Question

The maximum flux density in the core of a 250/3000 volts, 50Hz single phase
transformer is 1.2 Wb/m2. If the emf per turns is 8V, Determine(i) primary and
secondary turns (ii) area of the core

Plz say fast
If u don’t know don’t waste my time

Answer 1

Answer:

E= N1 × emf Induced/ turn

primary turns = N1 = 250/8 =32

secondary turns= N2 = 3000/8 = 375

above answer is right and but I don't know the answer of = area of the core

Answer 2

Answer:

The maximum flux density in the core of $250/3000$ volts, $50Hz$ single phase  transformer is $1.2Wb/m^{2}$. If the emf per turn is $8v$, the number of turns in the primary coil is $31$ and the number of turns in the secondary coil is, $375$.The area of the core is $0.03m^{2}$

Explanation:

The maximum flux density in the core of a single-phase transistor is

$B_{max}=1.2 Wb/m^{2}$

From the question, it is clear that the voltage of the primary coil is$v_{1}=250v$

and the voltage of the secondary coil is $v_{2}=3000v$

The frequency of the transformer is $f=50Hz$

The EMF per turn is represented as $v_{1}/n_{1}$ and is given as $8$

Thus, the number of turns in the primary coil is $n_{1}=v_{1}/8=250/8=31$ turns

and number of turns in the secondary coil is, $n_{2}=3000/8=375$ turns

We have an equation to find the maximum flux of the transformer

$v_{1}=4.44fn_{1}$Ф

rearranging to get maximum flux, Ф=$250/4.44*50*31 =0.036Wb$

The maximum flux density =maximum flux /area

So the area of the core is, $0.036/1.2=0.03m^{2}$