Question

A single-phase,50-HZ transformer has 100 turns on the primary winding and 400 turns on the secondary winding.the net cross-sectional area of the core is 250 cm✓.if the primary winding is connected to a 50-HZ, 230-V supply, calculate (a) the emf induced in the secondary winding, and (b) the maximum value of the flux density in the core.​

Answer 1

Answer:

The root mean square speed of a gas is given by

M

3RT

where T is the temperature of gas

M is the molecular weight of gas.

When oxygen molecules dissociate into oxygen atoms, the molecular weight of gas halves.

Hence T

′

=2T,M

′

=M/2

Hence

v

v

′

=

T/M

T

′

/M

′

=

1

4

=2

⟹v

′

=2v

Answer 2

The EMF induced in the secondary winding is 920 V, while the maximum value of the flux density is Bm = 0.042 Wb/m²

Step-by-step explanation:

Given: N₁ = 100 turns, N₂ = 400 turns, A = 250cm², f = 50Hz, V₁ = 230V

To Find: EMF induced in the secondary winding & maximum value of the flux density in the core.

Solution:

• Finding the EMF induced in the secondary winding

For a single-phase transformer, we have

$\frac{V_2}{V_1} = \frac{N_2}{N_1} \Rightarrow V_2 = \frac{V_1 \times N_2}{N_1}$

Putting the values and finding the voltage in the secondary winding, we have

$V_2 = \frac{230 \times 400}{100} = 920 V$

For an ideal transformer, $V_2 = E_2$. Therefore, the EMF induced is $E_2 = 920 V$

• Finding the maximum value of the flux density in the core

The relation between EMF induced and the maximum flux density is given by,

$E_2 = 4.44fN_2 B_m A$

where Bm is the maximum flux density

Putting the given values and calculating, we get,

$920 = 4.44 \times 50 \times 400 \times B_m \times 250 \times 10^{-4}$

$\Rightarrow B_m = 0.042 \ Wb/m^{2}$

Hence, the EMF induced in the secondary winding is 920 V, while the maximum value of the flux density is Bm = 0.042 Wb/m²