Transformer: Important Solved Examples Problems

Basic Electronics and Electrical Engineering

Chapter 3: Transformer

SOLVED EXAMPLES

EXAMPLE 1

A single phase transformer has 400 primary turns and 1000 secondary turns. The net cross sectional area of the core is 60 cm², if the primary winding is connected to 50 Hz supply at 500 V. Calculate the value of maximum flux density in the core and the EMF induced in the secondary winding

Given Data: N₁ =400, N₂ = 1000, A = 60 cm², f= 50 Hz, V₁ (or E₁) = 500 V,

Turn ratio, N₂/Ν₁ = E₂/Ε₁

To Find: B_m=?

SOLUTION:

E₂ = N₂E₁ / N₁

= 1000×500 / 400

= 1250 V

E₂ = 4.44ϕfN₂

ϕ = E₂ / (4.44fN₂)

=  1250 / 4.44×50×1000

ϕ = 0.0056 Wb

We know

ϕ = B_mA

Flux density, B_m = flux / area

= 0.0056 / 60×10⁴

B_m = 0.933 Wb/m²

EXAMPLE 2

The required no - load ratio in a single phase 50 Hz core type transformer is 6000/250 V. Find the number of turns in each winding, if the flux in the core is to be about 0.06 Wb.

Given Data: V₁ (or E₁) = 6000 V, E₂ = 250 V, the average value of flux in the core = 0.06 Wb, f= 50 Hz

To Find: N₁, N₂ = ?

SOLUTION

For a sinusoidal quantity, we have,

Maximum value = (π/2) (Average value)

Maximum value of flux ϕ_m = π/2 × 0.06 Wb

(i) E₁ = 4.44fϕ_mN₁

6000 = 4.44 × π/2 × 0.06 × 50 × N₁

N₁ = 288 turns

(ii) N₂/N₁ = E₂/E₁

N₂ = N₁ (E₂/E₁)

N₂ = 288 (250/6000)

= 12

N₂ = 12 turns

EXAMPLE 3

A single phase transformer has 500 primary and 1000 secondary turns. The area of the core is 75 cm². If the primary winding is connected to 400 V, 50 Hz supply, calculate the secondary voltage and the peak value of the flux density.

Given Data: N₁ = 500, N₂ = 1000, A = 75 cm², V₁ =400 V, f = 50 Hz

To Find: V₂ and B_m = ?

SOLUTION:

N₂/Ν₁ = V₂/V₁

= 1000 / 500 = K = 2

V₂=2V₁ = 2×400

V₂ = 800 V

Now, V₁ = 4.44fϕ_mN₁

400 = 4.44 × 50 × ϕ_m × 500

ϕ_m = 3.6036 mWb

Now, ϕ_m = B_mA

B_m = ϕ_m / A =  3.6036×10³ / 75×10^‒4

B_m = 0.4804 Wb/m²

EXAMPLE 4

A 20 kVA, single phase transformer has 200 turns on the primary and 40 turns on the secondary. The primary is connected to 1000 V, 50 Hz supply. Determine the secondary voltage on open circuit and the current flowing through the two windings on full load.

Given Data: 20 kVA, N₁ = 200, N₂ = 40, V₁ = 1000 V, f = 50 Hz

To Find: V₂, I₁ and I₂ =?

SOLUTION:

V₁/V₂ = N₁/Ν₂

1000 / V₂ = 200 / 40

V₂ = 200 V

I₁ (F.L) = kVA / V₁

= 20×10³ / 1000

I₁ (F.L) = 20 A

and

I₂ (F.L) = kVA / V₂

= 20×10³ / 200

I₂ (F.L) = 100 A

I₁(F.L) / I₂(F.L) = N₂/N₁

I₂(F.L) = 20×200 / 40

I₂(F.L) = 100 A

EXAMPLE 5

A transformer with 40 turns on the high voltage winding is to be used to step down the voltage from 240 V to 120 V. Find the number of turns in the low voltage winding.

SOLUTION:

By transformation ratio, we know that,

K = V_HV / V_LV = N_HV/N_LV

So, N_LV = (V_LV/V_HV) × N_HV

= 120/240 × 40

= 20

N_LV= 20 turns

EXAMPLE 6

Find the number of turns (N₂) in the secondary side of a transformer if V₁ = 10 V, N₁=1000 turns and V₂=5 V?

SOLUTION:

By transformation ratio,

 K = V₂/V₁ = N₂/Ν₁

 N₂ = (V₂/V₁) × N₁

 = (5/10) × 1000

 N₂ = 500 turns.

EXAMPLE 7

In a single phase transformer, N_p=350 turns, Ns = 1050 turns, E_p = 400 V. Find Es.

SOLUTION:

We know,

 E_S/E_P= N_S/Ν_P

 E_S = (N_S/Ν_P) × E_P

= (1050/350) × 400

E_S = 1200 V

EXAMPLE 8

A single phase, 2200 / 250 V, 50 Hz transformer has a net core area of 36 cm² and a maximum flux density of 6 wb/m². Calculate the number of turns of primary and secondary windings.

SOLUTION

Primary Voltage, E₁ = 4.44 fB_mAN₁

E₁ = 4.44 fB_mAN₁

⇒ N₁ = E₁ / 4.44fB_mA 

=  2200 / (4.44 × 50 × 6 × 36 × 10^‒4)

⇒ N₁ = 458.79

N₁ ≈ 459 turns

Secondary voltage, E₂ = 4.44fB_mAN₂

⇒ N₂ = E₂ / 4.44fB_mA 

=  250 / (4.44 × 50 × 6 × 36 × 10^‒4)

⇒ N₂ = 52.14

N₂ ≈ 52 turns

EXAMPLE 9

An ideal 25 KVA transformer has 500 turns on the primary winding and 40 turns on the secondary winding. The primary is connected to 3000 V, 50 Hz supply. Calculate (1) primary and secondary currents on full load (2) secondary emf and (3) the maximum core flux.

SOLUTION

1. Full load primary current, I₁= Rating / V₁

 = 25×10³ / 3000

⇒ I₁ = 18.33 A

By transformation ratio,

K = V₂/V₁ = N₂/N₁

V₂ = 3000 × 40/500

= 240 V

2. Full load secondary current,

I₂ = Rating / V₂

 = 25×10³ / 240

⇒I₂=104.17 A

3. Maximum core flux, ϕ_m

V₁ = 4.44f ϕ_mN₁

ϕ_m = V₁ / (4.44f N₁)

= 3000 / (4.44×50×500)

⇒ ϕ_m = 0.027 = 27 mwb

 ϕ_m = 27 mwb