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A 100 KVA, 6600/400 V , 50 Hz single phase transformer has 80 turns on low voltage side. At 25 Hz its flux increases by 10% . The high voltage and KVA rating at 25 Hz will be respectively (a) 3630 V, 55 KVA (b) 6600 V,100 KVA (c) 7260 V , 110 KVA (d) 3300 V , 50 KVA Correct answer :- (a) Can u explain this?
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A 100 KVA, 6600/400 V , 50 Hz single phase transformer has 80 turns on...
Solution:

Given data:

KVA rating, K = 100

Primary voltage, V1 = 6600 V

Secondary voltage, V2 = 400 V

Frequency, f1 = 50 Hz

Number of turns on secondary side, N2 = 80

Change in frequency, df = 25 – 50 = -25 Hz

Change in flux, dΦ/Φ = 10/100 = 0.1

Let's calculate the number of turns on the primary side as:

V1/V2 = N1/N2

N1 = (V1/V2) x N2 = (6600/400) x 80 = 1320 turns

1. Calculation of the induced emf at 50 Hz:

The induced emf on the secondary side of the transformer is given by:

E2 = 4.44fN2Φm

where Φm is the maximum value of flux linkage.

At 50 Hz, we have:

E2 = 4.44 x 50 x 80 x Φm

E2 = 17760 Φm volts

The induced emf on the primary side of the transformer is given by:

E1 = E2 x (N1/N2)

E1 = 17760 Φm x (1320/80)

E1 = 292920 Φm volts

2. Calculation of the induced emf at 25 Hz:

We know that the flux linkage is directly proportional to the frequency of the applied voltage. Therefore, at 25 Hz, the flux linkage will be:

Φ2’ = Φ2 x (25/50) = Φ2/2

The induced emf on the secondary side of the transformer is given by:

E2’ = 4.44fN2Φ2’

E2’ = 4.44 x 25 x 80 x (Φ2/2)

E2’ = 22200 Φ2 volts

The induced emf on the primary side of the transformer is given by:

E1’ = E2’ x (N1/N2)

E1’ = 22200 Φ2 x (1320/80)

E1’ = 365400 Φ2 volts

3. Calculation of high voltage and KVA rating at 25 Hz:

From the transformer equation, we have:

E1’/E1 = V1’/V1

where V1’ is the high voltage at 25 Hz.

Therefore, we can write:

V1’ = (E1’/E1) x V1

V1’ = (365400 Φ2/292920 Φm) x 6600

V1’ = 8250 Φ2 volts

The KVA rating of the transformer is given by:

K’ = (E1’ x I1’)/1000

where I1’ is the primary current at 25 Hz.

From the transformer equation, we have:

I1’/I1 = N2/N1

Therefore, we can write:

I1’ = (N2/N1) x I1

I1’ = (80/1320) x (K x 1000)/V1

I1’ = 0.0758 K amps

Substituting the values,
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A 100 KVA, 6600/400 V , 50 Hz single phase transformer has 80 turns on low voltage side. At 25 Hz its flux increases by 10% . The high voltage and KVA rating at 25 Hz will be respectively (a) 3630 V, 55 KVA (b) 6600 V,100 KVA (c) 7260 V , 110 KVA (d) 3300 V , 50 KVA Correct answer :- (a) Can u explain this?
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A 100 KVA, 6600/400 V , 50 Hz single phase transformer has 80 turns on low voltage side. At 25 Hz its flux increases by 10% . The high voltage and KVA rating at 25 Hz will be respectively (a) 3630 V, 55 KVA (b) 6600 V,100 KVA (c) 7260 V , 110 KVA (d) 3300 V , 50 KVA Correct answer :- (a) Can u explain this? for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Question and answers have been prepared according to the Electrical Engineering (EE) exam syllabus. Information about A 100 KVA, 6600/400 V , 50 Hz single phase transformer has 80 turns on low voltage side. At 25 Hz its flux increases by 10% . The high voltage and KVA rating at 25 Hz will be respectively (a) 3630 V, 55 KVA (b) 6600 V,100 KVA (c) 7260 V , 110 KVA (d) 3300 V , 50 KVA Correct answer :- (a) Can u explain this? covers all topics & solutions for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A 100 KVA, 6600/400 V , 50 Hz single phase transformer has 80 turns on low voltage side. At 25 Hz its flux increases by 10% . The high voltage and KVA rating at 25 Hz will be respectively (a) 3630 V, 55 KVA (b) 6600 V,100 KVA (c) 7260 V , 110 KVA (d) 3300 V , 50 KVA Correct answer :- (a) Can u explain this?.
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