Question

A 200 kVA, 3300/240 V, 50 Hz single-phase transformer has 80 turns on the secondary winding. Assuming an ideal transformer, the primary current I₁ and secondary current I₂ on full load are nearly

1. 60.6 A and 833 A
2. 72.2 A and 833 A
3. 60.6 A and 720 A
4. 72.2 A and 720 A

Answer 1

Correct Answer - Option 1 : 60.6 A and 833 A

Concept:

From the MMF balance equation of a transformer:

\(\frac{{{V_1}}}{{{V_2}}} = \frac{{{N_1}}}{{{N_2}}} = \frac{{{I_2}}}{{{I_1}}}\)

Also, the rated power can be written as:

P (KVA) = V × I

In a transformer, the current or the voltage steps up and down. But the power transferred is always equal.

Application:

Given power = 200 kVA. (Full load rated power)

We can write:

200 k = V₁ × I₁

Given voltage at the primary end, V₁ = 3300 V

∴ 200 k = 3300 × I₁

\({I_1} = \frac{{200\;K}}{{3300}}\)

I₁ = 60.6 A

The primary current is therefore 60.6 A

Similarly, we can write:

200 k = V₂ × I₂

Given V₂ = 240 V, the above equation becomes:

200 k = 240 × I₂

\({I_2} = \frac{{200\;k}}{{240}}\)

I₂ = 833.33 A

∴ The secondary current I₂ = 833.33 A