Capacitance And Capacitor Quiz Questions With Answers
What are capacitors and capacitance? Can you distinguish between the two? Here, we have a few capacitance and capacitor quiz questions with answers to evaluate your knowledge about electronic devices. A capacitor is an electronic device that stores an electrical charge and electric energy in the form of an electric field. The strong ability of a capacitor to store energy is known as capacitance. The quiz below consists of a few MCQ questions. Do you think you can answer them all? Try it now.
- 1.
Capacitance acts to store electrical energy as
- A.
Current
- B.
Voltage
- C.
A magnetic field
- D.
An electric field
Correct Answer
D. An electric fieldExplanation
Capacitance acts to store electrical energy as an electric field. When a voltage is applied across a capacitor, it creates an electric field between its plates. This electric field stores the electrical energy in the form of potential energy. The amount of energy stored is directly proportional to the capacitance value of the capacitor. Therefore, the correct answer is an electric field.Rate this question:
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- 2.
As capacitor plate area increases, all other things being equal,
- A.
The capacitance increases
- B.
The capacitance decreases
- C.
The capacitance does not change
- D.
The current-handling ability decreases
Correct Answer
A. The capacitance increasesExplanation
As the capacitor plate area increases, all other things being equal, the capacitance increases. This is because the capacitance of a capacitor is directly proportional to the area of its plates. A larger plate area provides more surface area for the electric field to interact with, resulting in a higher capacitance.Rate this question:
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- 3.
As the spacing between plates in a capacitor is made smaller, all other things being equal,
- A.
The capacitance increases
- B.
The capacitance decreases
- C.
The capacitance does not change
- D.
The resistance increases
Correct Answer
A. The capacitance increasesExplanation
The capacitance of a capacitor is directly proportional to the area of the plates and inversely proportional to the distance between them. As the spacing between plates in a capacitor is made smaller, the distance between the plates decreases, resulting in an increase in capacitance. This is because the closer the plates are, the stronger the electric field between them, allowing for more charge to be stored on the plates. Therefore, the capacitance increases.Rate this question:
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- 4.
A material with a high dielectric constant
- A.
Acts to increase capacitance per unit volume
- B.
Acts to decrease capacitance per unit volume
- C.
Has no effect on capacitance
- D.
Causes a capacitor to become polarized
Correct Answer
A. Acts to increase capacitance per unit volumeExplanation
A material with a high dielectric constant acts to increase capacitance per unit volume because it increases the ability of the material to store electrical energy in an electric field. The dielectric constant represents the material's ability to polarize in response to an applied electric field. When a high dielectric constant material is placed between the plates of a capacitor, it enhances the electric field, resulting in a higher capacitance value. This is because the increased polarization of the material allows for more charge to be stored per unit volume, increasing the capacitance.Rate this question:
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- 5.
A capacitance of 100 pF is the same as which of the following?
- A.
0.01 µF
- B.
0.001 µF
- C.
0.0001 µF
- D.
0.00001 µF
Correct Answer
C. 0.0001 µFExplanation
A capacitance of 100 pF is equivalent to 0.0001 µF because there are 1000 picofarads (pF) in 1 nanofarad (nF), and there are 1000 nanofarads (nF) in 1 microfarad (µF). Therefore, to convert from pF to µF, we divide by 1000 twice, resulting in 0.0001 µF.Rate this question:
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- 6.
A capacitance of 0.033 µF is the same as which of the following?
- A.
33 pF
- B.
330 pF
- C.
3300 pF
- D.
33,000 pF
Correct Answer
D. 33,000 pFExplanation
The given capacitance of 0.033 µF can be converted to picofarads (pF) by multiplying it by 1,000,000. Therefore, 0.033 µF is equal to 33,000 pF.Rate this question:
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- 7.
If five 0.050-µF capacitors are connected in parallel, what is the net capacitance of the combination?
- A.
0.010 µF
- B.
0.25 µF
- C.
0.50 µF
- D.
0.025 µF
Correct Answer
B. 0.25 µFExplanation
When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances. Therefore, when five 0.050-µF capacitors are connected in parallel, the net capacitance of the combination is 0.050 µF + 0.050 µF + 0.050 µF + 0.050 µF + 0.050 µF = 0.25 µF.Rate this question:
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- 8.
If five 0.050-µF capacitors are connected in series, what is the net capacitance of the combination?
- A.
0.010 µF
- B.
0.25 µF
- C.
0.50 µF
- D.
0.025 µF
Correct Answer
A. 0.010 µFExplanation
When capacitors are connected in series, the reciprocal of the net capacitance is equal to the sum of the reciprocals of the individual capacitances. In this case, the reciprocal of the net capacitance is equal to the sum of the reciprocals of the five 0.050 μF capacitors. Calculating this, we get:
1/net capacitance = 1/0.050 + 1/0.050 + 1/0.050 + 1/0.050 + 1/0.050
1/net capacitance = 20 + 20 + 20 + 20 + 20
1/net capacitance = 100
Taking the reciprocal of both sides, we find that the net capacitance is 1/100, which simplifies to 0.010 μF. Therefore, the correct answer is 0.010 μF.Rate this question:
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- 9.
Two capacitors, C1 and C2, are connected in series, with capacitances of 4 microfarads and 6 microfarads, respectively. What is the equivalent capacitance of the combination?
- A.
10 microfarads
- B.
2.4 microfarads
- C.
5 microfarads
- D.
3.75 microfarads
Correct Answer
B. 2.4 microfarads -
- 10.
Suppose that two capacitors are in parallel. Their values are 47.0 pF and 470 µF. What is thenet capacitance of this combination?
- A.
47.0 pF
- B.
517 pF
- C.
517 µF
- D.
470 µF
Correct Answer
D. 470 µFExplanation
When capacitors are connected in parallel, the total capacitance is the sum of the individual capacitances. In this case, the two capacitors have values of 47.0 pF and 470 µF. Since the values are in different units, they need to be converted to the same unit before adding. Converting 47.0 pF to µF, we get 0.047 µF. Adding this to the value of 470 µF gives a total capacitance of 470.047 µF, which is closest to 470 µF. Therefore, the correct answer is 470 µF.Rate this question:
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- 11.
Suppose that three capacitors are in parallel. Their values are 0.0200 µF, 0.0500 µF, and0.10000 µF. What is the net capacitance of this combination?
- A.
0.0125 µF
- B.
0.1700 µF
- C.
0.1000 µF
- D.
0.1250 µF
Correct Answer
B. 0.1700 µFExplanation
When capacitors are connected in parallel, the total capacitance is equal to the sum of the individual capacitances. In this case, the net capacitance is equal to 0.0200 µF + 0.0500 µF + 0.1000 µF = 0.1700 µF.Rate this question:
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- 12.
The main advantage of air as a dielectric material for capacitors is the fact that it
- A.
Has a high dielectric constant
- B.
Is not physically dense
- C.
Has low loss
- D.
Allows for large capacitance in a small volume
Correct Answer
C. Has low lossExplanation
Air has low loss as a dielectric material for capacitors. This means that it has a very low amount of energy loss when used as a dielectric, resulting in high efficiency. This is advantageous because it allows for better performance and less energy wastage in the capacitor.Rate this question:
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- 13.
Which of the following is not a characteristic of mica capacitors?
- A.
Excellent efficiency
- B.
Small size, even for large values of capacitance
- C.
High voltage-handling capacity
- D.
Low loss
Correct Answer
B. Small size, even for large values of capacitanceExplanation
Mica capacitors are known for their excellent efficiency, high voltage-handling capacity, and low loss. However, they are not characterized by small size, especially when it comes to large values of capacitance. Mica capacitors tend to be larger in size compared to other types of capacitors, making them less suitable for applications where space is limited.Rate this question:
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- 14.
Which of the following capacitance values is most typical of a disk-ceramic capacitor?(a) 100 pF
- A.
100 pF
- B.
33 µF
- C.
470 µF
- D.
10,000 µF
Correct Answer
A. 100 pFExplanation
A disk-ceramic capacitor is a type of capacitor that is commonly used in electronic circuits for its small size and high capacitance value. It is typically used for filtering and decoupling applications. The most typical capacitance value for a disk-ceramic capacitor is 100 pF, as it provides a good balance between size and capacitance. Capacitance values of 33 µF, 470 µF, and 10,000 µF are much larger and would be more suitable for applications that require higher capacitance values.Rate this question:
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- 15.
Which of the following capacitance values is most typical of a paper capacitor?
- A.
0.001 pF
- B.
0.01 µF
- C.
100 µF
- D.
3300 µF
Correct Answer
B. 0.01 µFExplanation
A paper capacitor is a type of capacitor that uses paper as the dielectric material. Paper capacitors are commonly used in low voltage and low frequency applications. The most typical capacitance value for a paper capacitor is 0.01 µF, which is also known as 10,000 pF. This value is suitable for filtering and coupling applications in electronic circuits. Capacitance values of 0.001 pF and 100 µF are not typical for paper capacitors, as they are either too small or too large for this type of capacitor. 3300 µF is a very large capacitance value and is typically used in applications that require high energy storage, such as power supply filtering.Rate this question:
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- 16.
Which of the following capacitance ranges is most typical of an air-variable capacitor?
- A.
0.01 µF to 1 µF
- B.
1 µF to 100 µF
- C.
1 pF to 100 pF
- D.
0.001 pF to 0.1 pF
Correct Answer
C. 1 pF to 100 pFExplanation
An air-variable capacitor is a type of capacitor that uses air as the dielectric material. Air-variable capacitors typically have capacitance values in the range of 1 pF to 100 pF. This range is considered most typical because it covers a wide range of values that are commonly used in electronic circuits. Capacitance values in the range of 1 pF to 100 pF are suitable for applications such as tuning circuits, filters, and impedance matching.Rate this question:
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- 17.
Which of the following types of capacitors is polarized?
- A.
Paper
- B.
Mica
- C.
Interelectrode
- D.
Electrolytic
Correct Answer
D. ElectrolyticExplanation
Electrolytic capacitors are polarized because they have a positive and negative terminal, meaning they can only be connected in a specific direction. This is due to the presence of an electrolyte inside the capacitor, which allows for the flow of current in one direction. In contrast, paper, mica, and interelectrode capacitors are non-polarized and can be connected in either direction.Rate this question:
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- 18.
If a capacitor has a negative temperature coefficient, then
- A.
Its capacitance decreases as the temperature rises
- B.
Its capacitance increases as the temperature rises.
- C.
Its capacitance does not change with temperature.
- D.
It will not work if the temperature is below freezing.
Correct Answer
A. Its capacitance decreases as the temperature risesExplanation
A capacitor with a negative temperature coefficient means that its capacitance decreases as the temperature rises. This is because as the temperature increases, the molecules in the dielectric material of the capacitor gain more energy and move more rapidly. This increased movement disrupts the alignment of the molecules, reducing the overall capacitance of the capacitor. Therefore, as the temperature rises, the capacitance of the capacitor decreases.Rate this question:
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- 19.
Suppose that a capacitor is rated at 33 pF 10%. Which of the following actual capacitancevalues is outside the acceptable range?
- A.
30 pF
- B.
37 pF
- C.
35 pF
- D.
31 pF
Correct Answer
B. 37 pFExplanation
The given capacitor is rated at 33 pF with a tolerance of 10%. This means that the actual capacitance can vary by 10% of 33 pF, which is 3.3 pF. Therefore, the acceptable range for the actual capacitance is between 29.7 pF (33 pF - 3.3 pF) and 36.3 pF (33 pF + 3.3 pF). Among the given options, 37 pF is outside this range, making it the answer.Rate this question:
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- 20.
Suppose that a capacitor, rated at 330 pF, shows an actual value of 317 pF. By how manypercent does its actual capacitance differ from its rated capacitance?
- A.
−0.039%
- B.
−3.9%
- C.
−0.041%
- D.
−4.1%
Correct Answer
B. −3.9%Explanation
The actual capacitance of the capacitor is 317 pF, which is 13 pF less than the rated capacitance of 330 pF. To find the percentage difference, we divide the difference by the rated capacitance and multiply by 100. So, (13/330) * 100 = 3.9%. Therefore, the actual capacitance differs from the rated capacitance by -3.9%.Rate this question:
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Aug 30, 2024Quiz Edited by
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Dec 13, 2010Quiz Created by
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