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


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Test: Voltage Regulation Of Transformer - Question 1

A 10 kVA, 400/200 V, 1-phase transformer with 2% resistance and 2% leakage reactance. It draws steady short circuit current at angle of

Detailed Solution for Test: Voltage Regulation Of Transformer - Question 1

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°

Test: Voltage Regulation Of Transformer - Question 2

While conducting open circuit test and short circuit test on a transformer, status of low-voltage and high-voltage windings will be such that in

Detailed Solution for Test: Voltage Regulation Of Transformer - Question 2

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

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Test: Voltage Regulation Of Transformer - Question 3

While conducting testing on the single phase transformer, one of the student tries to measure the resistance by putting an ammeter across one terminal of primary and other to secondary, the reading obtained will be

Detailed Solution for Test: Voltage Regulation Of Transformer - Question 3

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

Test: Voltage Regulation Of Transformer - Question 4

If the per unit leakage impedance for the primary of a transformer is ‘x’ on the given rated base value. If the voltage and volt-amperes are doubled, then the changed per unit impedance will be

Detailed Solution for Test: Voltage Regulation Of Transformer - Question 4

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

Test: Voltage Regulation Of Transformer - Question 5

The efficiency of two identical transformers under load conditions can be determined by

Test: Voltage Regulation Of Transformer - Question 6

 While estimating voltage regulation of a transformer, keeping

Detailed Solution for Test: Voltage Regulation Of Transformer - Question 6

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.

Test: Voltage Regulation Of Transformer - Question 7

A 200/400 V single phase transformer has leakage impedance z= r+jx. Then we can expect magnitude of load pf of ____ at zero voltage regulation.

Detailed Solution for Test: Voltage Regulation Of Transformer - Question 7

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

Test: Voltage Regulation Of Transformer - Question 8

 If the pu impedance of a single phase transformer is 0.01+j0.05, then its regulation at p.f. of 0.8 lagging will be

Detailed Solution for Test: Voltage Regulation Of Transformer - Question 8

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

Test: Voltage Regulation Of Transformer - Question 9

The transformer phasor diagram under the short circuit can be identified as

Detailed Solution for Test: Voltage Regulation Of Transformer - Question 9

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

Test: Voltage Regulation Of Transformer - Question 10

A transformer has resistance and reactance in per unit as 0.05 and 0.09 respectively. Its voltage regulation at full load for 0.8 power factor lagging and leading will be

Detailed Solution for Test: Voltage Regulation Of Transformer - Question 10

Concept:

In the transformer voltage regulation is given by,

Voltage regulation = x (Rpu cosθ ± Xpu sinθ)

Where x = fraction of load

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

Xpu = 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

Rpu = 0.05

Xpu = 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 %

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