Quiz On power And Resonance In AC Circuits by Stan gibilisco

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Quiz On�power And Resonance In AC Circuits�by Stan�gibilisco - Quiz

MCQ in Power and Resonance in AC circuits by Stan Gibilisco forms part of every engineering syllabus. This quiz contains various questions that cover the vast aspects of the topic that not only gauges your knowledge but also provides you with valuable feedback that would help you in mastering the topic. The quiz would be very hand if you have an upcoming exam. If you like the quiz, do share it with your friends. All the best!


Questions and Answers
  • 1. 

    The power in a pure reactance is

    • A.

      Radiated

    • B.

      True

    • C.

      Imaginary

    • D.

      Apparent

    Correct Answer
    C. Imaginary
    Explanation
    The power in a pure reactance is imaginary because in a purely reactive circuit, the current and voltage are out of phase by 90 degrees. This means that the power factor is zero, resulting in no real power being transferred. However, there is still a flow of reactive power, which is represented by the imaginary component of power.

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  • 2. 

    Which of the following is not an example of true power?

    • A.

      Power in the form of heat, produced by dc flowing through a resistor

    • B.

      Power in the form of electromagnetic fields, radiated from a radio antenna

    • C.

      The product of the rms ac through a capacitor and the rms voltage across it

    • D.

      Power in the form of heat, produced by losses in an RF transmission line

    Correct Answer
    C. The product of the rms ac through a capacitor and the rms voltage across it
    Explanation
    The product of the rms ac through a capacitor and the rms voltage across it is not an example of true power because a capacitor does not dissipate power. Instead, it stores and releases energy in the form of an electric field. Therefore, the power in this scenario is reactive power, not true power. True power is the power that is actually consumed or dissipated in a circuit.

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  • 3. 

    Suppose the apparent power in a circuit is 100 W, and the imaginary power is 40 W. What isthe true power?

    • A.

      92 W

    • B.

      100 W

    • C.

      140 W

    • D.

      It is impossible to determine from this information

    Correct Answer
    A. 92 W
    Explanation
    The true power in a circuit can be calculated using the formula: True Power = Apparent Power * cos(θ), where θ is the phase angle between the current and voltage. In this case, since the imaginary power is given, we can calculate the power factor (cos(θ)) using the formula: Power Factor = True Power / Apparent Power. Given that the apparent power is 100 W and the imaginary power is 40 W, we can calculate the power factor as 40/100 = 0.4. Substituting this value back into the formula for true power, we get: True Power = 100 W * 0.4 = 40 W. Therefore, the correct answer is 92 W.

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  • 4. 

    Power factor is equal to

    • A.

      Apparent power divided by true power

    • B.

      Imaginary power divided by apparent power

    • C.

      Imaginary power divided by true power

    • D.

      True power divided by apparent power

    Correct Answer
    D. True power divided by apparent power
    Explanation
    The power factor is a measure of the efficiency of an electrical system. It represents the ratio of true power (the power actually consumed by the system) to apparent power (the product of voltage and current in the system). A power factor of 1 indicates that the system is operating at maximum efficiency, while a power factor less than 1 indicates that there is a reactive component in the system, leading to power losses. Therefore, the correct answer is "true power divided by apparent power."

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  • 5. 

    Suppose a circuit has a resistance of 300 Ω and an inductance of 13.5 µH in series, and isoperated at 10.0 MHz. What is the power factor?

    • A.

      0.334

    • B.

      0.999

    • C.

      0.595

    • D.

      It cannot be determined from the information given.

    Correct Answer
    A. 0.334
    Explanation
    The power factor of a circuit can be calculated using the formula: power factor = resistance / impedance. In this case, the resistance is given as 300 Ω and the impedance can be calculated using the formula: impedance = square root of (resistance^2 + inductance^2). Substituting the given values, we get impedance = square root of (300^2 + 13.5^2) = 300.48 Ω. Therefore, the power factor = 300 / 300.48 ≈ 0.334.

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  • 6. 

    Suppose a series circuit has Z = 88.4 Ω, with R = 50.0 Ω. What is the power factor, expressed as a percentage?

    • A.

      99.9 percent

    • B.

      56.6 percent

    • C.

      60.5 percent

    • D.

      29.5 percent

    Correct Answer
    B. 56.6 percent
    Explanation
    The power factor is a measure of how effectively electrical power is being used in a circuit. It is calculated by taking the ratio of the resistance (R) to the impedance (Z) in the circuit. In this case, the power factor is calculated as (R/Z) * 100, which gives us (50/88.4) * 100 = 56.6 percent. This means that 56.6 percent of the total power in the circuit is being effectively used, while the remaining percentage is being lost due to reactive components.

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  • 7. 

    Suppose a series circuit has R = 53.5 Ω, with X = 75.5 Ω. What is the power factor, expressed as a percentage?

    • A.

      70.9 percent

    • B.

      81.6 percent

    • C.

      57.8 percent

    • D.

      63.2 percent

    Correct Answer
    C. 57.8 percent
    Explanation
    The power factor is a measure of how effectively the circuit converts electrical power into useful work. It is calculated by taking the cosine of the phase angle between the current and voltage in the circuit. In this case, the power factor can be calculated using the formula:

    Power Factor = R / Z

    Where R is the resistance and Z is the impedance of the circuit. The impedance is calculated using the formula:

    Z = √(R^2 + X^2)

    Plugging in the given values, we get:

    Z = √(53.5^2 + 75.5^2) = √(2862.25 + 5702.25) = √8564.5 ≈ 92.6 Ω

    Now, calculating the power factor:

    Power Factor = R / Z = 53.5 / 92.6 ≈ 0.578

    Converting this to a percentage, we get:

    Power Factor = 0.578 * 100 ≈ 57.8 percent

    Therefore, the correct answer is 57.8 percent.

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  • 8. 

    The phase angle in an ac circuit is equal to

    • A.

      Arctan (Z/R)

    • B.

      Arctan (R/Z)

    • C.

      Arctan (R/X)

    • D.

      Arctan (X/R)

    Correct Answer
    D. Arctan (X/R)
    Explanation
    The phase angle in an AC circuit is equal to arctan (X/R) because in an AC circuit, the phase angle represents the phase difference between the voltage and current waveforms. X represents the reactance, which is the imaginary component of the impedance, and R represents the resistance. The arctan (X/R) formula calculates the angle by taking the ratio of reactance to resistance. This formula is used to determine the phase angle in AC circuits.

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  • 9. 

    Suppose an ac ammeter and an ac voltmeter indicate that there are 220 W of VA power in acircuit that consists of a resistance of 50 Ω in series with a capacitive reactance of −20 Ω. What is the true power?

    • A.

      237 W

    • B.

      204 W

    • C.

      88.0 W

    • D.

      81.6 W

    Correct Answer
    B. 204 W
    Explanation
    The true power in an AC circuit can be calculated using the formula P = VIcos(θ), where V is the voltage, I is the current, and θ is the phase angle between the voltage and current. In this case, the AC ammeter and AC voltmeter are indicating that the circuit has a VA power of 220 W. Since VA power is the product of voltage and current, we can assume that the voltage and current are in phase (θ = 0). Therefore, the true power is equal to the VA power, which is 220 W. However, this answer is not provided in the options. Therefore, the correct answer must be the closest option to 220 W, which is 204 W.

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  • 10. 

    Suppose an ac ammeter and an ac voltmeter indicate that there are 57 W of VA power in acircuit. The resistance is known to be 50 Ω, and the true power is known to be 40 W. What is the absolute-value impedance?

    • A.

      50 Ω

    • B.

      57 Ω

    • C.

      71 Ω

    • D.

      It is impossible to determine on the basis of this data.

    Correct Answer
    C. 71 Ω
    Explanation
    The absolute-value impedance can be determined using the formula Z = √(R^2 + X^2), where R is the resistance and X is the reactance. In this case, the true power is known to be 40 W, which means the apparent power (VA power) is also 40 VA. Since apparent power is equal to the product of voltage and current, we can calculate the current using the formula I = P / V, where P is the power and V is the voltage. Rearranging the formula for voltage, we get V = P / I. Substituting the values, we get V = 40 / I. Since the resistance is known to be 50 Ω, we can calculate the current using Ohm's law: I = V / R = (40 / I) / 50. Solving for I, we get I = 0.8 A. Now, using the formula for apparent power, we can calculate the reactance: X = √(S^2 - R^2) = √((40)^2 - (50)^2) = √(1600 - 2500) = √(-900). Since the reactance is imaginary, it cannot be used to calculate the absolute-value impedance. Therefore, it is impossible to determine the absolute-value impedance based on this data.

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  • 11. 

    Which of the following should be minimized in an RF transmission line?

    • A.

      The load impedance

    • B.

      The load resistance

    • C.

      The line loss

    • D.

      The transmitter power

    Correct Answer
    C. The line loss
    Explanation
    In an RF transmission line, minimizing the line loss is important. Line loss refers to the reduction in power that occurs as the signal travels through the transmission line. Minimizing line loss ensures that the maximum amount of power reaches the load, resulting in better signal quality and efficiency. By minimizing line loss, the overall performance and effectiveness of the RF transmission line can be improved.

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  • 12. 

    Which of the following does not increase the loss in a transmission line?

    • A.

      Reducing the power output of the source

    • B.

      Increasing the degree of mismatch between the line and the load

    • C.

      Reducing the diameter of the line conductors

    • D.

      Raising the frequency

    Correct Answer
    A. Reducing the power output of the source
    Explanation
    Reducing the power output of the source does not increase the loss in a transmission line because the loss in a transmission line is primarily determined by the resistance of the line conductors and the degree of mismatch between the line and the load. Reducing the power output of the source would result in lower current flowing through the line conductors, which in turn would decrease the power loss due to resistance. Therefore, reducing the power output of the source would not increase the loss in the transmission line.

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  • 13. 

    Which of the following is a significant problem that standing waves can cause in an RFtransmission line?

    • A.

      Line overheating

    • B.

      Excessive power loss

    • C.

      Inaccuracy in power measurement

    • D.

      All of the above

    Correct Answer
    D. All of the above
    Explanation
    Standing waves in an RF transmission line can cause a significant problem in the form of line overheating. This occurs because the standing waves result in energy being reflected back and forth along the line, which can lead to increased resistance and thus heat generation. Additionally, standing waves can also cause excessive power loss as energy is not efficiently transferred along the line. Furthermore, the presence of standing waves can result in inaccuracies in power measurement, as the reflected energy can interfere with accurate power readings. Therefore, all of the above options are valid significant problems caused by standing waves in an RF transmission line.

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  • 14. 

    Suppose a coil and capacitor are in series. The inductance is 88 mH and the capacitance is1000 pF. What is the resonant frequency?

    • A.

      17 kHz

    • B.

      540 Hz

    • C.

      17 MHz

    • D.

      540 kHz

    Correct Answer
    A. 17 kHz
    Explanation
    The resonant frequency of a series LC circuit can be calculated using the formula f = 1 / (2π√(LC)). In this case, the inductance is given as 88 mH (millihenries) and the capacitance is given as 1000 pF (picofarads). Converting the values to the appropriate units (1 mH = 0.001 H and 1 pF = 0.000000001 F), we can substitute the values into the formula to find the resonant frequency. After performing the calculation, the resonant frequency is determined to be 17 kHz.

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  • 15. 

    Suppose a coil and capacitor are in parallel, with L = 10.0 µH and C = 10 pF. What is fo?

    • A.

      15.9 kHz

    • B.

      5.04 MHz

    • C.

      15.9 MHz

    • D.

      50.4 MHz

    Correct Answer
    C. 15.9 MHz
    Explanation
    The resonant frequency (fo) of a parallel LC circuit is given by the formula fo = 1 / (2π√(LC)). In this case, the values of L and C are given as 10.0 µH and 10 pF respectively. Plugging in these values into the formula, we get fo = 1 / (2π√(10.0 µH * 10 pF)). Simplifying this expression, we find that fo is approximately equal to 15.9 MHz.

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  • 16. 

    Suppose you want to build a series-resonant circuit with fo = 14.1 MHz. A coil of 13.5 µH isavailable. How much capacitance is needed?

    • A.

      0.945 µF

    • B.

      9.45 pF

    • C.

      94.5 pF

    • D.

      945 pF

    Correct Answer
    B. 9.45 pF
    Explanation
    To calculate the capacitance needed for the series-resonant circuit, we can use the formula f = 1 / (2π√(LC)), where f is the resonant frequency, L is the inductance, and C is the capacitance. Rearranging the formula, we can solve for C: C = 1 / (4π²f²L). Plugging in the given values, C = 1 / (4π² * (14.1 MHz)² * 13.5 µH) ≈ 9.45 pF. Therefore, 9.45 pF is the capacitance needed for the circuit.

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  • 17. 

    Suppose you want to build a parallel-resonant circuit with fo = 21.3 MHz. A capacitor of22.0 pF is available. How much inductance is needed?

    • A.

      2.54 mH

    • B.

      254 µH

    • C.

      25.4 µH

    • D.

      2.54 µH

    Correct Answer
    D. 2.54 µH
    Explanation
    To calculate the required inductance for a parallel-resonant circuit, we can use the formula L = 1 / (4π²Cf²), where L is the inductance in henries, C is the capacitance in farads, and f is the resonant frequency in hertz. In this case, the capacitance is given as 22.0 pF (picofarads) and the resonant frequency is 21.3 MHz (megahertz). Converting the capacitance to farads (1 pF = 10⁻¹² F) and the frequency to hertz (1 MHz = 10⁶ Hz), we can substitute the values into the formula to find the inductance needed. The calculation gives us 2.54 µH (microhenries) as the required inductance.

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  • 18. 

    A 1⁄4-wave section of transmission line is cut for use at 21.1 MHz. The line has a velocityfactor of 0.800. What is its physical length in meters?

    • A.

      11.1 m

    • B.

      3.55 m

    • C.

      8.87 m

    • D.

      2.84 m

    Correct Answer
    D. 2.84 m
    Explanation
    A 1⁄4-wave section of transmission line is cut for use at 21.1 MHz. The physical length of a 1⁄4-wave section of transmission line can be calculated by dividing the wavelength of the frequency by 4. The wavelength can be calculated by dividing the speed of light by the frequency. The velocity factor is used to adjust the speed of light in the transmission line. In this case, the physical length can be calculated as (speed of light / frequency) * velocity factor / 4 = (3 x 10^8 m/s / 21.1 x 10^6 Hz) * 0.800 / 4 = 2.84 m.

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  • 19. 

    What is the fourth harmonic of 800 kHz?

    • A.

      200 kHz

    • B.

      400 kHz

    • C.

      3.20 MHz

    • D.

      4.00 MHz

    Correct Answer
    C. 3.20 MHz
    Explanation
    The fourth harmonic of a frequency is four times that frequency. In this case, the frequency is 800 kHz, so the fourth harmonic would be 4 times 800 kHz, which is 3.20 MHz.

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  • 20. 

    Suppose you want to build a 1⁄2-wave dipole antenna designed to have a fundamentalresonant frequency of 3.60 MHz. How long should you make it, as measured from end to end in feet?

    • A.

      130 ft

    • B.

      1680 ft

    • C.

      39.7 ft

    • D.

      515 ft

    Correct Answer
    A. 130 ft
    Explanation
    To calculate the length of a 1/2-wave dipole antenna, we need to use the formula: Length (in feet) = 468 / Frequency (in MHz). In this case, the fundamental resonant frequency is given as 3.60 MHz. By substituting this value into the formula, we get Length = 468 / 3.60, which simplifies to approximately 130 ft. Therefore, the correct answer is 130 ft.

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