### Voltage Regulation of Transformer

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 (E2) to load value (V2) 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            E2 = Secondary terminal voltage on no load
V2 = Secondary terminal voltage on given load
then mathematically voltage regulation at given load can be expressed as,

The ratio (E2 - V2 / V2 ) 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 V2 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 V2 and drops more and more. In case of lagging power factor V2 < E2 and we get positive voltage regulation, while for leading power factor E2 < V2 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
The voltage regulation is defined as,
%R = (E2 - V2  /V2 ) x 100 = (Total voltage drop/V2) x 100
The expression for the total approximate voltage drop is already derived.
Total voltage drop = IR2e cos Φ ± I2 X2e 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 I1 , V1, R1e and X1e.
V2 /V1 =I1 /I2 = K
...                    V2 = KV1    ,    I2 = I1/K
while                R1e =R2e/K2 , X1e = X2e /K2
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 V2 < E2 and we get positive regulation. But as load becomes capacitive, V2 starts increasing as load increase. At a certain leading power factor we get E2 = V2 and the regulation becomes zero. If the load is increased further, E2 becomes less than V2 and we get negative regulation.
...    for zero voltage regulation,
E2 = V2
...          E2 - V2 = 0
or         VR cos Φ - Vx sin Φ = 0     .......... -ve sing as leading power factor
where   VR = I2 R2e /V2 = I1 R1e /V1         and Vx = I2 X2e /V2 = I1 X1e /V1
...          VR cos Φ = Vx sin Φ
...           tan Φ = VR /Vx
...           cos Φ = cos {tan-1(VR /Vx)}
This is the leading p.f. at which voltage regulation becomes zero while supplying the load.
1.3 Constants of a Transformer
From the regulation expression we can define constants of a transformer.
%R= (( I2 R2e cos Φ ± I2 X2e sin Φ )/ E2) x 100
= {(I2 R2e /E2) cos Φ ±  (I2 X2e/E2 ) sin Φ} x 100
The ratio (I2 R2e /E2) or (I1 R1e /E1) is called per unit resistive drop and denoted as VR.
The ratio (I2 X2e/E2 ) or (I1 X1e/E1) is called per unit reactive drop and is denoted as Vx.
The terms VR and Vx are called constants of a transformer because for the rated output I2, E2, R1e, X1e, R2e , X2e are constants. The regulation can be expressed interms of VR and Vx as,
%R = (VR cos Φ
iuu  ± Vx sin Φ ) x 100
On load condition, E2 = V2 and E1= V1
where V1 and V2 are the given voltage ratings of a transformer. Hence VR and  Vx can be expressed as,
VR = I2 R2e/ V= I1 R1e/ V1
and
Vx =I2 R2e/ V2   = I1 X1e/ V1
where V1and Vare no load primary and secondary voltages,
VR and Vx can be expressed on percentage basis as,
Percentage resistive drop = VR 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
i) Determine its regulation while supplying full load on 0.8 leading p.f.
ii) The secondary terminal voltage on full load 0.8 and leading p.f.
Solution : The given values are,
R1 = 0.2 Ω, X1 = 0.75 Ω, R2 = 0.05 Ω, X2 = 0.2 Ω, cos Φ = 0.8 leading
K= E2 /E1 = 125/250 = 1/2 = 0.5
(I2) F.L.= KVA/V2 = 5x103 /125 = 40 A        ...... full load
R2e = R2 + K2 R1 = 0.05 + (0.5)2 x 0.2 = 0.1 Ω
X2e = X2 + K2 X1 = 0.2 + (0.5)2 x 0.75 = 0.3875 Ω
sin Φ = 0.6
...         %R = ((I2 R2e cos Φ - I2 X2e sin Φ )/E2 ) x 100
where   I2 = 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 = ((E2 - V2 )/E2 ) x 100
...            -4.88 = ((125- V2) /125) x 100
...             -6.1 = 125 - V2
...              V2 = 131.1 V
It can be seen that for leading p.f. E2 <V2.

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
i) 0.9 lagging and ii) 0.9 leading.
Solution : Given values are,
%X = 5%
Now copper loss is,     Pcu = I22 R2e
and output is,                 Pout = V2 I2
...                                   % Copper loss = (Pcu/Pout) x100 = (I22 R2e /V2 I2 ) x 100
% VR = (I2 R2e / V2)x 100
= (I2 R2e /V2 ) x (I2 /I2 ) x 100 = (I22 R2e /V2 I2 ) x 100
= % copper loss
...                                    VR = 1% = 0.01 and Vx = 5% = 0.05
i)                                     cos Φ = 0.9 lagging
...                                     sin Φ  = 0.4358
...                                      % R = (VR cos Φ + Vx 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 = (VR cos Φ - Vx sin Φ ) x 100 = (0.01 x 0.9 - 0.05 x 0.4358) x 100
= -1.28%