**1. Voltage Regulation of Transformer**Because of the voltage drop across the primary and secondary impedances it is observed that the secondary terminal voltage drops from its no load value (E

_{2}) to load value (V

_{2}) as load and load current increases.

This decrease in the secondary terminal voltage expressed as a fraction of the no load secondary terminal voltage is called regulation of a transformer.

The regulation is defined as change in the magnitude of the secondary terminal voltage, when full load i.e. rated load of specified power factor supplied at rated voltage is reduced to no load, with primary voltage maintained constant expressed as the percentage of the rated terminal voltage.

Let E_{2 }= Secondary terminal voltage on no load

V

_{2 }= Secondary terminal voltage on given load

then mathematically voltage regulation at given load can be expressed as,

The ratio (E

_{2}- V

_{2 }/ V

_{2 }) is called per unit regulation.

The secondary terminal voltage does not depend only on the magnitude of the load current but also on the nature of the power factor of the load. If V

_{2 }is determined for full load and specified power factor condition the regulation is called full load regulation. As load current increases, the voltage drops tend to increase V

_{2 }and drops more and more. In case of lagging power factor V_{2 }< E_{2}and we get positive voltage regulation, while for leading power factor E_{2}< V_{2 }and we get negative voltage regulation. The voltage drop should be as small as possible hence less the regulation better is the performance of a transformer.

__1.1 Expression for Voltage Regulation__

%R = (E

_{2}- V

_{2 /}V

_{2 }) x 100 = (Total voltage drop/V

_{2}) x 100

The expression for the total approximate voltage drop is already derived.

Total voltage drop = I

_{2 }

_{}R

_{2e }cos Φ ± I

_{2 }X

_{2e }sin Φ

Hence the regulation can be expressed as,

'+' sing for lagging power factor while '-' sing for leading power factor loads.

The regulation van be further expressed interms of I

_{1 }, V

_{1}, R

_{1e }and X

_{1e}.

V

_{2 }/V

_{1 }=I

_{1 }/I

_{2 }= K

**.**V

^{.}._{2 }= KV

_{1 }, I

_{2 }= I

_{1}/K

while R

_{1e }=R

_{2e}/K

^{2}, X

_{1e }= X

_{2e }/K

^{2}

Substituting in the regulation expression we get,

__1.2 Zero Voltage Regulation__

We have seen that for lagging power factor and unity power factor condition V

_{2 }< E_{2}and we get positive regulation. But as load becomes capacitive, V_{2 }starts increasing as load increase. At a certain leading power factor we get E_{2}= V_{2 }and the regulation becomes zero. If the load is increased further, E_{2}becomes less than V_{2 }and we get negative regulation.**.**for zero voltage regulation,

^{.}.E

_{2}= V

_{2 }

**.**E

^{.}._{2}- V

_{2 }= 0

or V

_{R }cos Φ - V

_{x }sin Φ = 0 .......... -ve sing as leading power factor

where V

_{R }= I

_{2 }R

_{2e}/V

_{2 }= I

_{1 }R

_{1e }/V

_{1 }and V

_{x }= I

_{2 }X

_{2e }/V

_{2 }= I

_{1 }X

_{1e }/V

_{1 }

**.**V

^{.}._{R }cos Φ = V

_{x }sin Φ

**.**tan Φ = V

^{.}._{R }/V

_{x }

**.**cos Φ = cos {tan

^{.}.^{-1}(V

_{R }/Vx)}

This is the leading p.f. at which voltage regulation becomes zero while supplying the load.

__1.3 Constants of a Transformer__

%R= (( I

_{2 }R

_{2e}cos Φ ± I

_{2 }X

_{2e}sin Φ )/ E

_{2}) x 100

= {(I

_{2 }R

_{2e}/E

_{2}) cos Φ ± (I

_{2 }X

_{2e}/E

_{2}) sin Φ} x 100

The ratio (I

_{2 }R

_{2e}/E

_{2}) or (I

_{1 }R

_{1e}/E

_{1}) is called per unit resistive drop and denoted as V

_{R}.

The ratio (I

_{2 }X

_{2e}/E

_{2}) or (I

_{1 }X

_{1e}/E

_{1}) is called per unit reactive drop and is denoted as Vx.

The terms V

%R = (V_{R }and Vx are called constants of a transformer because for the rated output I_{2}, E_{2}, R_{1e}, X_{1e}, R_{2e}, X_{2e}are constants. The regulation can be expressed interms of V_{R }and Vx as,_{R }cos Φ

iuu ± Vx sin Φ ) x 100

On load condition, E

_{2 }= V

_{2 }and E

_{1}= V

_{1}

where V

_{1}and V

_{2 }are the given voltage ratings of a transformer. Hence V

_{R }and Vx can be expressed as,

V

_{R }= I

_{2 }R

_{2e}/ V

_{2 }= I

_{1 }R

_{1e}/ V

_{1}

and

Vx =I

_{2 }R

_{2e}/ V

_{2 }= I

_{1 }X

_{1e}/ V

_{1}

where V

_{1}and V

_{2 }are no load primary and secondary voltages,

V

_{R }and Vx can be expressed on percentage basis as,

Percentage resistive drop = V

_{R }x 100

Percentage reactive drop = Vx x 100

**Key Point**: Note that and are also called per unit resistance and reactance respectively.

**Example 1**: 250/125 V, 5 KVA single phase transformer has primary resistance of 0.2 Ω and reactance of 0.75Ω. The secondary resistance is 0.05 Ω and reactance of 0.2Ω

ii) The secondary terminal voltage on full load 0.8 and leading p.f.

**Solution**: The given values are,

R

_{1 }= 0.2 Ω, X

_{1 }= 0.75 Ω, R

_{2 }= 0.05 Ω, X

_{2 }= 0.2 Ω, cos Φ = 0.8 leading

K= E

_{2 }/E

_{1 }= 125/250 = 1/2 = 0.5

(I

_{2}) F.L.= KVA/V

_{2 }= 5x10

^{3 }/125 = 40 A ...... full load

R

_{2e }= R

_{2 }+ K

^{2 }R

_{1 }= 0.05 + (0.5)

^{2}x 0.2 = 0.1 Ω

X

_{2e }= X

_{2}

_{ }+ K

^{2 }X

_{1 }= 0.2 + (0.5)

^{2}x 0.75 = 0.3875 Ω

i) Regulation on full load, cos Φ = 0.8 leading

sin Φ = 0.6

**.**%R = ((I

^{.}._{2 }R

_{2e }cos Φ - I

_{2 }X

_{2e }sin Φ )/E

_{2 }) x 100

where I

_{2 }= Full load current

**.**% R = ((40 x 0.1 x 0.8 - 40 x 0.3875 x 0.6)/125) x 100 = -4.88%

^{.}.ii) For secondary terminal voltage, use basic expression of regulation

% R = ((E

_{2 }- V

_{2 })/E

_{2 }) x 100

**.**-4.88 = ((125- V

^{.}._{2})

_{ }/125) x 100

**.**-6.1 = 125 - V

^{.}._{2}

**.**V

^{.}._{2 }= 131.1 V

It can be seen that for leading p.f. E

_{2 }<V

_{2}.

**Example 2**: Calculate the regulation of a transformer in which the copper loss is 1% of output and the percentage reactance drop is 5% when load power factor is

**Solution**: Given values are,

%X = 5%

Now copper loss is, P

_{cu }= I

_{2}

^{2}

_{ }R

_{2e }

and output is, P

_{out }= V

_{2 }I

_{2 }

**.**% Copper loss = (P

^{.}._{cu}/P

_{out})

_{ }x100 = (I

_{2}

^{2}

_{ }R

_{2e }/V

_{2}I

_{2 }

_{ }) x 100

% V

_{R }= (I

_{2 }R

_{2e }/ V

_{2})x 100

= (I

_{2 }R

_{2e }

_{ }/V

_{2}) x (I

_{2 }/I

_{2 }) x 100 = (I

_{2}

^{2}

_{ }R

_{2e }/V

_{2}I

_{2 }

_{ }) x 100

= % copper loss

**.**V

^{.}._{R}= 1% = 0.01 and V

_{x}= 5% = 0.05

i) cos Φ = 0.9 lagging

**.**sin Φ = 0.4358

^{.}.**.**% R = (V

^{.}._{R}cos Φ + V

_{x}sin Φ ) x 100 = (0.01 x 0.9 + 0.05 x 0.4358) x 100

= + 3.08%

ii) cos Φ = 0.9 leading

**.**% R = (V

^{.}._{R}cos Φ - V

_{x}sin Φ ) x 100 = (0.01 x 0.9 - 0.05 x 0.4358) x 100

= -1.28%

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