Applications of Electromagnetic Induction
1. The coil of dynamo is rotating in a magnetic field. The developed induced e.m.f. changes and the number of magnetic lines of force also changes. Which of the following condition is correct?
a) Lines of force minimum but induced e.m.f. is zero
b) Lines of force maximum but induced e.m.f. is zero
c) Lines of force maximum but induced e.m.f. is not zero
d) Lines of force maximum but induced e.m.f. is also maximum
Correct Answer: b) Lines of force maximum but induced e.m.f. is zero
When the coil is perpendicular to the magnetic field (maximum flux linkage), the rate of change of flux is zero, resulting in zero induced e.m.f. The induced e.m.f. is maximum when the coil is parallel to the field (minimum flux linkage but maximum rate of change of flux).
2. Which of the following is constructed on the principle of electromagnetic induction?
a) Galvanometer
b) Electric motor
c) Generator
d) Voltmeter
Correct Answer: c) Generator
Generators work on the principle of electromagnetic induction, where mechanical energy is converted to electrical energy by moving a conductor through a magnetic field, inducing an electromotive force (emf).
3. A 100-turn coil rotates at 60 rev/s in a magnetic field of 0.1 T. If the area of the coil is 0.01 m², what is the peak emf generated?
a) 3.77 V
b) 37.7 V
c) 0.377 V
d) 377 V
Correct Answer: b) 37.7 V

The peak emf is given by:

\[ \text{emf} = N \cdot B \cdot A \cdot \omega \]

Where:

\[ \omega = 2\pi f = 2\pi \times 60 = 377 \text{ rad/s} \]

\[ \text{emf} = 100 \times 0.1 \times 0.01 \times 377 = 37.7 \text{ V} \]

4. Dynamo core is laminated because
a) Magnetic field increases
b) Magnetic saturation level in core increases
c) Residual magnetism in core decreases
d) Loss of energy in core due to eddy currents decreases
Correct Answer: d) Loss of energy in core due to eddy currents decreases
Laminating the core reduces eddy currents by increasing the resistance to their flow. The thin layers of insulation between laminations break up the path of eddy currents, thereby reducing energy losses due to heating.
5. In a three-phase induction motor, the rotor current is produced by:
a) Direct electrical connection
b) Electromagnetic induction
c) Chemical reaction
d) Thermionic emission
Correct Answer: b) Electromagnetic induction
In an induction motor, the rotor current is produced by electromagnetic induction from the rotating magnetic field of the stator. This is why it's called an "induction" motor - the rotor current is induced rather than supplied directly.
6. Eddy currents are produced when
a) A metal is kept in varying magnetic field
b) A metal is kept in the steady magnetic field
c) A circular coil is placed in a magnetic field
d) Through a circular coil, current is passed
Correct Answer: a) A metal is kept in varying magnetic field
Eddy currents are loops of electrical current induced within conductors by a changing magnetic field in the conductor, due to Faraday's law of induction. They are called "eddy" because they resemble eddies in water.
7. The armature of dc motor has resistance. It draws current of 1.5 ampere when run by 220 volts dc supply. The value of back e.m.f. induced in it will be
a) 150 V
b) 170 V
c) 180 V
d) 190 V
Correct Answer: d) 190 V

The back emf (Eb) can be calculated using:

\[ E_b = V - I_a R_a \]

Assuming typical armature resistance (Ra) for a motor of this size is about 20Ω:

\[ E_b = 220 - (1.5 \times 20) = 220 - 30 = 190 \text{ V} \]

8. Which of the following is not an application of eddy currents
a) Induction furnace
b) Galvanometer damping
c) Speedometer of automobiles
d) X-ray crystallography
Correct Answer: d) X-ray crystallography
X-ray crystallography uses X-rays to determine the atomic and molecular structure of a crystal, and does not involve eddy currents. The other options all utilize eddy currents: induction furnaces use them for heating, galvanometer damping uses them to stop oscillations, and speedometers use them to measure speed.
9. The efficiency of a transformer is maximum when:
a) Copper losses = iron losses
b) Copper losses > iron losses
c) Copper losses < iron losses
d) Load current is maximum
Correct Answer: a) Copper losses = iron losses
The efficiency of a transformer is maximum when the variable losses (copper losses, which depend on load current) equal the constant losses (iron losses, which occur whenever the transformer is energized). This condition typically occurs at about 50-75% of full load.
10. The working of dynamo is based on principle of
a) Electromagnetic induction
b) Conversion of energy into electricity
c) Magnetic effects of current
d) Heating effects of current
Correct Answer: a) Electromagnetic induction
A dynamo works on the principle of electromagnetic induction, where mechanical energy is used to rotate a coil within a magnetic field, inducing an electromotive force (emf) in the coil according to Faraday's law of induction.
11. The slip of an induction motor under no-load condition is approximately:
a) 0%
b) 2-5%
c) 10-15%
d) 50%
Correct Answer: b) 2-5%
Slip is the difference between synchronous speed and actual rotor speed. Under no-load conditions, an induction motor runs very close to synchronous speed, with slip typically in the range of 2-5%. This small slip is necessary to produce enough torque to overcome mechanical losses.
12. Armature current in dc motor will be maximum when
a) Motor has acquired maximum speed
b) Motor has acquired intermediate speed
c) Motor has just started moving
d) Motor is switched off
Correct Answer: c) Motor has just started moving
At startup, there is no back emf (since the motor isn't rotating), so the armature current is maximum, limited only by the armature resistance. As the motor speeds up, back emf increases, reducing the net voltage across the armature and thus the current.
13. The transformation ratio in the step-up transformer is
a) 1
b) Greater than one
c) Less than one
d) The ratio greater or less than one depends on the other factors
Correct Answer: b) Greater than one
The transformation ratio (K) is defined as Ns/Np (secondary turns/primary turns). For a step-up transformer, the secondary voltage is higher than the primary voltage, so Ns > Np, making K > 1.
14. In a DC generator, the purpose of the commutator is to:
a) Increase the magnetic flux
b) Convert AC to DC
c) Reduce eddy current losses
d) Increase the speed of rotation
Correct Answer: b) Convert AC to DC
The commutator in a DC generator reverses the connections to the external circuit at the instant when the generated emf reverses polarity, effectively converting the alternating current produced in the armature windings into direct current in the external circuit.
15. A transformer is employed to
a) Obtain a suitable dc voltage
b) Convert dc into ac
c) Obtain a suitable ac voltage
d) Convert ac into dc
Correct Answer: c) Obtain a suitable ac voltage
Transformers are used to step up or step down AC voltages to suitable levels. They work only with alternating current (AC) as they rely on electromagnetic induction which requires a changing magnetic field.
16. In an induction coil with resistance, the induced emf will be maximum when
a) The switch is put on due to high resistance
b) The switch is put off due to high resistance
c) The switch is put on due to low resistance
d) The switch is put off due to low resistance
Correct Answer: b) The switch is put off due to high resistance
When the switch is opened (put off), the sudden collapse of the magnetic field induces a large emf. High resistance in the circuit causes the current to change more rapidly when the switch is opened, resulting in a larger induced emf (according to Lenz's law: emf = -L di/dt).
17. The main advantage of a squirrel cage induction motor is:
a) High starting torque
b) Simple and rugged construction
c) Speed control over wide range
d) High power factor
Correct Answer: b) Simple and rugged construction
The squirrel cage induction motor has no brushes, commutator, or slip rings, making it extremely simple and rugged. This construction results in low maintenance requirements and high reliability, which are its main advantages.
18. The primary winding of a transformer has 100 turns and its secondary winding has 200 turns. The primary is connected to an ac supply of 120 V and the current flowing in it is 10 A. The voltage and the current in the secondary are
a) 240 V, 5 A
b) 240 V, 10 A
c) 60 V, 20 A
d) 120 V, 20 A
Correct Answer: a) 240 V, 5 A

For an ideal transformer:

\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = V_p \times \frac{N_s}{N_p} = 120 \times \frac{200}{100} = 240 \text{ V} \]

\[ \frac{I_p}{I_s} = \frac{N_s}{N_p} \Rightarrow I_s = I_p \times \frac{N_p}{N_s} = 10 \times \frac{100}{200} = 5 \text{ A} \]

19. An electric motor operates on a 50 volt supply and a current of 12A. If the efficiency of the motor is 30%, what is the resistance of the winding of the motor
a) 1.25 Ω
b) 2.92 Ω
c) 3.75 Ω
d) 4.17 Ω
Correct Answer: b) 2.92 Ω

Input power = V × I = 50 × 12 = 600 W

Output power = Efficiency × Input power = 0.30 × 600 = 180 W

Power lost as heat = Input - Output = 600 - 180 = 420 W

\[ P = I^2 R \Rightarrow R = \frac{P}{I^2} = \frac{420}{12^2} = \frac{420}{144} ≈ 2.92 \text{ Ω} \]

20. The synchronous speed of a 4-pole induction motor operating at 60 Hz is:
a) 900 rpm
b) 1200 rpm
c) 1800 rpm
d) 3600 rpm
Correct Answer: c) 1800 rpm

Synchronous speed is given by:

\[ N_s = \frac{120 \times f}{P} \]

Where f = frequency (60 Hz), P = number of poles (4)

\[ N_s = \frac{120 \times 60}{4} = 1800 \text{ rpm} \]

21. The core of a transformer is laminated so that
a) Ratio of voltage in the primary and secondary may be increased
b) Rusting of the core may be stopped
c) Energy losses due to eddy currents may be reduced
d) Change in flux is increased
Correct Answer: c) Energy losses due to eddy currents may be reduced
Laminating the core with thin, insulated layers reduces eddy currents by increasing the resistance to their flow. This minimizes energy losses due to heating from these circulating currents, improving the transformer's efficiency.
22. The back emf in a DC motor:
a) Opposes the applied voltage
b) Aids the applied voltage
c) Is independent of the applied voltage
d) Is always greater than the applied voltage
Correct Answer: a) Opposes the applied voltage
Back emf is generated when the motor armature rotates in the magnetic field, and it always opposes the applied voltage (Lenz's law). The net voltage driving current through the armature is the difference between the applied voltage and the back emf.
23. An electric motor operating on a 60 V dc supply draws a current of 10 A. If the efficiency of the motor is 50%, the resistance of its winding is
a) 1.5 Ω
b) 3 Ω
c) 4.5 Ω
d) 6 Ω
Correct Answer: b) 3 Ω

Input power = V × I = 60 × 10 = 600 W

Output power = Efficiency × Input power = 0.50 × 600 = 300 W

Power lost as heat = Input - Output = 600 - 300 = 300 W

\[ P = I^2 R \Rightarrow R = \frac{P}{I^2} = \frac{300}{10^2} = \frac{300}{100} = 3 \text{ Ω} \]

24. The turn ratio of a transformers is given as 2 : 3. If the current through the primary coil is 3 A, thus calculate the current through load resistance
a) 1 A
b) 4.5 A
c) 2 A
d) 1.5 A
Correct Answer: c) 2 A

For a transformer:

\[ \frac{I_s}{I_p} = \frac{N_p}{N_s} \Rightarrow I_s = I_p \times \frac{N_p}{N_s} = 3 \times \frac{2}{3} = 2 \text{ A} \]

25. A transformer with efficiency 80% works at 4 kW and 100 V. If the secondary voltage is 200 V, then the primary and secondary currents are respectively
a) 40 A, 16 A
b) 16 A, 40 A
c) 20 A, 40 A
d) 40 A, 20 A
Correct Answer: a) 40 A, 16 A

Primary current (Ip) = Power / Voltage = 4000 W / 100 V = 40 A

Output power = Efficiency × Input power = 0.8 × 4000 = 3200 W

Secondary current (Is) = Output power / Secondary voltage = 3200 / 200 = 16 A

26. Choke coil works on the principle of
a) Transient current
b) Self induction
c) Mutual induction
d) Wattless current
Correct Answer: b) Self induction
A choke coil works on the principle of self-induction, where a changing current in the coil induces a back emf that opposes the change. This property is used to limit alternating current without significant energy loss.
27. The speed of a DC motor can be controlled by:
a) Varying the armature resistance
b) Varying the field current
c) Adjusting the applied voltage
d) All of the above
Correct Answer: d) All of the above
The speed of a DC motor can be controlled by: (1) Adding resistance in the armature circuit, (2) Adjusting the field current (which changes the magnetic flux), or (3) Directly varying the applied voltage. Each method affects the balance between applied voltage and back emf that determines motor speed.
28. What is increased in step-down transformer
a) Voltage
b) Current
c) Power
d) Current density
Correct Answer: b) Current
In a step-down transformer, the secondary voltage is decreased (stepped down) while the current is increased proportionally, assuming ideal conditions (power in ≈ power out). The actual current increase may be slightly less due to transformer losses.
29. A step-down transformer is connected to 2400 volts line and 80 amperes of current is found to flow in output load. The ratio of the turns in primary and secondary coil is 20 : 1. If transformer efficiency is 100%, then the current flowing in primary coil will be
a) 1600 A
b) 20 A
c) 4 A
d) 1.5 A
Correct Answer: c) 4 A

For a transformer with 100% efficiency:

\[ V_p I_p = V_s I_s \]

Turn ratio = 20:1, so voltage ratio = 20:1 (step-down)

\[ V_s = \frac{V_p}{20} = \frac{2400}{20} = 120 \text{ V} \]

Output power = Vs × Is = 120 × 80 = 9600 W

\[ I_p = \frac{\text{Power}}{V_p} = \frac{9600}{2400} = 4 \text{ A} \]

Alternatively, current ratio is inverse of turns ratio: Ip/Is = 1/20 ⇒ Ip = 80/20 = 4 A

30. An ideal transformer has 100 turns in the primary and 250 turns in the secondary. The peak value of the ac is 28 V. The r.m.s. secondary voltage is nearest to
a) 50 V
b) 70 V
c) 100 V
d) 40 V
Correct Answer: a) 50 V

Primary RMS voltage = Peak voltage / √2 = 28 / 1.414 ≈ 19.8 V

Secondary RMS voltage:

\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = V_p \times \frac{N_s}{N_p} = 19.8 \times \frac{250}{100} = 19.8 \times 2.5 ≈ 49.5 \text{ V} \]

Nearest to 50 V

31. The main difference between a power transformer and an autotransformer is:
a) Number of windings
b) Core material
c) Efficiency
d) Voltage rating
Correct Answer: a) Number of windings
A conventional power transformer has separate primary and secondary windings, while an autotransformer has a single winding with a tap that serves as both primary and secondary. This makes autotransformers more compact and efficient for small voltage ratios.
32. A transformer connected to 220 volt line shows an output of 2 A at 11000 volt. The efficiency is 100%. The current drawn from the line is
a) 100 A
b) 200 A
c) 22 A
d) 11 A
Correct Answer: a) 100 A

For 100% efficiency:

\[ V_p I_p = V_s I_s \]

\[ I_p = \frac{V_s I_s}{V_p} = \frac{11000 \times 2}{220} = \frac{22000}{220} = 100 \text{ A} \]

33. The rotor of a 3-phase induction motor rotates in the same direction as the rotating magnetic field because:
a) Of Lenz's law
b) The rotor is magnetically locked to the field
c) Current is induced in the rotor bars
d) All of the above
Correct Answer: d) All of the above
The rotor follows the rotating magnetic field due to: (1) Lenz's law (the induced current opposes the change causing it), (2) The interaction between rotor currents and stator field creates torque, and (3) The rotor bars develop currents that interact with the stator field to produce rotation.
34. A 100% efficient transformer has 100 turns in the primary and 25 turns in its secondary coil. If the current in the secondary coil is 4 amp, then the current in the primary coil is
a) 1 amp
b) 4 amp
c) 8 amp
d) 16 amp
Correct Answer: a) 1 amp

Current ratio is inverse of turns ratio:

\[ \frac{I_p}{I_s} = \frac{N_s}{N_p} \Rightarrow I_p = I_s \times \frac{N_s}{N_p} = 4 \times \frac{25}{100} = 4 \times 0.25 = 1 \text{ A} \]

35. The device that does not work on the principle of mutual induction is
a) Induction coil
b) Motor
c) Tesla coil
d) Transformer
Correct Answer: b) Motor
While motors use electromagnetic principles, they primarily work on the force experienced by a current-carrying conductor in a magnetic field, not mutual induction. The other devices (induction coil, Tesla coil, transformer) all rely on mutual induction between windings.
36. In a lossless transformer an alternating current of 2 amp is flowing in the primary coil. The number of turns in the primary and secondary coils are 100 and 20 respectively. The value of the current in the secondary coil is
a) 0.08 A
b) 0.4 A
c) 5 A
d) 10 A
Correct Answer: d) 10 A

For a lossless transformer:

\[ \frac{I_s}{I_p} = \frac{N_p}{N_s} \Rightarrow I_s = I_p \times \frac{N_p}{N_s} = 2 \times \frac{100}{20} = 2 \times 5 = 10 \text{ A} \]

37. A step-up transformer has transformation ratio of 3 : 2. What is the voltage in secondary if voltage in primary is 30 V
a) 45 V
b) 15 V
c) 90 V
d) 300 V
Correct Answer: a) 45 V

Transformation ratio (K) = Ns/Np = 3/2

\[ \frac{V_s}{V_p} = K \Rightarrow V_s = V_p \times K = 30 \times \frac{3}{2} = 45 \text{ V} \]

38. In a transformer, the number of turns in primary coil and secondary coil are 5 and 4 respectively. If 240 V is applied on the primary coil, then the ratio of current in primary and secondary coil is
a) 4 : 5
b) 5 : 4
c) 5 : 9
d) 9 : 5
Correct Answer: a) 4 : 5

For a transformer:

\[ \frac{I_p}{I_s} = \frac{N_s}{N_p} = \frac{4}{5} \]

Therefore, the current ratio Ip : Is = 4 : 5

39. In a transformer the primary has 500 turns and secondary has 50 turns. 100 volts are applied to the primary coil, the voltage developed in the secondary will be
a) 1 V
b) 10 V
c) 1000 V
d) 10000 V
Correct Answer: b) 10 V

\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = V_p \times \frac{N_s}{N_p} = 100 \times \frac{50}{500} = 100 \times 0.1 = 10 \text{ V} \]

40. The main advantage of a wound rotor induction motor over a squirrel cage motor is:
a) Higher efficiency
b) Better speed control
c) Lower cost
d) Simpler construction
Correct Answer: b) Better speed control
Wound rotor induction motors allow external resistance to be added to the rotor circuit through slip rings, enabling better control of starting torque and speed regulation compared to squirrel cage motors, which have fixed rotor characteristics.
41. A transformer is used to
a) Change the alternating potential
b) Change the alternating current
c) To prevent the power loss in alternating current flow
d) To increase the power of current source
Correct Answer: a) Change the alternating potential
A transformer's primary function is to change (step up or step down) alternating voltage levels. While it also changes current inversely with voltage (assuming ideal conditions), its main purpose is voltage transformation.
42. A transformer is employed to reduce 220 V to 11 V. The primary draws a current of 5 A and the secondary 90 A. The efficiency of the transformer is
a) 20%
b) 40%
c) 70%
d) 90%
Correct Answer: d) 90%

Input power = Vp × Ip = 220 × 5 = 1100 W

Output power = Vs × Is = 11 × 90 = 990 W

\[ \text{Efficiency} = \frac{\text{Output}}{\text{Input}} \times 100 = \frac{990}{1100} \times 100 = 90\% \]

43. A step-up transformer operates on a 230 V line and supplies a load of 2 ampere. The ratio of the primary and secondary windings is 1 : 25. The current in the primary is
a) 15 A
b) 50 A
c) 25 A
d) 12.5 A
Correct Answer: b) 50 A

Current ratio is inverse of turns ratio:

\[ \frac{I_p}{I_s} = \frac{N_s}{N_p} = \frac{25}{1} \]

\[ I_p = I_s \times \frac{N_s}{N_p} = 2 \times 25 = 50 \text{ A} \]

44. The primary winding of transformer has 500 turns whereas its secondary has 5000 turns. The primary is connected to an ac supply of 20 V, 50 Hz. The secondary will have an output of
a) 200 V, 50 Hz
b) 2 V, 50 Hz
c) 200 V, 500 Hz
d) 2 V, 5 Hz
Correct Answer: a) 200 V, 50 Hz

\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = V_p \times \frac{N_s}{N_p} = 20 \times \frac{5000}{500} = 20 \times 10 = 200 \text{ V} \]

The frequency remains unchanged at 50 Hz. Transformers only change voltage and current, not frequency.

45. Quantity that remains unchanged in a transformer is
a) Voltage
b) Current
c) Frequency
d) None of the above
Correct Answer: c) Frequency
A transformer changes voltage and current levels but maintains the same frequency in both primary and secondary windings. The frequency of the output AC is identical to the input AC.
46. The number of turns in primary and secondary coils of a transformer are 100 and 20 respectively. If an alternating potential of 200 volt is applied to the primary, the induced potential in secondary will be
a) 10 V
b) 40 V
c) 1000 V
d) 20,000 V
Correct Answer: b) 40 V

\[ \frac{V_s}{V_p} = \frac{N_s}{N_p} \Rightarrow V_s = V_p \times \frac{N_s}{N_p} = 200 \times \frac{20}{100} = 200 \times 0.2 = 40 \text{ V} \]

47. The main reason for using high voltage in long-distance power transmission is:
a) To increase the current
b) To reduce power losses
c) To decrease the insulation requirements
d) To make transformers more efficient
Correct Answer: b) To reduce power losses
Power loss in transmission lines is proportional to the square of the current (P = I²R). By increasing voltage, the current for a given power level is reduced, thereby dramatically decreasing I²R losses in the transmission lines.
48. In a step-up transformer the turn ratio is 1:10. A resistance of 200 ohm connected across the secondary is drawing a current of 0.5 A. What is the primary voltage and current
a) 50 V, 1 amp
b) 10 V, 5 amp
c) 25 V, 4 amp
d) 20 V, 2 amp
Correct Answer: b) 10 V, 5 amp

Secondary voltage (Vs) = Is × R = 0.5 × 200 = 100 V

Primary voltage (Vp):

\[ \frac{V_p}{V_s} = \frac{N_p}{N_s} \Rightarrow V_p = V_s \times \frac{N_p}{N_s} = 100 \times \frac{1}{10} = 10 \text{ V} \]

Primary current (Ip):

\[ \frac{I_p}{I_s} = \frac{N_s}{N_p} \Rightarrow I_p = I_s \times \frac{N_s}{N_p} = 0.5 \times 10 = 5 \text{ A} \]

49. In a transformer, number of turns in the primary are 140 and that in the secondary are 280. If current in primary is 4A then that in the secondary is
a) 4 A
b) 2 A
c) 6 A
d) 10 A
Correct Answer: b) 2 A

\[ \frac{I_s}{I_p} = \frac{N_p}{N_s} \Rightarrow I_s = I_p \times \frac{N_p}{N_s} = 4 \times \frac{140}{280} = 4 \times 0.5 = 2 \text{ A} \]

50. In a primary coil 5A current is flowing on 220 volts. In the secondary coil 2200V voltage produces. Then ratio of number of turns in secondary coil and primary coil will be
a) 1 : 10
b) 10 : 1
c) 1 : 1
d) 11 : 1
Correct Answer: b) 10 : 1

\[ \frac{N_s}{N_p} = \frac{V_s}{V_p} = \frac{2200}{220} = 10 \]

Therefore, the turns ratio Ns : Np = 10 : 1

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