# For a single phase, two winding transformer, the supply frequency and voltage are both increased by 10%. The percentage changes in the hysteresis loss

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For a single phase, two winding transformer, the supply frequency and voltage are both increased by 10%. The percentage changes in the hysteresis loss and eddy current loss, respectively, are
1. 10 and 21
2. -10 and 21
3. 21 and 10
4. -21 and 10

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Correct Answer - Option 1 : 10 and 21

Concept:

​Hysteresis loss:

It is due to the reversal of magnetization of the transformer core whenever it is subjected to the alternating nature of the magnetizing force.

The power consumed by the magnitude domains to change their orientation after every half cycle whenever core is subjected to alternating nature of magnetizing force is called as hysteresis loss.

​Hysteresis loss can be determined by using the Steinmetz formula  given by

${W_h} = \eta B_{max}^2fV$

Where

x is the Steinmetz constant, Bm = maximum flux density

f = frequency of magnetization or supply frequency, V = volume of the core

Eddy current losses:

Eddy current loss is basically I2R loss present in the core due to the production of eddy currents in the core, because of its conductivity.

Eddy current losses are directly proportional to the conductivity of the core.

Eddy current loss ${W_e} = K{f^2}B_m^2{t^2}$

Where

K - coefficient of eddy current., Bm - maximum value of flux density

t - thickness of lamination in meters, f - frequency of eddy current

If (V / f) ratio is constant:

As we know,

Bmax ∝ (V / f)

⇒ Bmax = constant

Hysteresis loss Wh ∝ f

And, eddy current loss We ∝ f2

If V/f is not constant:

Hysteresis losses

$\Rightarrow {W_h} \propto V_1^{1.6}{f^{ - 0.6}}$

Eddy current losses

$\Rightarrow {W_e} \propto V_1^2$

Explanation:

Supply frequency (f) and voltage (V) increased by 10%.

${B_m} \propto \frac{V}{f}$

Hysteresis losses:

Ph ∝ f

Therefore, change in hysteresis loss

⇒ ΔPh = Δf = 10%

Eddy current losses:

Pe ∝ f2

$\Rightarrow \frac{{{p_{e1}}}}{{{p_{e2}}}} = \frac{{f_1^2}}{{f_2^2}}$

Now, f2 = 1.1 f1

$\Rightarrow \frac{{{p_{e1}}}}{{{p_{e2}}}} = \frac{1}{{1.1}} \Rightarrow {p_{e2}} = \left( {1.21} \right){p_{e1}}$

$\Rightarrow {\rm{\Delta }}{p_e} = \frac{{{p_{e2}} - {p_{e1}}}}{{{p_{e1}}}} = \frac{{1.21 - 1}}{1} \times 100\%$

⇒ Δpe = 21%