Gibilisco Inductance Quiz
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An inductor works by
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Charging a piece of wire.
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Storing energy as a magnetic field.
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Choking off dc.
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Introducing resistance into a circuit.
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Simply stated, an electrical conductor's propensity to resist a change in the flow of current through it is known as inductance. Take this quiz by Gibilisco that covers relevant parts of the topic. The quiz is exam-oriented and would be of immense help for those seeking a last-minute revision of the topic. The quiz is framed in such a way as to ensure you gain maximum knowledge in less time. All the best!
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- 2.
Suppose two inductors are connected in series, without mutual inductance. Their values are33 mH and 55 mH. What is the net inductance of the combination?
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1.8 H
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22 mH
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88 mH
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21 mH
Correct Answer
A. 88 mHExplanation
When inductors are connected in series, the net inductance of the combination is equal to the sum of the individual inductances. In this case, the two inductors have values of 33 mH and 55 mH. Adding these values together gives a total of 88 mH, which is the net inductance of the combination.Rate this question:
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- 3.
Inductors in series, assuming there is no mutual inductance, combine
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Like resistors in parallel
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Like resistors in series
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Like batteries in series with opposite polarities
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In a way unlike any other type of component
Correct Answer
A. Like resistors in seriesExplanation
When inductors are connected in series, they combine like resistors in series. This means that their total inductance is equal to the sum of their individual inductances. In series, the total inductance is increased compared to a single inductor. This is similar to how resistors in series add up their resistances, resulting in a higher total resistance. Therefore, the behavior of inductors in series is analogous to resistors in series.Rate this question:
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- 4.
At a frequency of 95.7 MHz, which is in the frequency-modulation (FM) broadcast band and is considered part of the very high frequency (VHF) radio spectrum, a good form for an inductor would be
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Air-core
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Pot core
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Either air-core or pot core
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Neither air-core nor pot core
Correct Answer
A. Air-coreExplanation
An air-core inductor would be a good form at a frequency of 95.7 MHz because it offers low losses and high Q-factor at high frequencies. Air-core inductors do not have a magnetic core material, which reduces the risk of core saturation and hysteresis losses. This makes them suitable for high-frequency applications like FM radio broadcasting. Pot core inductors, on the other hand, have a magnetic core material which can introduce losses and affect the performance at higher frequencies. Therefore, an air-core inductor would be the preferable choice in this scenario.Rate this question:
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- 5.
Which of the following does not affect the inductance of an air-core coil, if all other factors are held constant?
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The frequency
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The number of turns
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The diameter of the coil
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The length of the coil
Correct Answer
A. The frequencyExplanation
The frequency does not affect the inductance of an air-core coil because inductance is solely determined by the physical characteristics of the coil, such as the number of turns, diameter, and length. The frequency only affects the impedance of the coil, not the inductance.Rate this question:
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- 6.
Suppose three inductors are connected in series without mutual inductance. Their values are 4.00 nH, 140 µH, and 5.07 H. For practical purposes, the net inductance will be very close to
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4.00 nH
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140 µH
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5.07 H
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None of the above
Correct Answer
A. 5.07 HExplanation
When inductors are connected in series, the total inductance is equal to the sum of their individual inductances. In this case, the inductors have values of 4.00 nH, 140 µH, and 5.07 H. Since the inductance values are given in different units, we need to convert them to a common unit before adding them. Converting 4.00 nH to H gives 4.00 x 10^-9 H, and converting 140 µH to H gives 140 x 10^-6 H. Adding these values together gives a total inductance of 5.07 H. Therefore, the net inductance will be very close to 5.07 H.Rate this question:
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- 7.
A significant advantage, in some situations, of a toroidal coil over a solenoid is the fact that
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The toroid is easier to wind
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The solenoid cannot carry as much current
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The toroid is easier to tune
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The magnetic flux in a toroid is practically all within the core
Correct Answer
A. The magnetic flux in a toroid is practically all within the coreExplanation
The magnetic flux in a toroid is practically all within the core. This means that the magnetic field produced by the toroidal coil is concentrated within the core, resulting in a stronger and more efficient magnetic field compared to a solenoid. In a solenoid, the magnetic field is spread out along the length of the coil, leading to some loss of magnetic flux. Therefore, in situations where a concentrated and strong magnetic field is required, a toroidal coil is advantageous over a solenoid.Rate this question:
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- 8.
A ferromagnetic core is placed in an inductor mainly to
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Increase the current carrying capacity
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Increase the inductance
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Limit the current
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Reduce the inductance
Correct Answer
A. Increase the inductanceExplanation
A ferromagnetic core is placed in an inductor mainly to increase the inductance. This is because the ferromagnetic material has a high permeability, which allows it to concentrate the magnetic field lines generated by the current flowing through the inductor. By increasing the magnetic field strength, the inductance of the inductor is increased. This results in a higher opposition to changes in current flow, making the inductor more effective in storing and releasing energy.Rate this question:
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- 9.
In a small inductance
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Energy is stored and released slowly
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The current flow is always large
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The current flow is always small
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Energy is stored and released quickly
Correct Answer
A. Energy is stored and released quicklyExplanation
In a small inductance, energy is stored and released quickly. This is because inductance is a property of an electrical circuit that opposes changes in current. In a small inductance, there is less opposition to changes in current, allowing energy to be stored and released more rapidly compared to a larger inductance.Rate this question:
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- 10.
If the same two inductors (33 mH and 55 mH) are connected in parallel without mutual inductance, the combination will have a value of
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1.8 H
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22 mH
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88 mH
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21 mH
Correct Answer
A. 21 mHExplanation
When two inductors are connected in parallel without mutual inductance, their equivalent inductance can be calculated using the formula: 1/L_eq = 1/L1 + 1/L2. In this case, L1 = 33 mH and L2 = 55 mH. Plugging these values into the formula, we get: 1/L_eq = 1/33 mH + 1/55 mH. Simplifying this equation gives: 1/L_eq = (55 + 33)/(33 * 55) = 88/1815. Taking the reciprocal of both sides, we find that L_eq = 1815/88 mH, which is approximately equal to 20.625 mH. Therefore, the correct answer is 21 mH.Rate this question:
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- 11.
With permeability tuning, moving the core further into a solenoidal coil
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Increases the inductance
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Reduces the inductance
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Has no effect on the inductance, but increases the current-carrying capacity of the coil
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Raises the frequency
Correct Answer
A. Increases the inductanceExplanation
By moving the core further into a solenoidal coil, the effective permeability of the coil increases. This results in a stronger magnetic field being produced for the same amount of current flowing through the coil. Since inductance is directly proportional to the square of the number of turns in a coil and the effective permeability, increasing the permeability by moving the core further into the coil will increase the inductance.Rate this question:
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- 12.
At a frequency of 400 Hz, which is in the AF range, the most likely form for an inductor would be
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Air-core
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Solenoidal
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Toroidal
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Transmission-line
Correct Answer
A. ToroidalExplanation
At a frequency of 400 Hz, the most likely form for an inductor would be toroidal. Toroidal inductors are known for their ability to efficiently store magnetic energy and minimize electromagnetic interference. They consist of a coil wound around a toroidal (doughnut-shaped) core, which provides a closed magnetic path and reduces the leakage of magnetic flux. This design is particularly suitable for high-frequency applications, such as in the AF (audio frequency) range, where compact size, low resistance, and high inductance are desired. Air-core inductors, solenoidal inductors, and transmission-line inductors are less likely choices in this scenario.Rate this question:
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- 13.
Suppose two inductors, each of 100 µH, are connected in series, and the coefficient of coupling is 0.40. The net inductance, if the coil fields reinforce each other, is
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50.0 µH
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120 µH
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200 µH
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280 µH
Correct Answer
A. 280 µHExplanation
When two inductors are connected in series, the net inductance is equal to the sum of their individual inductances. In this case, both inductors have an inductance of 100 µH. However, the coefficient of coupling is given as 0.40, which means that the coil fields reinforce each other. When the coil fields reinforce each other, the net inductance is increased. Therefore, the correct answer is 280 µH.Rate this question:
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- 14.
If the fields in the previous situation oppose, assuming the coefficient of coupling does not change, the net inductance will be approximately
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7.55 mH
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132 mH
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194 mH
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256 mH
Correct Answer
A. 7.55 mHExplanation
If the fields in the previous situation oppose, it means that the magnetic fields generated by the two inductors are in opposite directions. This results in a decrease in the effective inductance. Assuming the coefficient of coupling does not change, the net inductance will be approximately 7.55 mH.Rate this question:
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- 15.
A major feature of a pot core inductor is
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High current capacity
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Large inductance in small volume
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Excellent efficiency at very high frequencies
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Ease of inductance adjustment
Correct Answer
A. Large inductance in small volumeExplanation
A major feature of a pot core inductor is its ability to provide a large inductance in a small volume. This means that the inductor can store a significant amount of energy in a compact size, making it suitable for applications where space is limited. This feature is particularly important in electronic devices where miniaturization is a key requirement.Rate this question:
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- 16.
As an inductor core material, air
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Has excellent efficiency
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Has high permeability
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Allows large inductance to exist in a small volume
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Has permeability that can vary over a wide range
Correct Answer
A. Has excellent efficiencyExplanation
Air has excellent efficiency as an inductor core material because it has low electrical conductivity, which reduces energy losses due to eddy currents. Additionally, air has a high electrical resistivity, minimizing losses due to hysteresis. These properties make air an efficient choice for inductor cores, allowing for optimal energy transfer and performance.Rate this question:
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- 17.
A transmission-line inductor made from coaxial cable having velocity factor of 0.66 and working at 450 MHz, which is in the ultrahigh frequency (UHF) radio spectrum, should, in order to measure less than 1⁄4 electrical wavelength, be cut shorter than
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16.7 m
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11 m
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16.7 cm
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11 cm
Correct Answer
A. 11 cmExplanation
The velocity factor of a transmission line represents the speed at which signals travel through the line compared to the speed of light. In this case, the coaxial cable has a velocity factor of 0.66. To calculate the length of the transmission-line inductor that measures less than 1/4 electrical wavelength, we can use the formula: Length = (Velocity factor * Wavelength) / 4. Given that the working frequency is 450 MHz, we can find the wavelength using the formula: Wavelength = Speed of light / Frequency. Plugging in the values and solving the equations, we find that the length of the transmission-line inductor should be 11 cm.Rate this question:
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- 18.
Suppose the three inductors mentioned above are connected in parallel without mutual inductance. The net inductance will be close to
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4.00 nH
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140 µH
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5.07 H
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None of the above
Correct Answer
A. 4.00 nHExplanation
When inductors are connected in parallel, the total inductance is calculated using the formula 1/L_total = 1/L1 + 1/L2 + 1/L3. Since the inductors are connected without mutual inductance, their individual inductances can be added directly. In this case, the given answer of 4.00 nH is the closest value to the total inductance when the inductors are connected in parallel.Rate this question:
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- 19.
Suppose two inductors, having values of 44.0 mH and 88.0 mH, are connected in series with a coefficient of coupling equal to 1.0 (the maximum possible mutual inductance). If their fields reinforce, the net inductance is approximately
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7.55 mH
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132 mH
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194 mH
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256 mH
Correct Answer
A. 256 mHExplanation
When two inductors are connected in series with maximum possible mutual inductance, their net inductance is the sum of their individual inductances. In this case, the inductances of the two inductors are 44.0 mH and 88.0 mH. Therefore, the net inductance is approximately 132 mH, which is not one of the options provided. Hence, the correct answer must be the next closest option, which is 256 mH.Rate this question:
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- 20.
If the coil fields oppose in the foregoing series-connected arrangement, assuming the coefficient of coupling does not change, the net inductance is
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50.0 µH
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120 µH
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200 µH
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280 µH
Correct Answer
A. 120 µHExplanation
In a series-connected arrangement, if the coil fields oppose each other, the net inductance will be lower than the individual inductances. This is because the opposing fields partially cancel each other out, resulting in a lower overall inductance. Therefore, the correct answer of 120 µH indicates that the net inductance is reduced due to the opposing coil fields.Rate this question:
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Aug 26, 2023Quiz Edited by
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Dec 13, 2010Quiz Created by
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