Question

A 6600/600V, 50Hz single phase transformer has a maximum flux density of 1.35T in its core. If the net cross sectional area of iron core is 200 cm ^ 2 . Calculate the number of turns in the primary and secondary windings of the transformer.​

Answer 1

Explanation:

Cross sectional area of core, A = 428 cm2 = 0.0428 m2

Maximum flux density in core, Bm = 1.5 Tesla

Maximum flux in core, Φm = Bm * A = 0.0642 Wb

Winding voltage, E = 4.44 * Φm * f * T

Frequency, f = 50 Hz

Therefore, winding turns, T1 = E1/(4.44 * Φm * f) = 463 turns

T2 = E2/(4.44 * Φm * f) = 28 turns

Answer 2

Answer:

So, number of primary turns will be approximately 1101, while the number of secondary turns will be approximately 100.

Explanation:

As per the data given in the question,

We have,

A 6600/600V, 50Hz single phase transformer

Maximum flux density = 1.35T

Net cross section area $=200\:cm^2$

We have to calculate the number of turn:

Step 1: Calculating the maximum flux

Cross sectional area of core, A $= 200 \:cm^2 = 200 \times 10^{-4}=0.02\: m^2$

Maximum flux density in core, $B_m = 1.5\: Tesla$

We know that,

Maximum flux in core, $\Phi _m = B_m \times A = 1.35 \times 0.02 = 0.027 Wb$

Step 2: Calculating the number of turns:

As we know,

Winding voltage, $E = 4.44 \times \Phi_m \times f \times T$

Frequency, f = 50 Hz

Therefore, primary winding turns, $T_1 = \frac{E_1}{(4.44 \times \Phi _m \times f)}$

So, the primary turn will be

$\frac{6600}{4.44 \times 0.027 \times 50}\\\\\approx 1101 \:turns$

So, similarly secondary turn will be  $T_2 = \frac{E_2}{(4.44 \times \Phi _m \times f)}$

So, secondary turns will be

$T_2=\frac{600}{4.44 \times 0.027 \times 50}\\\\\approx 100 \:turns$