# 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

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## Options:

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

### Correct Answer: Option 1 (Solution Below)

This question was previously asked in

ESE Electronics 2020: Official Paper

## Solution:

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