Test: Voltage Regulation Of Transformer - Electrical Engineering (EE) MCQ

Given Data:

• Transformer Rating: 10 kVA, 400/200 V, 1-phase
• Resistance (R): 2%
• Leakage Reactance (X): 2%

Steps to Calculate the Angle of Short-Circuit Current:

1. Determine the Impedance: The impedance ZZZ of the transformer can be calculated using the given percentage values for resistance and reactance:

   Z=R+jXZ = R + jXZ=R+jX

   where RRR and XXX are the resistance and reactance.

   Given:

   R=2% of Z (assuming Z as base impedance, Z = 100)R = 2\% \text{ of } Z \text{ (assuming Z as base impedance, Z = 100)}R=2% of Z (assuming Z as base impedance, Z = 100) X=2% of Z (assuming Z as base impedance, Z = 100)X = 2\% \text{ of } Z \text{ (assuming Z as base impedance, Z = 100)}X=2% of Z (assuming Z as base impedance, Z = 100)

2. Calculate the Impedance Angle: The impedance angle θ\thetaθ is determined by:

   θ=tan⁡−1(XR)\theta = \tan^{-1} \left(\frac{X}{R}\right)θ=tan−1(RX​)

   Substituting R=2R = 2R=2 and X=2X = 2X=2:

   θ=tan⁡−1(22)=tan⁡−1(1)=45°\theta = \tan^{-1} \left(\frac{2}{2}\right) = \tan^{-1}(1) = 45°θ=tan−1(22​)=tan−1(1)=45°

Angle of the Short-Circuit Current:

The angle of the short-circuit current is equal to the angle of the impedance because the current in a short-circuit condition is essentially the same as the impedance angle.

Conclusion:

The steady short-circuit current draws at an angle of:

1. 45°

For transformer testing:

• Open Circuit (OC) Test:

  • This test is used to determine the core losses and no-load characteristics of the transformer. In this test, the high-voltage (HV) winding is energized, and the low-voltage (LV) winding is kept open.

• Short Circuit (SC) Test:

  • This test is used to measure the copper losses and the impedance of the transformer. In this test, the low-voltage (LV) winding is short-circuited, and the high-voltage (HV) winding is energized.

So, the correct configuration is:

2. OC test – l.v. open, SC test – h.v. short-circuited

As the primary and secondary are physically isolated, the impedance will be infinite for not electrically connected circuit.

 pu(new base)=(x)*(MVA(new)/MVA(old))*(kV(old)/kV(new))^2
=x*2*(1/4)
=0.5x.

V.R. is calculated keeping the primary constant because then the core flux will change and the change of secondary voltage can not be fixed.

ZVR occurs at the leading pf of load at x/r.

V.R. = (r(pu)*cosθ+x(pu)*sinθ)*100 % = (0.01*0.8 + 0.05*0.6)*100 = 3.8%

For the short-circuit condition of a transformer, voltage across the secondary will be voltage drop across winding only.

Concept:

In the transformer voltage regulation is given by,

Voltage regulation = x (R_pu cosθ ± X_pu sinθ)

Where x = fraction of load

R_pu = resistance in pu (we can use % resistive drop in fraction)

X_pu = reactance in pu (we can use % reactive drop in fraction)

+ For lagging load

– For leading load

Calculation:

Given that, the transformer is working at 0.8 lagging power factor at full load

⇒ cos θ = 0.8, sin θ = 0.6

x = 1

R_pu = 0.05

X_pu = 0.09

%V.R. = 1 × (0.05 × 0.8 + 0.09× 0.6)  

= 0.094 

= 9.4 %

For leading pf = 0.8

%V.R. = 1 × (0.05 × 0.8 -  0.09× 0.6) = - 0.014

= -1.4 %