ACM-105 - BASIC ELEC (FAA) Flashcards
Study FAA Exam (127 cards)
8001
The working voltage of a capacitor in an AC circuit should be:
A. Equal to the highest applied voltage.
B. At least 20 percent greater than the highest applied voltage.
C. At least 50 percent greater than the highest applied voltage.
C
The working voltage of a capacitor is the highest voltage that can be steadily applied to it without the danger of the dielectric breaking down. The working voltage depends upon the material used as the dielectric and on its thickness~ A capacitor used in an AC circuit should have a working voltage at least 50 percent greater than the highest voltage that will be applied to it.
(AM.I.A.KS)- FAA-H-8083-30
8002
The term that describes the combined resistive forces in an AC circuit is
A. Resistance.
B. Reactance.
C. Impedance.
C
lmpedence (Z) is the vector sum of the resistance and the total reactance in a circuit. It is expressed in ohms, and found using the formula:
Z= ✓R2 +X2
There are three types of resistive forces in an AC circuit: inductive reactance, which causes the current to lag the voltage; capacitive reactance, which causes the current to lead the voltage; and resistance, which allows the current and voltage to remain in phase. Inductive and capacitive reactance are 180° out of phase and cancel each other.
(AM.I.A.KS) - FAA-H- 8083-30
8002-1
What is the opposition to the flow of AC produced by an electromagnetic field (EMF) with generated back voltage called?
A. Inductive reactance.
B. Capacitive reactance.
C. Mutual inductance.
A
Alternating current is in a constant state of change; EMF causes continuously inducted voltage opposition to the current in the circuit. This opposition is called inductive reactance (XL), and is measured in ohms just as resistance is measured. Inductance is the property of a circuit to oppose any change in current and is measured in henries.
Inductive reactance is a measure of how much the countering EMF in the circuit will oppose current variations.
(AM.I.A.KS) - FAA-H-8083-30
8002-2
Electrostatic fields are also known as
A. Dielectric fields
B. Electrostatic fields
C. Static fields
A
A field of force exists around a charged body. This is an electrostatic field (sometimes called a dielectric field) and it is represented by lines that extend in all directions from the charged body which terminate where there is an equal and opposite charge.
(AM.I.A.K24) - FAA-H-8083-30
8003
The basis for transformer operation in the use of alternating current is mutual:
A. Inductance.
B. Capacitance.
C. Reactance.
A
A transformer operates on the basis of mutual inductance.
The changing current in the primary windings produces a changing magnetic field whose flux cuts across the turns of the secondary winding and induces a voltage into it.
(AM.I.A.K16) - FAA-H-8083-30
8004
The opposition offered by a coil to the flow of alternating current (ignoring resistance) is called:
A. Impedance.
B. Reluctance.
C. Inductive reactance.
C
When alternating current flows in a coil of wire, the changing lines of flux cutting across the turns of wire in the coil induce a voltage in it. The polarity of this voltage (the counter EMF) is opposite to the polarity of the voltage that caused it. The counter EMF decreases the total voltage across the coil, and this decreases the current flowing through it. This opposition to the flow of alternating current is called inductive reactance (XL) and is measured in ohms. It opposes the flow of current, but it does not cause heat nor use any power.
(AM.I.A.K5) - FAA-H-8083-30
8004-1
What factors strengthen a coil inductor?
A. Limiting and separating the coils.
B. Adding and separating the coils.
C. Adding coils close together.
C
As more loops are added close together, the strength of them magnetic field increases.
Many loops close together will result in a strong electromagnet.
(AM.I.A.K2)- FAAH-8083-30
8005
An increase in which of the following factors will cause an increase in the inductive reactance of a circuit?
A-Inductance and frequency.
B-Resistance and voltage.
C-Resistance and capacitive reactance.
A
The inductive reactance (XL) in an AC circuit is increased when either the frequency of the alternating current or the inductance of the circuit is increased.
Resistance, voltage, or capacitive reactance have no effect on the inductive reactance of a circuit. The formula for inductive reactance is:
XL =2nfL
(AM.I.A.K4) - FAA-H-8083-30
- (Refer to Figure 1.) When different rated capacitors are connected in series in a circuit, the total capacitance is:
A. Less than the capacitance of the lowest rated capacitor.
B. Greater than the capacitance of the highest rated capacitor.
C. Equal to the sum of all the capacitances.
A
When capacitors of different values are connected in series, the total capacitance is less than that of the lowest rated capacitor. (AM.I.A.KS) - FAA-H-8083-30
8006-1
Capacitors are sometimes used in DC circuits to:
A. Counteract inductive reactance at specific locations.
B. Smooth out slight pulsations in current/voltage.
C. Assist in stepping voltage and current up and/or down.
B
Capacitors store electrical charges and are sometimes used in DC circuits to smooth out slight pulsations in current or voltage. Capacitors accept electrons when there is an excess and release them back into the circuit when the values decrease.
(AM.I.A.K3) - FAA-H-8083-30
8007
In an AC circuit, the effective voltage is:
A. Equal to the maximum instantaneous voltage.
B. Greater than the maximum instantaneous voltage.
C. Less than the maximum instantaneous voltage.
C
The effective voltage of sine wave alternating current is 0.707 time its peak voltage. The effective voltage, also called the root mean square (rms) voltage, is the voltage measured by most of the AC voltmeters. Peak voltage is measured with a special peak voltmeter or an oscilloscope.
(AM.I.A.K5) - FAA-H-8083-30
- The amount of electricity a capacitor can store is directly proportional to the:
A. Distance between the plates and inversely proportional to the plate area.
B. Plate area and is not affected by the distance between the plates.
C. Plate area and inversely proportional to the distance between the plates.
C
Three factors affect the amount of electricity a capacitor can store:
- The area of the plates. The larger the plate area, the greater the capacity.
- The thickness of the dielectric {the distance between the plates). The closer the plates are together, the stronger the electrical field will be and the greater the capacity.
- The material from which the dielectric is made (its dielectric constant). The higher the dielectric constant, the greater the capacity.
(AM.I.A.K3) - FAA-H-8083-30
- (Refer to Figure 2.) What is the total capacitance of a certain circuit containing three capacitors with capacitances of .02 microfarad, .05 microfarad, and .10 microfarad, respectively?
A-.170 μF.
B-0.125 pF.
C-.0125 μF.
C
When a 0.02-microfarad, a 0.05-microfarad, and a 0.1 a microfarad capacitor are connected in series, the total capacitance is 0.0125 microfarad.
(AM.I.A.K3) - FAA-H-8083-30
8009-1
What is the total capacitance of a circuit containing three capacitors in parallel with capacitances of .02 microfarad, .05 microfarad, and .10 microfarad, respectively?
A-.170μF.
B-0.125 μF.
C-.0125 μF.
A
Use the formula CT = C1 + C2 + C3. When capacitors are connected together in parallel, the plate area of all the capacitors add together and the total capacitance is the sum of the individual capacitances.
.02 + .05 + .10 = .170 uF
(AM.I.A.K3) - FAA-H-8083-30
8009-2
Convert farads to microfarads by:
A. Multiplying farads by 10 to the power of 6
B. Multiplying picofarads by 10 to the power of 6
C. Multiplying microfarads by 1 0 to the power of 6
A
One farad is equal to 106 microfarads.
Example: 2 farads is equal to 2 x 10(6) (2,000,000) microfarads.
(AM.I.A.KB) - FAA-H-8083-30
8009-3
Convert farads to picofarads by:
A. Multiplying farads by 10 to the power of 12
B. Multiplying microfarads by 10 to the power of-12
C. Multiplying picofarads by 10 to the power of 12
A
One farad is equal to 1012 picofarads. Example: 2 farads is equal to 2 x 1012 (2,000,000,000p) icofarads.
( AM.I.A.KB) - FAA-H-8083-30
8010
Unless otherwise specified, any values given for current or voltage in an AC circuit are assumed to be:
A. Instantaneous values.
B. Effective values.
C. Maximum values.
B
Almost all measuring instruments used for electrical system servicing measure the effective rms values of alternating current. Unless peak values, peak-to-peak values, or average values are specifically called out, effective rms values are assumed.
(AM.I.A.K5) - FAA-H-8083-30
8011
When different rated capacitors are connected in parallel in a circuit, the total capacitance is
(Note: Cr = C1 + C2 + C3 … )
A. less than the capacitance of the lowest rated capacitor.
B. Equal to the capacitance of the highest rated capacitor.
C. Equal to the sum of all the capacitances.
C
When capacitors are connected in parallel, the effective area of the plates combine, and the total capacitance is the sum of the individual capacitances.
(AM.I.A.K4) - FAA-H-8083-30
8012
When inductors are connected in series in a circuit, the total inductance (where the magnetic fields of each inductor do not affect the others) is:
(Note: Lr = L1 + L2 + L3 … )
A. Less than the inductance of the lowest rated inductor.
B. Equal to the inductance of the highest rated inductor.
C. Equal to the sum of the individual inductances.
C
When several inductors are connected together in such a way that there is no inductive coupling, the total inductance is the sum of the individual inductances.
(AM.I.A.K4) - FAA-H-8083-30
8013
(Refer to Figure 3.) When more than two inductors of different inductances are connected in parallel in a circuit, the total inductance is:
A. Less than the inductance of the lowest·rated inductor.
B. Equal to the inductance of the highest rated inductor.
C. Equal to the sum of the individual inductances.
A
When two or more inductors having different inductances are connected in parallel, the total inductance is less than the inductance of the lowest rated inductor.
(AM.I.A.K4) - FAA-H-8083-30
8014
What is the total capacitance of a certain circuit containing three capacitors with capacitances of .25 microfarad, .03 microfarad, and .12 microfarad, respectively?
(Note: CT = C1 + C2 + Ca)
A-.4μF.
B-.04pF.
C-.04μF.
A
When three capacitors are connected in parallel, their total capacitance is the sum of the individual capacitances.
Cr = C1 + C2 + C3
= 0.25 + 0.03 = 0.12
= 0.4 μF
(AM.I.A.K3) - FAA-H-8083-30
8015
Which requires the most electrical power during operation?
(Note: 1 horsepower = 746 watts)
A. A 12-volt motor requiring 8 amperes.
B. Four 30-watt lamps in a 12-volt parallel circuit.
C. Two lights requiring 3 amperes each in a 24-volt parallel system.
C
The 12-volt motor requires 96 watts of power. The four 30-watt lamps require 120 watts of power. The two 24-volt, 3-amp lights require 144 watts of power.
(AM.I.A.KB) - FAA-H-8083-30
8016
How much power must a 24-volt generator furnish to a system which.contains the following loads?
Unit Rating
One motor (75 percent efficient) ………………………. 1/5 hp
Three position lights …………………………… 20 watts each
One heating element ……………………………………… 5 amp
One anticollision light …………………………………….. 3 amp
(Note: 1 horsepower =· 746 watts)
A. 402 watts.
B. 385 watts.
C. 450 watts.
C
The motor is 1/5 hp, therefore 746 watts (1 hp) + 5 = 149 watts. This is the output of the engine. It takes more. energy to produce 149 watts; find this with the efficiency rating:
149/X = 75/100; X = 199. Therefore, the 1/5 hp motor that is 75 percent efficient requires 199 watts.
The three position lights require a total of 60 watts. The heating element requires 120 watts. The anti-collision light requires 72 watts. The total power the generator must produce is 451 watts. (AM.I.A.K12)- FAA-H-8083-30
8017
A 12-volt electric motor has 1,000 watts input and 1 horsepower output. Maintaining the same efficiency, how much input power will a 24-volt, 1-horsepower electric motor require?
(Note: 1 horsepower = 746 watts)
A. 1,000 watts.
B. 2,000 watts.
C. 500 watts.
A
The power produced by an electric motor is the product of its voltage and its current. A 12-volt motor will require 83.3 amps of current for its 1,000 watts of input power to produce 746 watts (1 horsepower) of output power.
A 24-volt motor operating at the same efficiency will also require 1,000 watts of input power for its 746 watts of output power, but it will need only 41.7 amps of current.
(AM.I.A.K12)- FAA-H-8083-30