Gibilisco:Chapter 2 - Electrical Units
(184).jpg "Gibilisco:Chapter 2 - Electrical Units - Quiz")
This is the MCQs for Gibilisco:CHAPTER 2 - ELECTRICAL UNITS
- 1.
A positive electric pole:
- A.
Has a deficiency of electrons.
- B.
Has fewer electrons than the negative pole.
- C.
Has an excess of electrons.
- D.
Has more electrons than the negative pole
Correct Answer
B. Has fewer electrons than the negative pole.Explanation
A positive electric pole has fewer electrons than the negative pole because electrons are negatively charged particles and a positive pole has a deficiency or lack of electrons compared to the negative pole.Rate this question:
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- 2.
An EMF of one volt:
- A.
Cannot drive much current through a circuit.
- B.
Represents a low resistance.
- C.
Can sometimes produce a large current.
- D.
Drops to zero in a short time.
Correct Answer
C. Can sometimes produce a large current.Explanation
An EMF of one volt can sometimes produce a large current because the amount of current flowing through a circuit is determined by the resistance in the circuit. If the resistance is low, a higher current can flow even with a lower EMF. Therefore, even though one volt may not typically drive much current through a circuit, it can still produce a large current if the resistance is low.Rate this question:
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- 3.
A potentially lethal electric current is on the order of:
- A.
0.01 mA.
- B.
0.1 mA.
- C.
1 mA.
- D.
0.1 A.
Correct Answer
D. 0.1 A.Explanation
A potentially lethal electric current is typically measured in amperes (A). A current of 0.1 A (ampere) is considered potentially lethal because it is a high enough current to cause severe harm or even death to a person. Currents in the milliampere (mA) range are generally not lethal, while currents in the order of amperes (A) can be extremely dangerous. Therefore, the correct answer is 0.1 A.Rate this question:
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- 4.
A current of 25 A is most likely drawn by:
- A.
A flashlight bulb.
- B.
A typical household.
- C.
A power plant.
- D.
A clock radio.
Correct Answer
B. A typical household.Explanation
A current of 25 A is most likely drawn by a typical household because household appliances such as refrigerators, air conditioners, and washing machines require a significant amount of current to operate. A flashlight bulb, clock radio, and power plant typically draw much less current compared to a typical household.Rate this question:
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- 5.
A piece of wire has a conductance of 20 siemens. Its resistance is:
- A.
20 Ω.
- B.
0.5 Ω.
- C.
0.05 Ω.
- D.
0.02 Ω.
Correct Answer
C. 0.05 Ω.Explanation
The conductance of a wire is the reciprocal of its resistance. In this case, the wire has a conductance of 20 siemens. Since conductance is the reciprocal of resistance, the resistance of the wire can be calculated by taking the reciprocal of the conductance. Therefore, the resistance of the wire is 1/20 siemens, which simplifies to 0.05 Ω.Rate this question:
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- 6.
A resistor has a value of 300 ohms. Its conductance is:
- A.
3.33 millisiemens.
- B.
33.3 millisiemens.
- C.
333 microsiemens.
- D.
0.333 siemens.
Correct Answer
A. 3.33 millisiemens.Explanation
The conductance of a resistor is the reciprocal of its resistance. In this case, the resistance of the resistor is 300 ohms. To find the conductance, we take the reciprocal of the resistance: 1/300 = 0.00333 siemens. Since the answer choices are given in millisiemens and microsiemens, we convert 0.00333 siemens to millisiemens by multiplying by 1000: 0.00333 * 1000 = 3.33 millisiemens. Therefore, the correct answer is 3.33 millisiemens.Rate this question:
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- 7.
A mile of wire has a conductance of 0.6 siemens. Then three miles of the same wire has a conductance of:
- A.
1.8 siemens.
- B.
1.8 Ω.
- C.
0.2 siemens.
- D.
Not enough information has been given to answer this.
Correct Answer
C. 0.2 siemens.Explanation
The conductance of a wire is inversely proportional to its length. Therefore, if a mile of wire has a conductance of 0.6 siemens, three miles of the same wire would have a conductance of 0.6 siemens divided by 3, which is equal to 0.2 siemens.Rate this question:
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- 8.
A 2-kW generator will deliver approximately how much current, reliably, at 117 V?
- A.
17 mA.
- B.
234 mA.
- C.
17 A.
- D.
234 A.
Correct Answer
C. 17 A.Explanation
A 2-kW generator can deliver a maximum power of 2,000 watts. To determine the current, we can use the formula P = IV, where P is power, I is current, and V is voltage. Rearranging the formula to solve for I, we get I = P/V. Plugging in the values, we have I = 2,000/117. This gives us approximately 17 amperes (A) of current. Therefore, the correct answer is 17 A.Rate this question:
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- 9.
A circuit breaker is rated for 15 A at 117 V. This represents approximately how many kilowatts?
- A.
1.76
- B.
1760
- C.
7.8
- D.
0.0078
Correct Answer
A. 1.76Explanation
The correct answer is 1.76. To calculate the power in kilowatts, we use the formula P = IV, where P is power in watts, I is current in amperes, and V is voltage in volts. In this case, the current is 15 A and the voltage is 117 V. Plugging these values into the formula, we get P = 15 A * 117 V = 1755 W. To convert watts to kilowatts, we divide by 1000, so the power in kilowatts is 1.755 kW, which can be rounded to 1.76.Rate this question:
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- 10.
You are told that a certain air conditioner is rated at 500 Btu. What is this in kWh?
- A.
147
- B.
14.7
- C.
1.47
- D.
0.147
Correct Answer
D. 0.147Explanation
The air conditioner is rated at 500 Btu. To convert Btu to kWh, we need to divide the Btu value by 3412.14 (the conversion factor). Therefore, 500 Btu divided by 3412.14 equals approximately 0.147 kWh.Rate this question:
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- 11.
Of the following energy units, the one most often used to define electrical energy is:
- A.
The Btu
- B.
The erg
- C.
The foot pound
- D.
The kilowatt hour
Correct Answer
D. The kilowatt hourExplanation
The kilowatt hour is the most commonly used unit to define electrical energy. It measures the amount of energy consumed or produced by an electrical device over a period of time. It is widely used by utility companies to measure and bill for electricity usage. The kilowatt hour is a practical unit as it combines both power (kilowatt) and time (hour) into a single measurement, making it easier to quantify and compare energy consumption.Rate this question:
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- 12.
The frequency of common household ac in the U.S. is:
- A.
60 Hz
- B.
120 Hz
- C.
50 Hz
- D.
100 Hz
Correct Answer
A. 60 HzExplanation
The frequency of common household AC in the U.S. is 60 Hz. This means that the alternating current changes direction 60 times per second. In the U.S., the standard frequency for residential electricity is 60 Hz, while in some other countries it may be 50 Hz. The frequency of the AC power supply is important because it determines the speed at which electric motors and other electrical devices operate.Rate this question:
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- 13.
Half-wave rectification means that:
- A.
Half of the ac wave is inverted
- B.
Half of the ac wave is chopped off
- C.
The whole wave is inverted
- D.
The effective value is half the peak value
Correct Answer
B. Half of the ac wave is chopped offExplanation
Half-wave rectification refers to the process of converting an alternating current (ac) waveform into a unidirectional current by allowing only one half of the ac waveform to pass through while blocking the other half. In this case, the correct answer states that half of the ac wave is chopped off, indicating that only one half of the waveform is allowed to pass through while the other half is eliminated. This process is commonly achieved using a diode, which acts as a one-way valve for the current.Rate this question:
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- 14.
In the output of a half-wave rectifier:
- A.
Half of the wave is inverted
- B.
The effective value is less than that of the original ac wave
- C.
The effective value is the same as that of the original ac wave
- D.
The effective value is more than that of the original ac wave
Correct Answer
B. The effective value is less than that of the original ac waveExplanation
In a half-wave rectifier, only one half of the input AC wave is allowed to pass through, while the other half is blocked. This means that the output waveform will only have positive half-cycles. Since the negative half-cycles are eliminated, the effective value of the output waveform will be less than that of the original AC wave, which includes both positive and negative half-cycles. Therefore, the effective value is less than that of the original AC wave.Rate this question:
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- 15.
In the output of a full-wave rectifier:
- A.
The whole wave is inverted
- B.
The effective value is less than that of the original ac wave
- C.
The effective value is the same as that of the original ac wave
- D.
The effective value is more than that of the original ac wave
Correct Answer
C. The effective value is the same as that of the original ac waveExplanation
In a full-wave rectifier, both the positive and negative halves of the AC wave are utilized, resulting in a rectified waveform that closely resembles the original AC waveform. This means that the effective value of the rectified waveform is the same as that of the original AC waveform.Rate this question:
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- 16.
A low voltage, such as 12 V:
- A.
Is never dangerous
- B.
Is always dangerous
- C.
Is dangerous if it is ac, but not if it is dc
- D.
Can be dangerous under certain conditions
Correct Answer
D. Can be dangerous under certain conditionsExplanation
A low voltage, such as 12 V, can be dangerous under certain conditions. While low voltage is generally considered to be less hazardous than higher voltages, it can still pose a risk depending on the circumstances. Factors such as the amount of current, duration of exposure, and the specific conditions of the situation can determine the level of danger. For example, if the low voltage is applied to sensitive body parts or if there is a risk of electric shock due to wet or conductive environments, it can be potentially dangerous.Rate this question:
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- 17.
Which of these can represent magnetomotive force?
- A.
The volt-turn
- B.
The ampere-turn
- C.
The gauss
- D.
The gauss-turn
Correct Answer
B. The ampere-turnExplanation
Magnetomotive force is a measure of the magnetic field strength produced by an electric current. It is represented by the product of the current (in amperes) and the number of turns in a coil (ampere-turns). The volt-turn is not a valid unit for magnetomotive force. The gauss is a unit of magnetic flux density, not magnetomotive force. The gauss-turn is not a recognized unit in the context of magnetomotive force.Rate this question:
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- 18.
Which of the following units can represent magnetic flux density?
- A.
The volt-turn
- B.
The ampere-turn
- C.
The gauss
- D.
The gauss-turn
Correct Answer
C. The gaussExplanation
The gauss is a unit that represents magnetic flux density. It is named after the German mathematician and physicist Carl Friedrich Gauss. The gauss is defined as one maxwell per square centimeter. It is commonly used in the field of magnetism to measure the strength of a magnetic field.Rate this question:
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- 19.
A ferromagnetic material:
- A.
Concentrates magnetic flux lines within itself
- B.
Increases the total magnetomotive force around a current-carrying wire
- C.
Causes an increase in the current in a wire
- D.
Increases the number of ampere-turns in a wire
Correct Answer
A. Concentrates magnetic flux lines within itselfExplanation
A ferromagnetic material concentrates magnetic flux lines within itself due to its unique atomic structure. This material has aligned magnetic domains that allow it to retain a strong magnetic field even after an external magnetic field is removed. As a result, the magnetic flux lines tend to stay within the material, making it an excellent conductor of magnetic flux. This property is utilized in various applications such as magnetic cores in transformers and magnetic storage devices.Rate this question:
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- 20.
A coil has 500 turns and carries 75 mA of current. The magnetomotive force will be:
- A.
37,500 At
- B.
375 At
- C.
37.5 At
- D.
3.75 At
Correct Answer
C. 37.5 AtExplanation
The magnetomotive force is determined by multiplying the number of turns in the coil by the current passing through it. In this case, the coil has 500 turns and carries 75 mA of current. Therefore, the magnetomotive force is calculated as 500 turns * 75 mA = 37,500 At.Rate this question:
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Mar 22, 2023Quiz Edited by
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Oct 23, 2013Quiz Created by
Froydwess
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