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A one-phase transformer has 400 and 1000 turns in primary and secondary, respectively. The cross-sectional area of the core is 60 cm2. The primary of the transformer is connected to a supply of one-phase, 50 Hz, 500 V. Determine the secondary voltage of the transformer.


1. 800 V
2. 1250 V
3. 125 V
4. 8000 V

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Correct Answer - Option 2 : 1250 V

Concept:

Transformation ratio: It is defined as the ratio of the secondary voltage to the primary voltage. It is denoted by 'K'.

\(K = \frac{{{N_2}}}{{{N_1}}} = \frac{{{V_2}}}{{{V_1}}} = \frac{{{I_1}}}{{{I_2}}}\)  ----- (1)

Turns ratio or Voltage ratio: It is defined as the ratio of primary winding turns to secondary winding turns. It is denoted by 'a'.

\(a = \frac{{{1}}}{{{K}}} = \frac{{{N_1}}}{{{N_2}}} = \frac{{{V_1}}}{{{V_2}}} = \frac{{{I_2}}}{{{I_1}}}\)   ----- (2)

N1 = primary winding turns

N2 = secondary winding turns

V1 = primary winding voltage

V2 = secondary winding voltage

I1 = current through the primary winding

I2 = current through the secondary winding

Calculation:

Given:

N1 = 400, N2 = 1000, V1 = 500, V2 =?

from eq (2), we can write

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

\(\frac{{500}}{{{V_2}}} = \frac{{400}}{{1000}}\)

V2 = 1250 V

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