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A 415 V/1.1 kV, 60 Hz transformer has 1675 primary turns. Calculate the number of secondary turns and the maximum flux in the core of the transformer.
1. 4440, 0.0009 Wb
2. 3000, 0.0001 Wb
3. 4343, 0.0073 Wb
4. 3745, 0.0005 Wb

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Correct Answer - Option 1 : 4440, 0.0009 Wb

Concept:

In a transformer, an alternating current is applied to the primary winding, a current in the primary winding (magnetizing current) produces alternating flux in the core of the transformer. This alternating flux gets linked with the secondary winding, by mutual induction, so an emf gets induced in the secondary winding. This induced emf is given by the EMF equation of the transformer.

 E1 = 4.44f N1 ϕm     ……….(1)

 E2 = 4.44f N2 ϕm     ……….(2)

Where, N1 = Number of turns in primary winding (high voltage side)

N2 = Number of turns in secondary winding (low voltage side)

ϕm  = Maximum flux in the core (Wb)

Φm = Bm ×  A

B= Max flux density (T)

A = Area of core (m2)

f = Frequency of the AC supply (Hz)

E1 = Induced emf on the primary side (high voltage side) (V)

E2 = Induced emf on the secondary side (low voltage side) (V)

From equations 1 and 2. we have 

\(\frac{E_1}{E_2}=\frac{N_1}{N_2}\)

Calculation:

Given: 

E= 415 V

E= 1.1 kV = 1100 V

N= 1675 

f = 60 Hz

Number of secondary turns = \(N_2=\frac{E_2}{E_1}\times N_1= \frac{1100}{415}\times 1675= 4439.76\)

∴ Number of secondary turns ≈ 4440 turns

From equation (1) maximum flux (ϕm) of the transformer can be calculated as 

415 = 4.44 × 60 × 1675 × ϕm

 ϕm = 9.3 × 104 Wb

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