# Gibilisco: MCQ in RLC and GLC Circuit Analysis | ECE Board Exam

(Last Updated On: April 26, 2022) This is the Multiple Choice Questions (MCQs) in Chapter 16: RLC and GLC Circuit Analysis 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

Choose the letter of the best answer in each questions.

1. A coil and capacitor are connected in series. The inductive reactance is 250 Ω, and the capacitive reactance is -300 Ω. What is the net impedance vector, R + jX?

A. 0 + j550.

B. 0 – j50.

C. 250 – j300

D. -300 – j250.

Solution:

2. A coil of 25.0 μH and capacitor of 100 pF are connected in series. The frequency is 5.00 MHz. What is the impedance vector, R + jX?

A 0 + j467.

B. 25 + j100.

C. 0 – j467.

D. 25 – j100.

Solution:

3. When R = 0 in a series RLC circuit, but the net reactance is not zero, the impedance vector:

A. Always points straight up.

B. Always points straight down.

C. Always points straight towards the right.

D. None of the above.

Solution:

4. A resistor of 150 Ω, a coil with reactance 100 Ω and a capacitor with reactance -200 Ω are connected in series. What is the complex impedance R + jX?

A. 150 + j100.

B. 150 – j200.

C. 100 – j200.

D. 150 – j100.

Solution:

5. A resistor of 330 Ω, a coil of 1.00 μH and a capacitor of 200 pF are in series. What is R + jX at 10.0 MHz?

A. 330 – j199.

B. 300 + j201.

C. 300 + j142.

D. 330 – j16.8.

Solution:

6. A coil has an inductance of 3.00 μH and a resistance of 10.0 Ω in its winding. A capacitor of 100 pF is in series with this coil. What is R + jX at 10.0 MHz?

A. 10 + j3.00.

B. 10 + j29.2.

C. 10 – j97.

D. 10 + j348.

Solution:

7. A coil has a reactance of 4.00 Ω. What is the admittance vector, G + jB, assuming nothing else is in the circuit?

A. 0 + j0.25.

B. 0 + j4.00.

C. 0 – j0.25.

D. 0 + j4.00.

Solution:

8. What will happen to the susceptance of a capacitor if the frequency is doubled, all other things being equal?

A. It will decrease to half its former value.

B. It will not change.

C. It will double.

Solution:

9. A coil and capacitor are in parallel, with jBL = –j0.05 and jBC = j0.03. What is the admittance vector, assuming that nothing is in series or parallel with these components?

A. 0 – j0.02.

B. 0 – j0.07.

C. 0 + j0.02.

D. -0.05 – j0.03.

Solution:

10. A coil, resistor, and capacitor are in parallel. The resistance is 1 Ω ; the capacitive susceptance is 1.0 siemens; the inductive susceptance is -1.0 siemens. Then the frequency is cut to half its former value. What will be the admittance vector, G + jB, at the new frequency?

A. 1 + j0.

B. 1 + jl.5.

C. 1 – jl.5.

D. 1 – j2.

Solution:

11. A coil of 3.50 μH and a capacitor of 47.0 pF are in parallel. The frequency is 9.55 MHz. There is nothing else in series or parallel with these components. What is the admittance vector?

A. 0 + j0.00282.

B. 0 – j0.00194.

C. 0 + j0.00194.

D. 0 – j0.00758.

Solution:

12. A vector pointing “southeast” in the GB plane would indicate the following:

A. Pure conductance, zero susceptance.

B. Conductance and inductive susceptance.

C. Conductance and capacitive susceptance.

D. Pure susceptance, zero conductance.

Solution:

13. A resistor of 0.0044 siemens, a capacitor whose susceptance is 0.035 Siemens, and a coil whose susceptance is -0.011 siemens are all connected in parallel. The admittance vector is:

A. 0.0044 + j0.024.

B. 0.035 – j0.011.

C. -0.011 – j0.035.

D. 0.0044 + j0.046.

Solution:

14. A resistor of 100 Ω, a coil of 4.50 μH, and a capacitor of 220 pF are in parallel. What is the admittance vector at 6.50 MHz?

A. 100 + j0.00354.

B. 0.010 + j0.00354.

C. 100 – j0.0144.

D. 0.010 + j0.0144.

Solution:

15. The admittance for a circuit, G + jB, is 0.02 + j0.20. What is the impedance, R + jX?

A. 50 + j5.0.

B. 0.495 – j4.95.

C. 50 – j5.0.

D. 0.495 + j4.95.

Solution:

16. A resistor of 51.0 Ω an inductor of 22.0 μH and a capacitor of 150 pF are in parallel. The frequency is 1.00 MHz. What is the complex impedance, R + jX?

A. 51.0 – j14.9.

B. 51.0 + j14.9.

C. 46.2 – j14.9.

D. 46.2 + j14.9.

Solution:

17. A series circuit has 99.0 Ω of resistance and 88.0 Ω of inductive reactance. An ac RMS voltage of 117 V is applied to this series network. What is the current?

A. 1.18 A.

B. 1.13 A.

C. 0.886 A.

D. 0.846 A.

Solution:

18. What is the voltage across the reactance in the above example?

A. 78.0 V.

B. 55.1 V.

C. 99.4 V.

D. 74.4 V.

Solution:

19. A parallel circuit has 10 ohms of resistance and 15 Ω of reactance. An ac RMS voltage of 20 V is applied across it. What is the total current?

A. 2.00 A.

B. 2.40 A.

C. 1.33 A.

D. 0.800 A.

Solution:

20. What is the current through the resistance in the above example?

A. 2.00 A.

B. 2.40 A.

C. 1.33 A.

D. 0.800 A.

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

PinoyBIX Engineering. © 2014-2021 All Rights Reserved | How to Donate? |  