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PTCUL AE E&M 2017 Official Paper (Set B)

Option 3 : 1350 Watts

__Concept:__

Separation of Iron loss in Transformer:

In transformer iron loss (Pi) has two components namely Hysteresis loss (Ph) and Eddy current loss (Pe).

Pi = Ph + Pe

\({P_i} = {k_h}fB_m^n + {k_e}{f^2}B_m^2\) …. (1)

Where,

Kh and ke are loss coefficient constant and their value depends on the type of material

Bm is the maximum value of flux density

f is supply frequency

\({P_h} \propto f\)

\({P_h} = af\)

\({P_e} \propto {f^2}\)

\({P_e} = b{f^2}\)

Where a and b are constants.

\({P_i} = af + b{f^2}\)

\(\frac{{{P_i}}}{f} = a + bf\)

Note: For the separation of these two losses, the no-load test is performed on the transformer.

__Calculation:__

Given,

V1 = 2200 V

f1 = 40 Hz

V2 = 3300 V

f2 = 60 Hz

\(\frac{{{V_1}}}{{{f_1}}} = \frac{{{V_2}}}{{{f_2}}} = \frac{2200}{40}=\frac{3300}{60}=55\)

Hence, (V / f) ratio is constant,

At frequency 40 Hz

P_{i} = 800 W = P_{h} + P_{e}

P_{h} = 600 W = af

a = 600/40 = 15

Now,

P_{e} = 800 - 600 = 200 W = bf^{2}

∴ b = \(\frac{200}{f^2}=\frac{200}{1600}=0.125\)

At, supply 3300 V and 60 Hz,

P_{i} = af + bf^{2} = (15 × 60) + (0.125 × 60^{2}) = 900 + 450

**∴ P _{i} = 1350 W**