# Electrical Machines Questions and Answers – Voltage Regulation of Transformer

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This set of Electrical Machines Multiple Choice Questions & Answers (MCQs) focuses on “Voltage Regulation of Transformer”.

1. A 10 kVA, 400/200 V, 1-phase transformer with 2% resistance and 2% leakage reactance. It draws steady short circuit current at which angle?
a) 45°
b) 75°
c) 135°
d) 0°

Explanation: The power factor angle will be atan(x/r) = 45°.

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 ____________
a) OC test – h.v. open, SC test-l.v. short-circuited
b) OC test – l.v. open, SC test-h.v. short-circuited
c) OC test – l.v. open, SC test-l.v. short-circuited
d) OC test – h.v. open, SC test-h.v. short-circuited

Explanation: In conducting short circuit test, l.v. winding is short circuited. In OC test h.v. is open circuited.

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 ___________
a) infinite
b) zero
c) finite
d) negative finite

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

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 ___________
a) 0.5x
b) 2x
c) 4x
d) x

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

5. 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 halved, then the changed per unit impedance will be ___________
a) 0.5x
b) 2x
c) 4x
d) x

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

6. The voltage regulation for transformer is given by ___________
a) (E2-V2)/E2
b) (E2-V2)/V2
c) (V2-E2)/E2
d) (V2-E2)/V2

Explanation: Voltage regulation is the change in secondary voltage with secondary rated voltage.

7. While estimating voltage regulation of a transformer, keeping ___________
a) primary voltage constant
b) secondary voltage constant
c) voltage changes constant at primary
d) all of the mentioned

Explanation: 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.
Explanation: Negative V.R. is achieved at leading power factor.
Explanation: When E2=V2, then V.R. will be zero.
Explanation: Maximum voltage regulation occurs when load power factor angle and leakage impedance angle are equal.
Explanation: For the short-circuit condition of a transformer, voltage across the secondary will be voltage drop across winding only.

12. With the reference of the diagram below, Choose the most appropriate.
A: B: a) A-lagging pf, B-leading pf
b) B-lagging pf, A-leading pf
c) A-lagging pf, B-lagging pf

Explanation: Maximum voltage regulation occurs at lagging pf while zero or minimum V.R. occurs at leading pf.

13. A 200/400 V single phase transformer has leakage impedance z= r+jx. Then we can expect zero voltage regulation at power factor of ___________
b) x/r lagging
d) r/x lagging

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

14. 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.
a) cosθ = x/z
b) cosθ = r/z
c) cosθ = x/r
d) cosθ = r/x

Explanation: Cosθ = x/z for the transformer at zero V.R.

15. 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?
a) 3.8%
b) 2.2%
c) -3.8%
d) -2.2%

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

16. If the pu impedance of a single phase transformer is 0.01+j0.05, then its regulation at p.f. of 0.8 leading will be?
a) 3.8%
b) 2.2%
c) -3.8%
d) -2.2%

Explanation: V.R. = (r(pu)*cosθ-x(pu)*sinθ)*100 % = (0.01*0.8 – 0.05*0.6)*100 = -2.2%.

Sanfoundry Global Education & Learning Series – Electrical Machines.

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