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°
View Answer

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

Answer: a
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
View Answer

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

Answer: a
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
View Answer

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

Answer: a
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
View Answer

Answer: a
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.
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8. Identify the phasor diagram for the negative voltage regulation from the below diagrams.
a) The phasor diagram for the negative voltage regulation - option a
b) The phasor diagram for the negative voltage regulation - option b
c) The phasor diagram for the negative voltage regulation - option c
d) The phasor diagram for the negative voltage regulation - option d
View Answer

Answer: a
Explanation: Negative V.R. is achieved at leading power factor.

9. Identify the phasor diagram for the zero voltage regulation from the below diagrams.
a) The phasor diagram for the negative voltage regulation - option c
b) The phasor diagram for the zero voltage regulation - option d
c) The phasor diagram for the negative voltage regulation - option b
d) The phasor diagram for the zero voltage regulation - option d
View Answer

Answer: a
Explanation: When E2=V2, then V.R. will be zero.
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10. Identify the phasor diagram for the maximum voltage regulation from the below diagrams.
a)The phasor diagram for the negative voltage regulation - option d
b)The phasor diagram for the zero voltage regulation - option d
c)The phasor diagram for the negative voltage regulation - option a
d)The phasor diagram for the negative voltage regulation - option c
View Answer

Answer: a
Explanation: Maximum voltage regulation occurs when load power factor angle and leakage impedance angle are equal.

11. The transformer phasor diagram under the short circuit can be identified by?
a) The transformer phasor diagram under the short circuit - option a
b) The phasor diagram for the negative voltage regulation - option c
c) The phasor diagram for the negative voltage regulation - option d
d) The phasor diagram for the zero voltage regulation - option d
View Answer

Answer: a
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: Find the appropriate option with the reference of the diagram
B:The phasor diagram for the negative voltage regulation - option a
a) A-lagging pf, B-leading pf
b) B-lagging pf, A-leading pf
c) A-lagging pf, B-lagging pf
d) A-leading pf, B-leading pf
View Answer

Answer: a
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 ___________
a) x/r leading
b) x/r lagging
c) r/x leading
d) r/x lagging
View Answer

Answer: a
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
View Answer

Answer: a
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%
View Answer

Answer: a
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%
View Answer

Answer: d
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.

To practice all areas of Electrical Machines, here is complete set of 1000+ Multiple Choice Questions and Answers.

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