# Transformers Questions and Answers – Transformers as Magnetically Coupled Circuits

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) R1i1 + L1*di1/dt – M*di2/dt
b) R1i1 + L1*di1/dt + M*di2/dt
c) R1i1 – L1*di1/dt + M*di2/dt
d) R1i1 – L1*di1/dt – M*di2/dt

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) R2i2 + L2*di2/dt – M*di1/dt
b) R2i2 + L2*di2/dt + M*di1/dt
c) -R2i2 + L2*di2/dt – M*di1/dt
d) -R2i2 – L2*di2/dt + M*di1/dt

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) i1-i2/a
b) i1+i2/a
c) i1/a-i2/a
d) i1/a+i2/a

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 N1/N2?
a) L1/L2
b) L2/L1
c) √ (L1/L2)
d) √ (L2/L1)

Explanation: When tight coupling is considered, k=1. Because, leakage is equal to 0. So, φC1 = φ1 and φC2 = φ2. Thus, it follows from equations N1/N2 (M/L1) = N2/N1(M/L2). 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

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.
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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

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

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

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.