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Gibilisco: MCQ in Inductance | ECE Board Exam

Gibilisco: MCQ in Inductance | ECE Board Exam

This is the Multiple Choice Questions (MCQs) in Chapter 10: Inductance from the book Teach Yourself Electricity and Electronics, 5th edition by Stan Gibilisco. If you are looking for a reviewer in Electronics Engineering this will definitely help you before taking the Board Exam.

Start Practice Exam Test Questions

Gibilisco: MCQ in Alternating Current Basic | ECE Board Exam

Choose the letter of the best answer in each questions.

1. An inductor works by:

A. Charging a piece of wire.

B. Storing energy as a magnetic field.

C. Choking off high-frequency ac.

D. Introducing resistance into a circuit.

View Answer:

Answer: Option B

Solution:

2. Which of the following does not affect the inductance of a coil?

A. The diameter of the wire.

B. The number of turns.

C. The type of core material.

D. The length of the coil.

View Answer:

Answer: Option A

Solution:

3. In a small inductance:

A. Energy is stored and released slowly.

B. The current flow is always large.

C. The current flow is always small.

D. Energy is stored and released quickly.

View Answer:

Answer: Option D

Solution:

4. A ferromagnetic core is placed in an inductor mainly to:

A. Increase the current carrying capacity.

B. Increase the inductance.

C. Limit the current.

D. Reduce the inductance.

View Answer:

Answer: Option B

Solution:

5. Inductors in series, assuming there is no mutual inductance, combine:

A. Like resistors in parallel.

B. Like resistors in series.

C. Like batteries in series with opposite polarities.

D. In a way unlike any other type of component.

View Answer:

Answer: Option B

Solution:

6. Two inductors are connected in series, without mutual inductance. Their values are 33 mH and 55 mH. The net inductance of the combination is:

A. 1.8 H.

B. 22 mH.

C. 88 mH.

D. 21 mH.

View Answer:

Answer: Option C

Solution:

7. If the same two inductors (33 mH and 55 mH) are connected in parallel without mutual inductance, the combination will have a value of:

A. 1.8 H.

B. 22 mH.

C. 88 mH.

D. 21 mH.

View Answer:

Answer: Option D

Solution:

8. Three inductors are connected in series without mutual inductance. Their values are 4 nH, 140 μH, and 5 H. For practical purposes, the net inductance will be very close to:

A. 4 nH.

B. 140 μH.

C. 5 H.

D. None of these.

View Answer:

Answer: Option C

Solution:

9. Suppose the three inductors mentioned above are connected in parallel without mutual inductance. The net inductance will be close to:

A. 4 nH.

B. 140 μH.

C. 5 H.

D. None of these.

View Answer:

Answer: Option A

Solution:

10. Two inductors, each of 100 μH, are in series. The coefficient of coupling is 0.40. The net inductance, if the coil fields reinforce each other, is:

A. 50 μH.

B. 120 μH.

C. 200 μH.

D. 280 μH.

View Answer:

Answer: Option D

Solution:

11. If the coil fields oppose in the foregoing series-connected arrangement, the net inductance is:

A. 50 μH.

B. 120 μH.

C. 200 μH.

D. 280 μH.

View Answer:

Answer: Option B

Solution:

12. Two inductors, having values of 44 mH and 88 mH, are connected in series with a coefficient of coupling equal to 1.0 (maximum possible mutual inductance). If their fields reinforce, the net inductance (to two significant digits) is:

A. 7.5 mH.

B. 132 mH.

C. 190 mH.

D. 260 mH.

View Answer:

Answer: Option D

Solution:

13. If the fields in the previous situation oppose, the net inductance will be:

A. 7.5 mH.

B. 132 mH.

C. 190 mH.

D. 260 mH.

View Answer:

Answer: Option A

Solution:

14. With permeability tuning, moving the core further into a solenoidal coil:

A. Increases the inductance.

B. Reduces the inductance

C. Has no effect on the inductance, but increases the current-carrying capacity of the coil.

D. Raises the frequency.

View Answer:

Answer: Option A

Solution:

15. A significant advantage, in some situations, of a toroidal coil over a solenoid is:

A. The toroid is easier to wind.

B. The solenoid cannot carry as much current.

C. The toroid is easier to tune.

D. The magnetic flux in a toroid is practically all within the core.

View Answer:

Answer: Option D

Solution:

16. A major feature of a pot-core winding is:

A. High current capacity.

B. Large inductance in small volume.

C. Efficiency at very high frequencies.

D. Ease of inductance adjustment.

View Answer:

Answer: Option B

Solution:

17. As an inductor core material, air:

A. Has excellent efficiency.

B. Has high permeability.

C. Allows large inductance in a small volume.

D. Has permeability that can vary over a wide range.

View Answer:

Answer: Option A

Solution:

18. At a frequency of 400 Hz, the most likely form for an inductor would be:

A. Air-core.

B. Solenoidal.

C. Toroidal.

D. Transmission-line.

View Answer:

Answer: Option C

Solution:

19. At a frequency of 95 MHz, the best form for an inductor would be:

A. Air-core.

B. Pot core.

C. Either of the above.

D. Neither of the above.

View Answer:

Answer: Option A

Solution:

20. A transmission-line inductor made from coaxial cable, having velocity factor of 0.66, and working at 450 MHz, would be shorter than:

A. 16.7 m.

B. 11 m.

C. 16.7 cm.

D. 11 cm.

View Answer:

Answer: Option D

Solution:

Topics Included in the Test from Chapter 9 to Chapter 18

Part 2: Alternating Current
Chapter 10: MCQ in Inductance
Chapter 11: MCQ in Capacitance
Chapter 12: MCQ in Phase

Complete List of Multiple Choice Questions from this Book

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