This set of Transformers Quiz focuses on “Transformers as Magnetically Coupled Circuits”.

1. Considering transformer as a magnetically coupled circuit, the voltage at the primary of the transformer is _________________

a) R_{1}i_{1} + L_{1}*di_{1}/dt – M*di_{2}/dt

b) R_{1}i_{1} + L_{1}*di_{1}/dt + M*di_{2}/dt

c) R_{1}i_{1} – L_{1}*di_{1}/dt + M*di_{2}/dt

d) R_{1}i_{1} – L_{1}*di_{1}/dt – M*di_{2}/dt

View Answer

Explanation: If transformer is considered as a magnetically coupled circuit, then by fundamental Kirchoff’s law one can find equations for applied voltage, current and inductance of each coil of a transformer.

2. Considering transformer as a magnetically coupled circuit, the voltage at the secondary of the transformer is _________________

a) R_{2}i_{2} + L_{2}*di_{2}/dt – M*di_{1}/dt

b) R_{2}i_{2} + L_{2}*di_{2}/dt + M*di_{1}/dt

c) -R_{2}i_{2} + L_{2}*di_{2}/dt – M*di_{1}/dt

d) -R_{2}i_{2} – L_{2}*di_{2}/dt + M*di_{1}/dt

View Answer

Explanation: If transformer is considered as a magnetically coupled circuit, then by fundamental Kirchoff’s voltage law one can find equations for applied voltage, current and inductance of each coil of a transformer, considering the direction of current, which implies negative sign.

3. Core magnetizing current is __________________

a) i_{1}-i_{2}/a

b) i_{1}+i_{2}/a

c) i_{1}/a-i_{2}/a

d) i_{1}/a+i_{2}/a

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Explanation: The current flowing through the primary is divided into two parts, one of which flows through series parameters of the circuit which is equal to i2/a and the remaining current flows through the parallel branch.

4. Which of the following ratio is equal to N_{1}/N_{2}?

a) L_{1}/L_{2}

b) L_{2}/L_{1}

c) √ (L_{1}/L_{2})

d) √ (L_{2}/L_{1})

View Answer

Explanation: When tight coupling is considered, k=1. Because, leakage is equal to 0. So, φC

_{1}= φ

_{1}and φC

_{2}= φ

_{2}. Thus, it follows from equations N

_{1}/N

_{2}(M/L

_{1}) = N

_{2}/N

_{1}(M/L

_{2}). By solving this equality, we get the answer.

5. A transformer has turn ratio of a = 10. The primary on application of 200 V draws 4 A with secondary open circuited which is found to have a voltage of 1950 V. Then, L1 is equal to ______________

a) 0.19 H

b) 0.159 H

c) 0.9 H

d) 0.259 H

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Explanation: Xm = 200/4 = 50 Ω, we also know that Xm = 2πfL1.

Here frequency denoted by f is equal to 50 Hz. While Xm is calculated previously by taking ratio of voltage with currents.

6. 5. A transformer has turn ratio of a = 10. The primary on application of 200 V draws 4 A with secondary open circuited which is found to have a voltage of 1950 V. Then, M is equal to ______________

a) 1.9 H

b) 1.55 H

c) 9 H

d) 2.59 H

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Explanation: Xm = 200/4 = 50 Ω, we also know that Xm = 2πfL1. So, by calculations at f= 50 Hz we get, value of L1 equal to 0.159 H. now, 1950 = √2π N2 φmax =√2π ψmax.

ψmax = 1950/ √2π = 8.78 Wb-T.

M= ψmax/ i1(max)= 8.78/ (√2*4) = 1.55 H.

7. What is the coupling factor if self-inductances and mutual inductance are 5.096 H, 0.05098 H and 0.5096 H respectively?

a) 1

b) 1.499

c) 1.2

d) 0.699

View Answer

Explanation: Coupling factor is calculated by the ratio of mutual inductance with square root of product of respective self-inductances. That is, k = M/ √(L1L2) = 0.5096 /√(5.096 * 0.05098) = 0.9999 =1.

8. Total cross-sectional area of earthing conductor depends on _____________

a) Minimum fault current

b) Maximum fault current

c) System output voltage

d) System input voltage

View Answer

Explanation: The chief points to be borne in mind when installing an earthing equipment are, that it must possess sufficient total cross-sectional area to carry the maximum fault current, and it must have a very low resistance in order to keep down to a safe value the potential gradient in the earth surrounding the plates, etc., under fault conditions.

**Sanfoundry Global Education & Learning Series – Transformers.**

To practice all areas of Transformers for Quizzes, __here is complete set of 1000+ Multiple Choice Questions and Answers__.