Gibilisco - Alternating-current Basics

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• 1.

Which of the following can vary with ac, but never with dc?

• A.

Power

• B.

Voltage

• C.

Frequency

• D.

Amplitude

C. Frequency
Explanation
Frequency is the correct answer because it refers to the number of cycles or oscillations per second in an alternating current (AC) waveform. In AC, the direction of current flow constantly changes, causing the frequency to vary. However, in direct current (DC), the current flows in only one direction, so the frequency remains constant at zero. Therefore, frequency can vary with AC but never with DC.

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• 2.

The length of time between a point in one cycle and the same point in the next cycle of an acwave is the

• A.

Frequency.

• B.

Magnitude.

• C.

Period.

• D.

Polarity.

C. Period.
Explanation
The length of time between a point in one cycle and the same point in the next cycle of an AC wave is referred to as the period. It represents the time taken for the wave to complete one full cycle. The frequency, on the other hand, refers to the number of cycles that occur in a given time period. Magnitude relates to the size or amplitude of the wave, while polarity indicates the direction or orientation of the wave.

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• 3.

On a spectrum analyzer, an ac signal having only one frequency component looks like

• A.

A single pip.

• B.

A sine wave.

• C.

A square wave.

• D.

A sawtooth wave.

A. A single pip.
Explanation
An ac signal with only one frequency component appears as a single pip on a spectrum analyzer. This is because a spectrum analyzer displays the frequency content of a signal, and when there is only one frequency component, it will be represented as a single point or pip on the display. The other options, such as a sine wave, square wave, or sawtooth wave, would show multiple frequency components, resulting in a different waveform representation on the spectrum analyzer.

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• 4.

The period of an ac wave, in seconds, is

• A.

The same as the frequency in hertz.

• B.

Not related to the frequency in any way.

• C.

Equal to 1 divided by the frequency in hertz.

• D.

Equal to the peak amplitude in volts divided by the frequency in hertz.

C. Equal to 1 divided by the frequency in hertz.
Explanation
The period of an AC wave is the time it takes for one complete cycle of the wave to occur. The frequency of an AC wave is the number of cycles that occur in one second. Since the period represents the time for one cycle and the frequency represents the number of cycles in one second, they are inversely related. Therefore, the period of an AC wave is equal to 1 divided by the frequency in hertz.

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• 5.

The sixth harmonic of an ac wave whose period is 1.000 millisecond (1.000 ms) has afrequency of

• A.

0.006 Hz.

• B.

167.0 Hz.

• C.

7.000 kHz.

• D.

6.000 kHz.

D. 6.000 kHz.
Explanation
The frequency of an ac wave is determined by the number of cycles it completes in one second. The period of the wave is given as 1.000 millisecond (1.000 ms), which means it completes 1 cycle in 1 ms. The sixth harmonic of the wave would complete 6 cycles in the same time period. To find the frequency, we divide the number of cycles (6) by the time period (1 ms), which gives us a frequency of 6 kHz.

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• 6.

A degree of phase represents

• A.

6.28 cycles.

• B.

57.3 cycles.

• C.

1⁄60 of a cycle.

• D.

1⁄360 of a cycle.

D. 1⁄360 of a cycle.
Explanation
A degree of phase represents 1/360 of a cycle. In a cycle, there are 360 degrees. This means that each degree represents 1/360th of the total cycle. Therefore, the correct answer is 1/360 of a cycle.

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• 7.

Suppose that two ac waves have the same frequency but differ in phase by exactly 1⁄20 of a cycle. What is the phase difference between these two waves?

• A.

18°

• B.

20°

• C.

36°

• D.

5.73°

A. 18°
Explanation
The phase difference between two waves is determined by the fraction of a cycle that they differ by. In this case, the waves differ by exactly 1/20 of a cycle. To convert this fraction into degrees, we multiply it by 360 (since a full cycle is 360°). Therefore, 1/20 * 360 = 18°. Hence, the phase difference between these two waves is 18°.

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• 8.

Suppose an ac signal has a frequency of 1770 Hz. What is its angular frequency?

• A.

• B.

• C.

• D.

Impossible to determine from the data given

Explanation
The angular frequency of an AC signal is equal to 2π times the frequency. In this case, the given frequency is 1770 Hz. Therefore, the angular frequency can be calculated by multiplying 1770 by 2π, which gives us 11,120 rad/s.

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• 9.

A triangular wave exhibits

• A.

An instantaneous rise and a defined decay.

• B.

A defined rise and an instantaneous decay.

• C.

A defined rise and a defined decay, and the two are equal.

• D.

An instantaneous rise and an instantaneous decay.

C. A defined rise and a defined decay, and the two are equal.
Explanation
A triangular wave exhibits a defined rise and a defined decay, and the two are equal. This means that the waveform increases from its lowest point to its highest point in a defined manner, and then decreases from its highest point back to its lowest point in the same defined manner. The rise and decay are symmetrical, resulting in equal durations for both phases of the waveform.

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• 10.

Three-phase ac

• A.

Has sawtooth waves that add together in phase.

• B.

Consists of three sine waves in different phases.

• C.

Is a sine wave with exactly three harmonics.

• D.

Is of interest only to physicists.

B. Consists of three sine waves in different phases.
Explanation
The correct answer is that three-phase AC consists of three sine waves in different phases. In three-phase AC, there are three separate waveforms that are 120 degrees out of phase with each other. These waveforms are typically represented as sine waves and are used in power distribution systems to provide a more efficient and balanced power supply. Each waveform carries the same frequency but has a different phase relationship, resulting in a more stable and reliable power transmission.

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• 11.

If two perfect sine waves have the same frequency and the same amplitude, but are inopposite phase, the composite wave

• A.

Has twice the amplitude of either input wave alone.

• B.

Has half the amplitude of either input wave alone.

• C.

Is complex, but has the same frequency as the originals.

• D.

Has zero amplitude (that is, it does not exist), because the two input waves cancel each other out.

D. Has zero amplitude (that is, it does not exist), because the two input waves cancel each other out.
Explanation
When two perfect sine waves have the same frequency and the same amplitude, but are in opposite phase, they will interfere destructively. This means that the positive peaks of one wave will coincide with the negative peaks of the other wave, resulting in cancellation. As a result, the composite wave will have zero amplitude, meaning it does not exist.

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• 12.

If two perfect sine waves have the same frequency and the same phase, the composite wave

• A.

Is a sine wave with an amplitude equal to the difference between the amplitudes of the two input waves.

• B.

Is a sine wave with an amplitude equal to the sum of the amplitudes of the two original waves.

• C.

Is not a sine wave, but has the same frequency as the two input waves.

• D.

Has zero amplitude (that is, it does not exist), because the two input waves cancel each other out.

B. Is a sine wave with an amplitude equal to the sum of the amplitudes of the two original waves.
Explanation
When two perfect sine waves with the same frequency and phase are combined, they constructively interfere, resulting in a composite wave. The amplitude of the composite wave is determined by adding the amplitudes of the two original waves together. Therefore, the correct answer is that the composite wave is a sine wave with an amplitude equal to the sum of the amplitudes of the two original waves.

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• 13.

In a 117-V rms utility circuit, the positive peak voltage is approximately

• A.

+82.7 V.

• B.

+165 V.

• C.

+234 V.

• D.

+331 V

B. +165 V.
Explanation
The positive peak voltage in a utility circuit is equal to the peak value of the voltage waveform, which is √2 times the rms value. In this case, the rms voltage is 117 V, so the positive peak voltage can be calculated as √2 * 117 V ≈ 165 V. Therefore, the correct answer is +165 V.

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• 14.

• A.

82.7 V.

• B.

165 V.

• C.

234 V.

• D.

331 V.

D. 331 V.
• 15.

In a perfect sine wave, the peak-to-peak amplitude is equal to

• A.

Half the peak amplitude.

• B.

The peak amplitude.

• C.

1.414 times the peak amplitude.

• D.

Twice the peak amplitude.

D. Twice the peak amplitude.
Explanation
In a perfect sine wave, the peak-to-peak amplitude is equal to twice the peak amplitude. This means that the distance between the highest point (peak) and the lowest point (trough) of the wave is twice the height of the peak. It is important to note that the peak-to-peak amplitude is a measure of the total variation of the wave, while the peak amplitude refers to the maximum value of the wave.

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• 16.

Which one of the following does not affect the power output available from a particular acgenerator?

• A.

The strength of the magnet

• B.

The number of turns in the coil

• C.

The type of natural energy source used

• D.

The speed of rotation of the coil or magnet

C. The type of natural energy source used
Explanation
The type of natural energy source used does not affect the power output available from a particular ac generator. The power output of an ac generator is determined by factors such as the strength of the magnet, the number of turns in the coil, and the speed of rotation of the coil or magnet. The type of natural energy source used, such as wind, water, or fossil fuels, may determine the initial source of energy to rotate the coil or magnet, but it does not directly impact the power output once the generator is in operation.

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• 17.

If a 175-V dc source were connected in series with the utility mains from a standard walloutlet, the result would be

• A.

Smooth dc at a constant voltage.

• B.

Pure ac with equal peak voltages.

• C.

Ac with one peak voltage greater than the other.

• D.

Fluctuating dc.

D. Fluctuating dc.
Explanation
If a 175-V dc source were connected in series with the utility mains from a standard wall outlet, the result would be fluctuating dc. This is because the utility mains provide alternating current (ac), which constantly changes direction. When the dc source is connected in series, it would cause the current to fluctuate between the dc voltage and the ac voltage from the mains. As a result, the output would be a fluctuating direct current (dc), with the voltage constantly changing between the dc source voltage and the ac voltage from the mains.

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• 18.

An advantage of ac over dc in utility applications is the fact that

• A.

Ac is easier to transform from one voltage to another.

• B.

Ac is transmitted with lower loss in wires.

• C.

Ac can be easily obtained from dc generators.

• D.

Ac can be generated with less-dangerous by-products.

A. Ac is easier to transform from one voltage to another.
Explanation
AC (alternating current) is easier to transform from one voltage to another compared to DC (direct current). This is because AC voltage can be easily stepped up or stepped down using transformers. Transformers work based on the principle of electromagnetic induction, which is more efficient for AC than DC. On the other hand, transforming DC voltage requires the use of more complex and less efficient methods such as choppers or inverters. Therefore, the ability to easily transform AC voltage makes it advantageous in utility applications.

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• Current Version
• Mar 21, 2023
Quiz Edited by
ProProfs Editorial Team
• Dec 14, 2010
Quiz Created by
BATANGMAGALING

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