𝑡 𝑃𝑜𝑢𝑡 +𝑃𝑐𝑜𝑟𝑒 +𝑃𝑐𝑢 𝐹𝐿

300(0.8)

× 100 = (300)(0.8)+1.5+4.5 × 100

ɳ𝑭𝑳 = 𝟗𝟕. 𝟓𝟔 %

REE – September 2002 24. A 20 kV/7.87 kV autotransformer has 200 A current in the common winding. What is the secondary line current? A. 143.52 B. 200 C. 56.48 D. 329 Solution: 𝐼𝑐 𝐼𝑝

=𝑎−1

𝑎=

20,000 7,870

𝑎 = 2.54 200 𝐼𝑝

= 2.54 − 1

𝐼𝑝 = 129.76 𝐴𝑚𝑝𝑒𝑟𝑒𝑠 𝐼𝑠 = 𝐼𝑝 + 𝐼𝑐 = 129.76 + 200 𝑰𝒔 = 𝟑𝟐𝟗. 𝟕𝟔 𝑨𝒎𝒑𝒆𝒓𝒆𝒔 25. An autotransformer is adjusted for an output voltage of 85.3 Volts when operated from a 117 Volts line. The variable power load draws 3.63 kW at unity power factor at this setting. Determine the transformed power and the connected power from the source to the load. A. 980 Watts and 2, 650 Watts B. 1, 343 Watts and 2, 287 Watts C. 1, 815 Watts and 1, 815 Watts D. 1, 210 Watts and 2, 420 Watts Solution:

𝑎=

117 85.3

𝑎 = 1.37 1

1

𝑎

1.37

𝑃𝑡𝑟𝑎𝑛𝑠 = 𝑃𝑖𝑛 (1 − ) = (3, 630) (1 −

)

𝑷𝒕𝒓𝒂𝒏𝒔 = 𝟗𝟖𝟎. 𝟑𝟔 𝑾𝒂𝒕𝒕𝒔 𝑃𝑐𝑜𝑛 =

𝑃𝑖𝑛 𝑎

=

3,630 1.37

𝑷𝒄𝒐𝒏 = 𝟐, 𝟔𝟒𝟗. 𝟔𝟒 𝑾𝒂𝒕𝒕𝒔 REE – April 2006 26. What would happen if you connect a transformer to a dc circuit with a voltage of 20% of nameplate ratings after a steady state condition is reached? A. No voltage is registered at the secondary B. Rated no-load current flows to the secondary C. primary current is equal to voltage over equivalent primary impedance D. Voltage is established at secondary

27. A short-circuit test was performed upon a 10 kVA, 2, 300/230-Volt transformer with the following results: 𝐸𝑠𝑐 = 137 𝑉𝑜𝑙𝑡𝑠; 𝑃𝑠𝑐 = 192 𝑊𝑎𝑡𝑡𝑠; 𝐼𝑠𝑐 = 4.34 𝐴𝑚𝑝𝑒𝑟𝑒𝑠. Calculate in secondary terms the transformer equivalent. A. 29.88 Ω B. 2.988 Ω C. 0.2988 Ω D. 298.8 Ω Solution:

𝑅=

𝑃𝑠𝑐

𝑍=

=

𝐼𝑠𝑐 2

192 4.34 2

𝑅 = 10.19 Ω 𝐸𝑠𝑐 𝐼𝑠𝑐

=

137 4.34

𝑍 = 31.57 Ω

𝑋 = √𝑍 2 − 𝑅 2 = √31.572 − 10.192

𝑿 = 𝟐𝟗. 𝟖𝟖 Ω REE – April 2007 28. A transformer is rated 500 kVA, 4, 800/480 V, 60 Hz when it is operated as a conventional two winding transformer. This transformer is to be used as a 5280/4800 V stepdown autotransformer in a power distribution system. In the autotransformer, what is the transformer rating when used in this manner? A. 5 MVA B. 6 MVA C. 5.5 MVA D. 6.5 MVA Solution: 𝑆

500,000

𝐼𝑝 = 𝑉 =

4,800

𝑝

𝐼𝑝 = 104.17 𝐴𝑚𝑝𝑒𝑟𝑒𝑠 𝑆

500,000

𝑠

4,80

𝐼𝑠 = 𝑉 =

𝐼𝑠 = 1, 041.7 𝐴𝑚𝑝𝑒𝑟𝑒𝑠 𝐼𝑝 ′ = 𝐼𝑠 𝐼𝑝 ′ = 1, 041.7 𝐴𝑚𝑝𝑒𝑟𝑒𝑠 𝑆 = 𝑉𝑝 ′𝐼𝑝 ′ = (5, 280)(1, 041.7)

𝑺 = 𝟓. 𝟓 𝑴𝑽𝑨 REE – September 2008 29. Two identical transformers bank on open delta serve a balanced three-phase load of 26 kVA at 240 V, 60 Hz. What is the minimum size of each in kVA needed to serve this load? A. 25 B. 10 C. 30 D. 15 Solution: 𝑆∅ 𝑟𝑎𝑡𝑒𝑑 =

𝑆∅ 𝑙𝑜𝑎𝑑 (√3)

=

26,000 √3

𝑺∅ 𝒓𝒂𝒕𝒆𝒅 = 𝟏𝟓 𝒌𝑽𝑨

30. Two single-phase, 100 kVA transformers are connected in V (open delta) bank supplying a balanced three-phase load. If the balanced three-phase load is 135 kW at 0.82 p.f lagging and 0.823 efficiency, determine the overload kVA on each transformer. A. 10.5 B. 5.5 C. 15.5 D. 20.5 Solution:

𝑆𝐿 =

𝑃 𝑝.𝑓

=

135,000 0.82

𝑆𝐿 = 164.634 𝑘𝑉𝐴 ɳ=

𝑃𝑜𝑢𝑡 𝑃𝑖𝑛 135

𝑃𝑖𝑛 =

0.823

𝑃𝑖𝑛 = 164.034 𝑘𝑊 𝑆𝑖𝑛 =

𝑃𝑖𝑛 𝑝.𝑓

=

164.034 𝑘𝑊 0.82

𝑆𝑖𝑛 = 200.041 𝑘𝑉𝐴 𝑆∅ =

𝑆𝑖𝑛 √3

=

200.041 √3

𝑆∅ = 115.494 𝑘𝑉𝐴 𝑆∅ 𝑟𝑎𝑡𝑒𝑑 = 100 𝑘𝑉𝐴 𝑆𝑂.𝐿 = 𝑆𝐿∅ − 𝑆∅𝑟𝑎𝑡𝑒𝑑 = 115.494 − 100 𝑺𝑶.𝑳 = 𝟏𝟓. 𝟒𝟗𝟒 𝒌𝑽𝑨 31. In problem No. 30, determine the p.f of each transformer secondary. A. 0.820 lagging and 0.820 lagging B. 0.996 lagging and 0.424 leading C. 0.996 lagging and 0.424 lagging D. 0.410 lagging and 0.410 lagging Solution:

𝜃 = 𝑐𝑜𝑠 −1 (0.82) 𝜃 = 34.92° 𝑝. 𝑓1 = 𝑐𝑜 𝑠(30 + 𝜃) = 𝑐𝑜 𝑠(30 + 34.92) 𝒑. 𝒇𝟏 = 𝟎. 𝟒𝟐𝟒 𝒍𝒂𝒈𝒈𝒊𝒏𝒈 𝑝. 𝑓2 = 𝑐𝑜 𝑠(30 − 𝜃) = 𝑐𝑜 𝑠(30 − 34.92) 𝒑. 𝒇𝟐 = 𝟎. 𝟗𝟗𝟔 𝒍𝒂𝒈𝒈𝒊𝒏𝒈 REE – April 2005 32. What is the normal secondary circuit current of a current transformer? A. 15 A B. 20 A C. 5 A D. 10 A

33. In Problem No.30, what is the minimum size in kVAR of a capacitor bank to be connected across the load so that each transformer is loaded 96% of its rated capacity? A. 87 kVAR B. 114 kVAR C. 27 kVAR D. 66 kVAR Solution:

𝑆∅ = (0.96)(𝑆∅𝑟𝑎𝑡𝑒𝑑 ) = (0.96)(100) 𝑆∅ = 96 𝑘𝑉𝐴 𝑃𝑇 = 164.034 𝑘𝑊 𝑆 = 200.041 𝑘𝑉𝐴 𝑄 = √𝑆 2 − 𝑃𝑇 2 = √(200.041)2 − (164.034)2 𝑄 = 114.5 𝑘𝑉𝐴𝑅 𝑆𝑛𝑒𝑤 = (96 𝑘𝑉𝐴)(√3) 𝑆𝑛𝑒𝑤 = 166.277 𝑘𝑉𝐴 𝑆 = 𝑃2 + 𝑄2 2

𝑄 = √𝑆𝑛𝑒𝑤 2 − 𝑃𝑇 2 = √(166.277)2 − (164.034)2 𝑸 = 𝟐𝟕. 𝟐𝟐 𝒌𝑽𝑨𝑹 34. A polarity test is performed upon a 1, 150/115 V transformer. If the input voltage is 116, calculate the voltmeter reading if the polarity is subtractive. A. 127.6 V B. 106 V C. 126 V D. 104.4 V Solution: 𝑠𝑢𝑏𝑡𝑟𝑎𝑐𝑡𝑖𝑣𝑒 𝑎=

1,150

115

𝑎 = 10

𝑉𝑟𝑒𝑎𝑑𝑖𝑛𝑔 = 116 −

116 10

𝑽𝒓𝒆𝒂𝒅𝒊𝒏𝒈 = 𝟏𝟎𝟒. 𝟒 𝑽𝒐𝒍𝒕𝒔

35. A 20:1 potential transformer is used with a 150-V voltmeter. If the instrument deflection is 118 Volts, calculate the line voltage. A. 3, 000 V B. 2, 850 V C. 2, 360 V D. 2, 242 V Solution:

𝑎= 20 1

=

𝑉𝐿 𝑉𝑟𝑒𝑎𝑑𝑖𝑛𝑔 𝑉𝐿 118

𝑽𝑳 = 𝟐, 𝟑𝟔𝟎 𝑽𝒐𝒍𝒕𝒔

REE –September 2010 36. A three-phase wye-delta connected, 50 MVA, 345/34.5 kV transformer is protected by differential protection. The current transformer on the high side for differential protection is 150:5. What is the current on the secondary side of CT’s? A. 3.83 A B. 2.53 A C. 4.50 A D. 4.83 A Solution: 𝑆

𝐼∅ 𝑝 =

=

3𝑉∅𝑝

50,000,000 3(

345,000 ) √3

𝐼∅ 𝑝 = 83.67 𝐴𝑚𝑝𝑒𝑟𝑒𝑠 𝑠𝑖𝑛𝑐𝑒 𝑤𝑦𝑒 − 𝑐𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑜𝑛 𝐼∅ 𝑝 = 𝐼𝐿 𝑝 𝐼𝐶𝑇 𝑝 = 𝐼𝐶𝑇 𝑝 = 83.67 𝐴𝑚𝑝𝑒𝑟𝑒𝑠 𝐿

𝑎= 150 5

𝐼𝑝 𝐼𝑠

=

=

∅

𝑁𝑝

𝑁𝑠 83.67 𝐼𝐶𝑇 𝑠

∅

𝐼𝐶𝑇 𝑠 =

(83.67)(5) 150

∅

𝐼𝐶𝑇 𝑠 = 2.789 𝐴𝑚𝑝𝑒𝑟𝑒𝑠 ∅

𝐼𝐶𝑇 𝑠𝑒𝑐 = √3 (2.789) 𝑰𝑪𝑻 𝒔𝒆𝒄 = 𝟒. 𝟖𝟑𝟏 𝑨𝒎𝒑𝒆𝒓𝒆𝒔 REE – October 2000 37. The CT ratio and PT ratio used to protect a line are 240 and 2,000, respectively. If the impedance of each line is 10 Ω, what is the relay impedance to protect the line from fault? A. 83.33 ohms B. 1.2 ohms C. 48, 000 ohms D. 12 ohms Solution:

𝑍𝑟 =

𝑃𝑇𝑟 𝐶𝑇𝑟

=

2,000 240

𝑍𝑟 = 8.33 𝑜ℎ𝑚𝑠 𝑍𝑟 =

𝑍𝑙𝑖𝑛𝑒

𝑍𝑟𝑒𝑙𝑎𝑦 10 𝑍𝑟𝑒𝑙𝑎𝑦 = 8.33

𝒁𝒓𝒆𝒍𝒂𝒚 = 𝟏. 𝟐 𝒐𝒉𝒎𝒔

38. Two transformers 1 and 2 are connected in parallel supplying a common load of 120 kVA. Transformer 1 is rated 50 kVA, 7, 620/240-V single-phase and has an equivalent impedance of 8.5 Ω while transformer 2 is rated 75 kVA, 7, 620/240-V single-phase and has an equivalent impedance of 5.1 Ω. The two transformers operate with the same power factors. What is the kVA load of each transformer? A. 48 & 72 B. 45 & 75 C. 42 & 78 D. 40 & 80 Solution: 𝑍𝑒 1 𝑍𝑒 2 8.5 5.1

= =

𝑆2 𝑆1 𝑆2 𝑆1

𝑆2 = 1.67𝑆1 → 𝑒𝑞′𝑛 1 𝑆𝐿 = 𝑆1 + 𝑆2 → 𝑒𝑞′ 𝑛 2 𝑆𝑢𝑏𝑠𝑡𝑖𝑡𝑢𝑡𝑒 𝑒𝑞’𝑛 1 𝑖𝑛 2 120 = 𝑆1 + 1.67𝑆1 𝑺𝟏 = 𝟒𝟒. 𝟗𝟒 𝒌𝑽𝑨 𝑆2 = 1.67(44.94) 𝑺𝟐 = 𝟕𝟓. 𝟎𝟓 𝒌𝑽𝑨 39. Two single-phase transformers are connected in parallel at no-load. One has a turns ratio of 5, 000/440 and rating of 200 kVA, the other has a ratio of 5, 000/480 and rating of 350 kVA the leakage reactance of each is 3.5%. The no-load circulating current is . A. 207 A B. 702 A C. 720 A D. 270 A Solution:

𝑋𝑏𝑎𝑠𝑒1 =

(𝑉𝑏𝑎𝑠𝑒1 )

2

=

𝑆𝑏𝑎𝑠𝑒1

(440)2 200,000

𝑋𝑏𝑎𝑠𝑒1 = 0.968 Ω 𝑋𝑏𝑎𝑠𝑒2 =

(𝑉𝑏𝑎𝑠𝑒2 ) 𝑆𝑏𝑎𝑠𝑒2

2

=

(480)2 350,000

𝑋𝑏𝑎𝑠𝑒2 = 0.658 Ω 𝑋𝑒−𝑠1 = 𝑋𝑒 𝑋𝑏𝑎𝑠𝑒1 = (0.035)(0.968) 𝑿𝒆−𝒔𝟏 = 𝟎. 𝟎𝟑𝟑𝟗 Ω 𝑋𝑒−𝑠2 = 𝑋𝑒 𝑋𝑏𝑎𝑠𝑒2 = (0.035)(0.658) 𝑿𝒆−𝒔𝟐 = 𝟎. 𝟎𝟐𝟑 Ω

REE – October 1997 40. A power transformer rated 50, 000 kVA, 34.5/13.8 kV is connected Y-grounded primary and delta on the secondary. Determine the full load phase current at the secondary side. A. 2, 092 A B. 1, 725 A C. 1, 449 A D. 1, 208 A Solution: 𝑆 = √3 𝑉𝐿 𝐼𝐿 50,000,000

𝐼𝐿 𝑝 =

(34,500)(√3)

𝐼𝐿 𝑝 = 836.74 𝐴𝑚𝑝𝑒𝑟𝑒𝑠 𝐼𝐿 𝑝 = 𝐼∅ 𝑝 𝑉𝑝 𝑉𝑠

𝐼

= 𝐼𝑠

𝑝

𝐼∅ 𝑠 =

(34,500)(836.74) 13,800 (√3)

𝑰𝒔 ∅ = 𝟏, 𝟐𝟎𝟕. 𝟕𝟑 𝑨𝒎𝒑𝒆𝒓𝒆𝒔

𝑎𝑛𝑜𝑡ℎ𝑒𝑟 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛: 𝐼𝐿 𝑠 = ( 𝐼∅ 𝑠 =

50,000,000 √3) (13,800)

𝐼𝐿 𝑠 = 2, 091.849 𝐼𝐿 𝑠 2,091.849 √3

=

𝐴𝑚𝑝𝑒𝑟𝑒𝑠

√3

𝑰∅ 𝒔 = 𝟏, 𝟐𝟎𝟕. 𝟕𝟑 𝑨𝒎𝒑𝒆𝒓𝒆𝒔

REE – April 2006 41. A 2, 000 kW, 2, 400-V, 75% p.f load is to be supplied from a 34, 5000-V, 3-phase line through a single bank of transformers. Give the primary and secondary line currents in amperes for the wye-wye connections. A. 50/700 B. 48/650 C. 60/800 D. 45/642 Solution: 𝑌 − ∆ 𝐶𝑜𝑛𝑛𝑒𝑐𝑡𝑖𝑜𝑛

𝐼𝐿 𝑝 = [

∗ 2 𝑀𝑊 ∠𝑐𝑜𝑠 −1 (0.75) 0.75 √3 (34,500)

]

𝐼𝐿 𝑝 = 44.626∠44.41°

𝑰𝑳 𝒑 = 𝟒𝟒. 𝟔𝟐𝟔 𝑨𝒎𝒑𝒆𝒓𝒆𝒔 ∗ 2 𝑀𝑊 ∠𝑐𝑜𝑠 −1 (0.75) 0.75

𝐼𝐿 𝑠 = [

√3 (2,400)

]

𝐼𝐿 𝑠 = 641.5∠44.41°

𝑰𝑳 𝒔 = 𝟔𝟒𝟏. 𝟓 𝑨𝒎𝒑𝒆𝒓𝒆𝒔

REE – April 2005 42. A 3, 000 kVA, 2, 400 V, 75% power factor load is to be supplied from a 34, 500-V, three-phase line through a single bank of transformers. What is the voltage rating of each transformer if the connection is wye-wye? A. 20, 000/1, 380 B. 18, 500/1, 350 C. 18, 000/1, 850 D. 19, 000/1, 350 Solution:

𝑉∅ 𝑝 =

34,500 √3

𝑽∅ 𝒑 = 𝟏𝟗, 𝟗𝟏𝟖. 𝟓𝟖𝟒 𝑽𝒐𝒍𝒕𝒔

𝑉∅ 𝑠 =

2,400 √3

𝑽∅ 𝒔 = 𝟏, 𝟑𝟖𝟓. 𝟔𝟒𝟏 𝑽𝒐𝒍𝒕𝒔

REE – March 1998 43. A 13.8 kV/480 V, 10 MVA three-phase transformer has 5% impedance. What is the impedance in ohms referred to the primary? A. 0.952 ohm B. 0.03 ohm C. 5.125 ohm D. 9.01 ohm Solution:

𝑍𝑝𝑢 = 𝑍𝑏 =

𝑍𝑎𝑐𝑡𝑢𝑎𝑙

𝑍𝑏 (𝑉𝑏 )2 𝑆𝑏

=

(13,800)2 10,000,000

𝑍𝑒 𝑝

𝑍𝑏 = 19.044 𝑜ℎ𝑚𝑠 = 𝑍𝑝𝑢 𝑍𝑏 = (0.05)(19.044)

𝒁𝒆 𝒑 = 𝟎. 𝟗𝟓𝟐 𝒐𝒉𝒎𝒔

REE – May 2009 44. A three-phase transformer is rated 15 MVA, 69/13.2 kV has a series impedance of 5%. What is the new per unit impedance if the system study requires a 100 MVA base and 67 kV base? A. 0.354 B. 0.347 C. 0.372 D. 0.333 Solution 𝑆

𝑉

𝑆𝑜

𝑉𝑁

𝑍𝑝𝑢 𝑛𝑒𝑤 = 𝑍𝑝𝑢 𝑜𝑙𝑑 ( 𝑛 ) ( 𝑜 ) 𝑍𝑝𝑢 𝑛𝑒𝑤 = (0.05) ( 𝒁𝒑𝒖

𝒏𝒆𝒘

100 15

2

69 2

)( )

= 𝟎. 𝟑𝟓𝟒

67

REE – April 2004 45. A transformer rated 2, 000 kVA, 34, 500/240 volts has 5.75% impedance. What is the per unit impedance? A. 0.0635 B. 0.0656 C. 0.0575 D. 34.2 Solution:

𝑍𝑝𝑢 = 57.5% ∴ 𝑍𝑝𝑢 = 0.0575 46. Three 5:1 transformers are connected in delta-wye to step up the voltage at the beginning of a 13, 200-Volt three-phase transmission line. Calculate the line voltage on the high side of the transformers. A. 114, 300 V B. 66, 000 V C. 132, 000 V D. 198, 000 V Solution: 5 1

=

𝑉∅ 2 𝑉∅ 1

=

𝑉∅ 2 13,200 𝑉

𝑉∅ 2 = 𝑎𝑉𝑝 𝑉∅ 2 = (5)(13, 200) 𝑉∅ 2 = 66, 000 𝑉𝑜𝑙𝑡𝑠 𝑉𝐿 = √3 𝑉∅ 2 = √3 (66, 000) 𝑽𝑳 = 𝟏𝟏𝟒, 𝟑𝟏𝟓. 𝟑𝟓𝟑 𝑽𝒐𝒍𝒕𝒔 47. A 150 kVA, 2, 400/480-V, three-phase transformer with an equivalent impedance of 4%is connected to an infinite bus and without load. If a three-phase fault occurs at the secondary terminals, the fault current in amperes is . A. 4, 512 A B. 3, 908 A C. 7, 815 A D. 1, 504 A Solution: 3 ∅ 𝑓𝑎𝑢𝑙𝑡 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑒𝑐𝑜𝑛𝑑𝑎𝑟𝑦 𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑙𝑠

𝐼𝐹 3∅ = 𝐼𝐹 3∅ =

𝑆𝐵 √3 𝑉𝐵 𝑋𝑝𝑢 150,000 √3 (480)(0.04)

𝑰𝑭 𝟑∅ = 𝟒, 𝟓𝟏𝟎. 𝟓𝟒𝟗 𝑨𝒎𝒑𝒆𝒓𝒆𝒔

48. Transformer 1 is in parallel with Transformer 2 Transformer 1 Transformer 2 150 kVA, single-phase 300 kVA, single-phase 6, 600/240 V 6, 600/240 V 𝑍𝑒−𝑠1 = 0.02425∠62.9° Ω 𝑍𝑒−𝑠2 = 0.01067∠62.9° Ω Determine the maximum kVA load the bank can carry without overloading any of the two transformers, assuming that the two transformers operate at the same power factors. A. 450 kVA B. 432 kVA C. 420 kVA D. 412 kVA Solution:

𝑆1 + 𝑆2 = 𝑆𝑙𝑜𝑎𝑑 𝐶𝑎𝑠𝑒 𝐼: 𝑇1 @ 150 𝑘𝑉𝐴 𝑍𝑒−𝑠1 𝑍𝑒−𝑠2 𝑍𝑒−𝑠1 𝑍𝑒−𝑠2

= =

𝑆2 𝑆1 0.02425∠62.9°

0.01067∠62.9° 𝑍𝑒−𝑠1 𝑍𝑒−𝑠2

2.273 =

= 2.273

𝑆2 𝑆1

𝑆2 = 2.273 (150, 000) 𝑆2 = 340.95 𝑘𝑉𝐴 ∴ 𝑜𝑣𝑒𝑟𝑙𝑜𝑎𝑑 𝑎𝑡 𝑇2 𝐶𝑎𝑠𝑒 𝐼𝐼: 𝑇2 @ 300 𝑘𝑉𝐴

𝑆1 =

300,000 2.273

𝑆1 = 131.984 𝑘𝑉𝐴 𝑆𝑙𝑜𝑎𝑑 = 𝑆1 + 𝑆2 = (131.984) + (300) 𝑺𝒍𝒐𝒂𝒅 = 𝟒𝟑𝟏. 𝟗𝟖𝟏 𝒌𝑽𝑨

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