Basic Electricity Flashcards
(122 cards)
1
Q
- 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.
A
C— at least 50 percent greater than the highest applied voltage.
Explanation:
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.
2
Q
- The term that describes the combined resistive forces in an AC circuit is
A— resistance.
B— reactance.
C— impedance
A
C— impedance
Explanation:
Impedance (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:
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.
3
Q
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
A— Inductive reactance
Explanation:
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 (X L ), 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.
4
Q
8002-2.
Electrostatic fields are also known as
A— dielectric fields.
B— electrostatic fields.
C— static fields.
A
A— dielectric fields.
Explanation:
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.
5
Q
- The basis for transformer operation in the use of alternating current is mutual
A— inductance.
B— capacitance.
C— reactance.
A
A— inductance.
Explanation:
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.
6
Q
- The opposition offered by a coil to the flow of alternating current (ignoring resistance) is called
A— impedance.
B— reluctance.
C— inductive reactance.
A
C— inductive reactance.
Explanation:
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 (X L ) and is measured in ohms. It opposes the flow of current, but it does not cause heat nor use any power.
7
Q
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.
A
C— Adding coils close together.
Explanation:
As more loops are added close together, the strength of the magnetic field increases. Many loops close together will result in a strong electromagnet.
8
Q
- 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
A— Inductance and frequency.
Explanation: The inductive reactance (X L ) 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:
9
Q
- (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
A— less than the capacitance of the lowest rated capacitor.
Explanation:When capacitors of different values are connected in series, the total capacitance is less than that of the lowest rated capacitor.
10
Q
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.
A
B— smooth out slight pulsations in current/voltage.
Explanation: 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.
11
Q
- 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.
A
C— less than the maximum instantaneous voltage.
Explanation: 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.
12
Q
- 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.
A
C— plate area and inversely proportional to the distance between the plates.
Explanation: 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.
13
Q
- (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.
A
C— .0125 μF.
Explanation: When a 0.02-microfarad, a 0.05-microfarad, and a 0.10-microfarad capacitor are connected in series, the total capacitance is 0.0125 microfarad
14
Q
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— .170 μF.
B— 0.125 μF.
C— .0125 μF.
A
A— .170 μF.
Explanation: Use the formula C T = C 1 + C 2 + C 3. 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 µF
15
Q
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 10 to the power of 6
A
A— multiplying farads by 10 to the power of 6
One farad is equal to 10 6 microfarads. Example: 2 farads is equal to 2 × 10 6 (2,000,000) microfarads.
16
Q
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
A— multiplying farads by 10 to the power of 12
Explanation: One farad is equal to 10 12 picofarads. Example: 2 farads is equal to 2 × 10 12 (2,000,000,000) picofarads.
17
Q
- 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.
A
B— effective values.
Explanation: 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.
18
Q
- When different rated capacitors are connected in parallel in a circuit, the total capacitance is (Note: C T = C 1 + C 2 + C 3 . . .)
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.
A
C— equal to the sum of all the capacitances.
Explanation: When capacitors are connected in parallel, the effective area of the plates combine, and the total capacitance is the sum of the individual capacitances.
19
Q
- 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: L T = L 1 + L 2 + L 3 . . .)
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
C— equal to the sum of the individual inductances.
Explanation: 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.
20
Q
- (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
A— less than the inductance of the lowest rated inductor.
Explanation: 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.
21
Q
- What is the total capacitance of a certain circuit containing three capacitors with capacitances of .25 microfarad, .03 microfarad, and .12 microfarad, respectively? (Note: C T = C 1 + C 2 + C 3 )
A— .4 μF.
B— .04 pF.
C— .04 μF.
A
A— .4 μF.
When three capacitors are connected in parallel, their total capacitance is the sum of the individual capacitances.
22
Q
- 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.
A
C— Two lights requiring 3 amperes each in a 24-volt parallel system.
Explanation:
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.
23
Q
- 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.
A
C— 450 watts.
Explanation: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 anticollision light requires 72 watts.
The total power the generator must produce is 451 watts.
24
Q
- 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
A— 1,000 watts.
Explanation: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.
25
8018.
How many amperes will a 28-volt generator be required to supply to a circuit containing five lamps in parallel, three of which have a resistance of 6 ohms each and two of which have a resistance of 5 ohms each?
A— 1.11 amperes.
B— 1 ampere.
C— 25.23 amperes.
C— 25.23 amperes.
A current of 4.67 amps flows through each of the three six-ohm lamps. And a current of 5.6 amps flows through each of the 5-ohm lamps. Since all of these lamps are in parallel, the total current is the sum of the currents flowing through each lamp. The total current is 25.21 amps.
26
8019.
A 24-volt, 1-horsepower DC electric motor that is 80 percent efficient requires 932.5 watts. How much power will a 12-volt, 1-horsepower DC electric motor that is 75 percent efficient require? (Note: 1 horsepower = 746 watts)
A— 932.5 watts.
B— 1,305.5 watts.
C— 994.6 watts.
C— 994.6 watts.
Explanation: When we know the horsepower output and the efficiency of an electric motor, the voltage does not enter into the computation. To find the number of watts required, divide the wattage for the total horsepower by the decimal equivalent of the efficiency. 746 ÷ 0.75 = 994.6 watts
27
8020.
The potential difference between two conductors which are insulated from each other is measured in
A— volts.
B— amperes.
C— coulombs.
A— volts.
The potential difference between two conductors is a measure of the electrical pressure difference between the conductors. Electrical pressure is measured in volts.
28
8020-1.
Which effect does not apply to the movement of electrons flowing in a conductor?
A— Magnetic energy.
B— Thermal energy.
C— Static energy.
C— Static energy.
Explanation:Current flowing through a conductor produces a magnetic field and also dissipates thermal energy.
29
8021.
A 24-volt source is required to furnish 48 watts to a parallel circuit consisting of four resistors of equal value. What is the voltage drop across each resistor?
A— 12 volts.
B— 6 volts.
C— 24 volts.
C— 24 volts.
Explanation: Since the resistors are all in parallel across the 24-volt power source, each resistor has the entire 24 volts dropped across it.
30
8022.
When calculating power in a reactive or inductive AC circuit, the true power is
A— more than the apparent power.
B— less than the apparent power in a reactive circuit and more than the apparent power in an inductive circuit.
C— less than the apparent power.
C— less than the apparent power.
Explanation:True power in an AC circuit is the product of the circuit voltage and only that part of the current in phase with the voltage. Apparent power is the circuit voltage multiplied by all of the current. True power is always less than the apparent power in a reactive circuit which is any AC circuit containing either inductance or capacitance.
31
8023. (Refer to Figure 4.) How much power is being furnished to the circuit?
A— 575 watts.
B— 2,875 watts.
C— 2,645 watts.
C— 2,645 watts.
Explanation: This is a resistive circuit. The power is the product of the square of the current times the resistance. P = I 2 × R = 23 2 × 5 = 2,645 watts
32
8024.
(Refer to Figure 5.)
What is the impedance of an AC-series circuit consisting of an inductor with a reactance of 10 ohms, a capacitor with a reactance of 4 ohms, and a resistor with a resistance of 8 ohms?
A— 22 ohms.
B— 5.29 ohms.
C— 10 ohms.
C— 10 ohms.
Explanation:The total reactance in this circuit is the difference between the inductive reactance and the capacitive reactance. Total reactance is 10 – 4 = 6 ohms. The impedance is the square root of the resistance squared plus the reactance squared. This is the square root of 64 plus 36, or the square root of 100. The circuit impedance is 10 ohms.
33
8025.
(Refer to Figure 6.) If resistor R 5 is disconnected at the junction of R 4 and R 3 as shown, what will the ohmmeter read?
A— 2.76 ohms.
B— 3 ohms.
C— 12 ohms.
B— 3 ohms.
With resistor R 5 disconnected, the ohmmeter reads the parallel resistance of R 1 and R 2 in parallel with R 4 and R 3 , which are in series. The total resistance is found by the formula:
34
8026.
(Refer to Figure 7.) If resistor R 3 is disconnected at terminal D, what will the ohmmeter read?
A— Infinite resistance.
B— 10 ohms.
C— 20 ohms.
A— Infinite resistance.
Explanation: When resistor R 3 is disconnected at terminal D, it is isolated from the rest of the circuit, and the ohmmeter will read only the resistance of R 3 . Because R 3 is open (it has a break in it), its resistance is infinite.
35
8027.
(Refer to Figure 8.) With an ohmmeter connected into the circuit as shown, what will the ohmmeter read?
A— 20 ohms.
B— Infinite resistance.
C— 10 ohms.
C— 10 ohms.
Explanation:The ohmmeter will read the resistance of R 1 and R 2 in parallel; this is 10 ohms. The open circuit (break) in resistor R 3 gives it an infinite resistance, and it does not affect the reading of the ohmmeter.
36
8028.
(Refer to Figure 9.) How many instruments (voltmeters and ammeters) are installed correctly?
A— Three.
B— One.
C — Two.
C — Two.
Explanation: The first ammeter is installed across the voltage. This is wrong; the ammeter will burn out. The first voltmeter will measure the source voltage (the voltage of the battery), but its polarity is wrong. It will read backward. The voltmeter across the light bulb is installed correctly. The ammeter in series with the light bulb and the battery is correct. It will read the current flowing through the light bulb.
37
8029.
The correct way to connect a test voltmeter in a circuit is
A— in series with a unit.
B— between the source voltage and the load.
C— in parallel with a unit.
C— in parallel with a unit
Explanation:A voltmeter must always be connected in a circuit in parallel with the unit whose voltage is to be measured.
38
8029-1.
What will a voltmeter read if properly connected across a closed switch in a circuit with electrical power on?
A— Voltage drop in the component(s) the switch is connected to.
B— System voltage.
C— Zero voltage.
C— Zero voltage.
Explanation: When a voltmeter is connected across a closed switch in perfect condition or a good fuse, it will read zero voltage. A voltage drop of up to 0.2 volts is acceptable with maximum circuit current flow through the switch.
39
8029-2.
What does the letter Q symbolize when measuring electrical charge?
A— Farad.
B— Electron.
C— Coulomb.
C— Coulomb.
Explanation:The general formula for capacitance in terms of charge and voltage is C = Q/E, where:
C = Capacitance measured in farads
E = Applied voltage measured in volts
Q = Charge measured in coulombs
40
8030.
Which term means .001 ampere?
A— Microampere.
B— Kiloampere.
C— Milliampere.
C— Milliampere.
Explanation:The metric prefix “milli-” means one thousandth. 0.001 ampere is one milliampere.
41
8031. A cabin entry light of 10 watts and a dome light of 20 watts are connected in parallel to a 30-volt source. If the voltage across the 10-watt light is measured, it will be
A— equal to the voltage across the 20-watt light.
B— half the voltage across the 20-watt light.
C— one-third of the input voltage.
A— equal to the voltage across the 20-watt light.
Explanation: When lights are connected in parallel across a voltage source, the voltage across each of the lights will be the same as the voltage of the source.
42
8032. A 14-ohm resistor is to be installed in a series circuit carrying .05 ampere. How much power will the resistor be required to dissipate?
A— At least .70 milliwatt.
B— At least 35 milliwatts.
C— Less than .035 watt.
B— At least 35 milliwatts
Explanation: Power dissipated in a resistor is found by multiplying its resistance by the square of the current. P = I 2 × R = 0.05 2 × 14 = 0.035 watts 0.035 watts is 35 milliwatts.
43
8033.
.002 kV equals
A— 20 volts.
B— 2.0 volts.
C— .2 volt.
B— 2.0 volts.
Explanation: One kilovolt (kV) is 1,000 volts. Two thousandths (.002) of a kilovolt is equal to 2.0 volts.
44
8034.
(Refer to Figure 10.) What is the measured voltage of the series-parallel circuit between terminals A and B?
A— 1.5 volts.
B— 3.0 volts.
C— 4.5 volts.
B— 3.0 volts.
Explanation: The two batteries on the left side are connected in series and the two batteries on the right side are connected in series. The two pairs of batteries are connected in parallel. The series connections between terminals A and B give this circuit a voltage of 3.0 volts.
45
8035.
(Refer to Figure 64.) A 24-volt source is required to furnish 48 watts to a parallel circuit consisting of two resistors of equal value. What is the value of each resistor?
A— 24 ohms.
B— 12 ohms.
C— 6 ohms.
A— 24 ohms.
Explanation: To solve this problem, first find the total resistance of the circuit.
R T = E 2 ÷ P
= 24 2 ÷ 48
= 12 Ω
There are two resistors of equal value in parallel that provide this resistance, therefore each resistor must have a resistance of twice this value, or 24 ohms.
46
8036.
Which requires the most electrical power? (Note: 1 horsepower = 746 watts)
A— A 1/5-horsepower, 24-volt motor which is 75 percent efficient.
B— Four 30-watt lamps arranged in a 12-volt parallel circuit.
C— A 24-volt anticollision light circuit consisting of two light assemblies which require 3 amperes each during operation.
A— A 1/5-horsepower, 24-volt motor which is 75 percent efficient.
Explanation: The 1/5-horsepower motor operating at 75 percent efficiency uses 198.93 watts of power. The four 30-watt lamps use 120 watts of power. The anticollision light circuit uses 144 watts of power.
47
8037. What unit is used to express electrical power?
A— Volt.
B— Watt.
C— Ampere.
B— Watt.
Explanation:A watt is a measure of electrical power. A volt is a measure of electrical pressure. An ampere is a measure of electrical current flow
48
8037-1.
What is the basic unit of electrical quantity?
A— Electromotive force.
B— Ampere.
C— Coulomb.
C— Coulomb.
The coulomb is the basic unit of electrical quantity. One coulomb is equal to 6.28 × 10 18 electrons.
49
8038.
What is the operating resistance of a 30-watt light bulb designed for a 28-volt system?
A— 1.07 ohms.
B— 26 ohms.
C— 0.93 ohm.
B— 26 ohms.
Explanation: A 30-watt light bulb operating in a 28-volt electrical system has a hot resistance (operating resistance) of 26.13 ohms.
50
8039.
When referencing resistance in a parallel DC cir- cuit, which of the following statements is true?
A— The current is equal in all portions of the circuit.
B— The total current is equal to the sum of the currents through the individual branches of the circuit.
C— The current in amperes can be found by dividing the source voltage in volts by the sum of the resistors in ohms.
B— The total current is equal to the sum of the currents through the individual branches of the circuit.
Explanation: According to Kirchhoff’s current law, the current flowing in a parallel circuit is equal to the sum of the currents flow- ing through each of the individual branches of the circuit.
51
8039-1.
Which of the following is commonly used as a rectifier in electrical circuits?
A— Diodes and anodes.
B— Diodes and cathodes.
C— Diodes.
C— Diodes.
Explanation:Diodes are two element electronic components that act as electron check valves. They allow electrons to pass freely in one direction but block their flow in the opposite direction. They are used as rectifiers in electrical circuits
52
8040.
Diodes are used in electrical power supply circuits primarily as
A— switches.
B— rectifiers.
C— relays.
B— rectifiers.
Explanation: A diode (either a semiconductor diode or an electron-tube diode) is an electron check valve. A diode allows electrons to pass in one direction, but blocks their flow in the reverse direction. This is the action of a rectifier.
53
8041.
Transfer of electric energy from one circuit to another without the aid of electrical connections
A— is called induction.
B— is called capacitance.
C— can cause excessive arcing and heat, and as a result is practical for use only with low voltages/amperages.
A— is called induction.
Explanation:The continually changing current in an AC circuit causes a changing magnetic field to cut across conductors in an adjacent circuit. When the changing field cuts across a conductor, it induces a voltage in it. Induction allows electrical energy to be transferred from one circuit to another without the aid of electrical connections.
54
8042.
If three resistors of 3 ohms, 5 ohms, and 22 ohms are connected in series in a 28-volt circuit, how much current will flow through the 3-ohm resistor?
A— 9.3 amperes.
B— 1.05 amperes.
C— 0.93 ampere.
C— 0.93 ampere.
Explanation: In a series circuit, the total resistance is the sum of the individual resistances. In this circuit, the total resistance is 30 ohms. All of the current flows through each resistor. Therefore, the current through each resistor is 0.93 amp.
55
8043.
A circuit has an applied voltage of 30 volts and a load consisting of a 10-ohm resistor in series with a 20-ohm resistor. What is the voltage drop across the 10-ohm resistor?
A— 10 volts.
B— 20 volts.
C— 30 volts.
A— 10 volts.
In a series circuit, the voltage drop across each resistor is determined by its resistance. In this circuit, the total resistance is 30 ohms and the total voltage is 30 volts. One amp of current flows through each resistor and this gives a 10-volt drop across the 10-ohm resistor.
56
8044.
(Refer to Figure 11.) Find the total current flowing in the wire between points C and D.
A— 6.0 amperes.
B— 2.4 amperes.
C— 3.0 amperes.
C— 3.0 amperes.
57
8045.
(Refer to Figure 11.) Find the voltage across the 8- ohm resistor.
A— 8 volts.
B— 20.4 volts.
C— 24 volts.
C— 24 volts.
Explanation:The 8-ohm resistor, R 1 , is across the full 24 volts of the battery
58
8045-1.
In a parallel circuit with four 6-ohm resistors across a 24-volt battery, what is the total voltage across resistor-three (VR3) in the circuit?
A— 6 volts.
B— 18 volts.
C— 24 volts.
C— 24 volts.
Explanation: In a parallel circuit the source or battery voltage is applied to each of the individual resistors.
59
8045-2.
If each cell, connected in series, equals 2 volts, how would a 12-cell lead acid battery be rated?
A— 24 volts.
B— 12 volts.
C— 6 volts.
A— 24 volts.
Explanation: The voltage of a battery is determined by the number of cells connected in series to form the battery. Although the voltage of one lead-acid cell just removed from a charger is approximately 2.2 volts, a lead-acid cell is normally rated at approximately 2 volts. A battery rated at 12 volts consists of 6 lead-acid cells connected in series, and a battery rated at 24 volts is composed of 12 cells.
60
8046.
(Refer to Figure 12.) Find the total resistance of the circuit.
A— 16 ohms.
B— 2.6 ohms.
C— 21.2 ohms.
C— 21.2 ohms.
Explanation: This problem can be solved in four steps:
1. Combine resistors R 4 and R 5 in parallel. R 4–5 = R 4 × R 5 R 4 + R 5 = 12 × 6 12 + 6 = 72 18 = 4Ω
2. Combine resistor R 2 with R 4–5 in series. R 2–4–5 = R 2 + R 4–5 = 12 + 4 = 16Ω
3. Combine resistors R 2–4–5 with R 3 in parallel. R 2–3–4–5 = R 2–4–5 × R 3 R 2–4–5 + R 3 = 16 × 4 16 + 4 = 64 20 = 3.2Ω
4. Combine resistors R 2–3–4–5 with R 1 in series. R T = R 2–3–4–5 + R 1 = 3.2 + 18 = 21.2Ω
61
8047.
Which is correct in reference to electrical resis- tance?
A— Two electrical devices will have the same combined resistance if they are connected in series as they will have if connected in parallel.
B— If one of three bulbs in a parallel lighting circuit is removed, the total resistance of the circuit will become greater.
C— An electrical device that has a high resistance will use more power than one with a low resistance with the same applied voltage.
B— If one of three bulbs in a parallel lighting circuit is removed, the total resistance of the circuit will become greater.
Explanation: In a parallel electrical circuit, each bulb provides a path for current to flow. The more paths there are, the less the circuit resistance will be. When one bulb is removed, the circuit resistance increases.
62
8048.
What happens to the current in a voltage step-up transformer with a ratio of 1 to 4?
A— The current is stepped down by a 1 to 4 ratio.
B— The current is stepped up by a 1 to 4 ratio.
C— The current does not change.
A— The current is stepped down by a 1 to 4 ratio.
Explanation: The power (voltage times current) in the secondary of a transformer is equal to the power in the primary. When the voltage in the secondary is four times that in the pri- mary, the current in the secondary is one fourth of that in the primary.
63
8049.
(Refer to Figure 13.) Determine the total current flow in the circuit.
A— 0.2 ampere.
B— 1.4 amperes.
C— 0.8 ampere.
B— 1.4 amperes.
64
8049-1.
In a parallel circuit with three 6-ohm resistors across a 12-volt battery, what is the total current (I t ) value in the circuit?
A— 2 amps.
B— 6 amps.
C— 12 amps.
B— 6 amps.
Explanation: The total resistance of this circuit is 2 ohms. The total current flowing in this circuit is 6 amps.
65
8050.
(Refer to Figure 14.) The total resistance of the circuit is
A— 25 ohms.
B— 35 ohms.
C— 17 ohms.
C— 17 ohms.
66
8051.
Which of these will cause the resistance of a conductor to decrease?
A— Decrease the length or the cross-sectional area.
B— Decrease the length or increase the cross-sectional area.
C— Increase the length or decrease the cross-sectional area.
B— Decrease the length or increase the cross-sectional area.
Explanation: The resistance of a conductor varies directly as its length, inversely as its cross-sectional area, and directly with the resistivity of its material. Either decreasing the length or increasing the cross-sectional area of a conductor will cause its resistance to decrease.
67
8052.
Through which material will magnetic lines of force pass the most readily?
A— Copper.
B— Iron.
C— Aluminum.
B— Iron.
Explanation: The permeability of a material is a measure of the ease with which lines of magnetic force can pass through it. Iron has the highest permeability of all the metals listed in this question.
68
8053.
(Refer to Figure 64.) A 48-volt source is required to furnish 192 watts to a parallel circuit consisting of three resistors of equal value. What is the value of each resistor?
A— 36 ohms.
B— 4 ohms.
C— 12 ohms.
A— 36 ohms.
Explanation: The total resistance needed to dissipate 192 watts of electrical power from a 48-volt source is 12 ohms. R T = E 2 ÷ P = 48 2 ÷ 192 = 12Ω Since three resistors in parallel give 12 ohms, each resistor must have a resistance three times this value.
69
8054.
Which is correct concerning a parallel circuit?
A— Total resistance will be smaller than the smallest resistor.
B— Total resistance will decrease when one of the resistances is removed.
C— Total voltage drop is the same as the total resistance.
A— Total resistance will be smaller than the smallest resistor.
Explanation: In a parallel circuit each resistor forms a path for the current to follow and the total resistance is always smaller than that of the smallest resistor.
70
8055.
The voltage drop in a circuit of known resistance is dependent on
A— the voltage of the circuit.
B— only the resistance of the conductor and does not change with a change in either voltage or amperage.
C— the amperage of the circuit.
C— the amperage of the circuit.
Explanation: The voltage drop across a circuit is determined by two things: the resistance of the circuit and the amount of current flowing through it (the amperage). In this question, the resistance of the circuit is fixed; therefore, the voltage drop is determined by the amperage in the circuit.
71
8056.
A thermal switch, or thermal protector, as used in an electric motor, is designed to
A— close the integral fan circuit to allow cooling of the motor.
B— open the circuit in order to allow cooling of the motor.
C— reroute the circuit to ground.
B— open the circuit in order to allow cooling of the motor.
Explanation: A thermal switch is another name for a built-in thermal circuit breaker. This is a circuit protection device that opens the circuit when the windings of the motor get too hot. If the motor overheats for any reason, the thermal switch will open the power circuit to the motor and allow the motor to cool.
72
8057.
(Refer to Figure 15.) With the landing gear retracted, the red indicator light will not come on if an open occurs in wire.
A— 19.
B — 7.
C — 17.
A— 19.
Explanation: With the landing gear retracted, the red indicator light will not come on if there is an open in wire 19. Wire 19 supplies power from the bus to the red indicator light, through the up-limit switch and through wire 8. Wire 7 supplies power to the red indicator light for the press-to-test circuit. Wire 17 supplies power to the green indicator light for its press-to-test circuit.
73
8058.
(Refer to Figure 15.) Wire 7 is used to
A— close the PUSH-TO-TEST circuit.
B— open the UP indicator light circuit when the landing gear is retracted.
C— close the UP indicator light circuit when the landing gear is retracted.
A— close the PUSH-TO-TEST circuit.
Explanation: Wire 7 supplies power to both the red and green indicator lights from the 5-amp circuit breaker. When either of the press-to-test lamps is pushed, the circuit is completed to ground, allowing the lamp to light up as long as it remains pressed. Press-to-test lamps are used to allow the pilot to ascertain that the bulb in the lamp is good.
74
8059.
(Refer to Figure 15.) When the landing gear is down, the green light will not come on if an open occurs in wire
A — 7.
B— 6.
C — 17.
B— 6.
Explanation: The green light will not come on when the landing gear is down if there is an open in wire 6. (Note: The switches in Figure 15 are shown in the position for the gear down and the airplane on the ground.) Wire 7 supplies power to both the red and green indicator lights from the 5-amp circuit breaker for the push-to-test circuit. Wire 17 supplies power to the green indicator light for its push-to-test circuit.
75
8060.
(Refer to Figure 16.) What will be the effect if the PCO relay fails to operate when the left-hand tank is selected?
A— The fuel pressure crossfeed valve will not open.
B— The fuel tank crossfeed valve open light will illuminate.
C— The fuel pressure crossfeed valve open light will not illuminate.
C— The fuel pressure crossfeed valve open light will not illuminate
Explanation: PCO relay provides 24-volt DC power from the bus through a 5-amp circuit breaker to the fuel pressure crossfeed valve open caution warning light in the cockpit. If PCO relay fails to operate, contacts 13 will not complete the circuit and the light will not illuminate, but the rest of the system will operate normally
76
8061.
(Refer to Figure 16.) The TCO relay will operate if 24-volts DC is applied to the bus and the fuel tank selector is in the
A— right-hand tank position.
B— crossfeed position.
C— left-hand tank position
B— crossfeed position.
Explanation: The fuel-selector switch must be in the crossfeed position for the relay TCO to close contacts 14. When the tank selector is put in the crossfeed position, contacts 17 of the FCF relay close, and current flows from the 24-volt DC bus, through a 5-amp circuit breaker, into the “open” coils of the fuel tank crossfeed valve motor. When the motor opens this valve, contacts 19 close and current flows to theTCO relay closing contacts 14.
77
8062.
(Refer to Figure 16.) With power to the bus and the fuel selector switched to the right-hand tank, how many relays in the system are operating?
A— Three.
B — Two.
C— Four.
A— Three.
Explanation: When the fuel selector switch is in the right-hand tank position and power is supplied from the bus, three relays (RTS, PCO, and TCC) will actuate.
Current flowing through the fuel-selector switch energizes the coil for relay RTS. This closes contacts 8 and opens contacts 7.
Current from the bus flows through closed contacts 5 of relay LTS, through contacts 8 of relay RTS that has just closed, and through the “open” winding of the fuel pressure crossfeed valve motor.
As soon as this motor fully opens the valve, contacts 12 close and energize the PCO relay to close contacts 13 in the circuit for the fuel pressure crossfeed valve open light.
Current flows from the bus through the right-hand 5-amp circuit breaker and through the normally closed contacts 18 of the FCF relay, through contacts 20 in the fuel tank crossfeed valve to the coil of relay TCC which opens its contacts 16.
78
8063.
(Refer to Figure 16.) When electrical power is applied to the bus, which relays are energized?
A— PCC and TCC.
B— TCC and TCO.
C— PCO and PCC.
A— PCC and TCC.
As soon as electrical power is supplied to the 24-volt bus, two relays, PCC and TCC, are energized. Current flows from the bus through the left-hand, 5-amp circuit breaker and through contacts 5, 7, and 9 of the relays controlled by the fuel selector switch. It flows from contacts 9, through contacts 11 in the fuel pressure crossfeed valve, to the coil of the PCC relay.
At the same time, current flows through the right-hand 5-amp circuit breaker, through contacts 18 and then through contacts 20 in the fuel tank crossfeed valve to the coil of the TCC relay.
79
8064.
(Refer to Figure 16.) Energize the circuit with the fuel tank selector switch selected to the left-hand position. Using the schematic, identify the switches that will change position.
A— 5, 9, 10, 11, 12, 13, and 15.
B— 3, 5, 6, 7, 11, and 13.
C— 5, 6, 11, 12, 13, 15, and 16.
C— 5, 6, 11, 12, 13, 15, and 16.
When power is supplied to the bus with the fuel selector switch in the left-hand position, seven switches change their position (5, 6, 11, 12, 13, 15, and 16).
Current flows from the bus through the left-hand, 5-amp circuit breaker, through the left-hand tank position of the fuel selector switch, into the coil of the LTS relay. This changes the position of switches 5 and 6.
Since current can no longer flow through switch 5, the coil of the PCC relay is de-energized and this changes the position of switch 15.
Current can now flow through switch 6 into the “open” motor coil of the fuel-pressure crossfeed valve. When this valve is open, switches 11 and 12 change their position.
Current now flows through switch 12 to the coil of the PCO relay. This closes switch 13.
At the same time, all of this is happening, current is flowing from the right-hand, 5-amp circuit breaker, through switches 18 and 20, to the coil of the TCC relay. When the TCC relay is energized, it changes the position of switch 16.
80
8065.
(Refer to Figure 17.) Which of the components is a potentiometer?
A— 5.
B— 3.
C — 11.
B— 3.
Explanation:
The component identified as 3 is a potentiometer.
The component identified as 5 is a variable capacitor.
The component identified as 11 is an inductor (choke).
81
8066.
(Refer to Figure 17.) The electrical symbol represented at number 5 is a variable
A— inductor.
B— resistor.
C— capacitor.
C— capacitor.
Explanation: A variable capacitor is shown by symbol 5. An inductor is shown by symbol 11. A variable resistor (potentiometer) is shown by symbol 3.
82
8067.
(Refer to Figure 18.) When the landing gear is up and the throttles are retarded, the warning horn will not sound if an open occurs in wire
A— 4.
B— 2.
C— 9.
A— 4.
Explanation: Wire 4 is in the landing gear warning horn ground circuit. If it is open, the warning horn cannot sound when the landing gears are up and the throttles are retarded. Wire 2 is in the red warning-light circuit and not in the warning horn circuit. Wire 9 is in the green light circuit.
83
8068.
(Refer to Figure 18.) The control valve switch must be placed in the neutral position when the landing gear is down to
A— permit the test circuit to operate.
B— prevent the warning horn from sounding when the throttles are closed.
C— remove the ground from the green light.
B— prevent the warning horn from sounding when the throttles are closed.
If the control-valve switch were not in the neutral position, as shown here, the landing gear warning horn would sound when either throttle is closed and both landing gears are down. The ground for the horn would be supplied through wire 6, the closed throttle switch, wires 5 and 10, the control valve switch, wire 11, the right gear switch, wire 3, the left gear switch, and wire 14 to ground.
84
8069.
(Refer to Figure 19.) Under which condition will a ground be provided for the warning horn through both gear switches when the throttles are closed?
A— Right gear up and left gear down.
B— Both gears up and the control valve out of neutral.
C— Left gear up and right gear down.
C— Left gear up and right gear down.
Explanation: A ground for the landing gear warning horn is provided through both of the gear switches only when the right landing gear is down and the left landing gear is up. Ground current from the warning horn flows through wire 11, through a closed throttle switch, through wire 12, through the left gear switch in its UP position, through wire 5, through the right gear switch in its DOWN position, to ground.
85
8070.
(Refer to Figure 19.) When the throttles are retarded with only the right gear down, the warning horn will not sound if an open occurs in wire
A— 5.
B— 13.
C— 6.
A— 5.
Explanation: Wire 5 is in the landing gear warning horn ground circuit. If there is an open in this wire, the warning horn cannot get a ground when the right landing gear is down and the left gear is not down. Wire 13 is not in the warning horn circuit if the left landing gear is not down. Wire 6 is in the warning horn test circuit.
86
8071.
(Refer to Figure 19.) When the landing gears are up and the throttles are retarded, the warning horn will not sound if an open occurs in wire
A— 5.
B — 7.
C— 6.
C— 6.
Explanation: Wire 6 is in the ground circuit for the landing gear warning horn. The horn will not sound if there is an open in this wire when the throttles are retarded and the landing gears are up. Wire 5 is not in the warning horn ground circuit when the right landing gear is up. Wire 7 is not in the warning horn ground circuit.
87
8072.
When referring to an electrical circuit diagram, what point is considered to be at zero voltage?
A— The circuit breaker.
B— The switch.
C— The ground reference.
C— The ground reference.
Explanation: The ground reference, shown on a schematic diagram as a triangular-shaped series of parallel lines, is the point considered to be at zero voltage. All voltages, both positive and negative, are measured from this ground reference.
88
8072-1.
What is the purpose of the ground symbol used in electrical circuit diagrams?
A— To show that there is common bus for connection of the source of electrical energy to the load.
B— To show the source of electrical energy for the load.
C— To show that there is a return path for the current between the source of electrical energy and the load.
C— To show that there is a return path for the current between the source of electrical energy and the load.
Explanation: The ground symbol used on electrical schematic diagrams indicates that there is a return path for the current between the source of electrical energy and the load.
89
8073.
(Refer to Figure 20.) Troubleshooting an open circuit with a voltmeter as shown in this circuit will
A— permit current to flow and illuminate the lamp.
B— create a low resistance path and the current flow will be greater than normal.
C— permit the battery voltage to appear on the voltmeter.
C— permit the battery voltage to appear on the voltmeter.
Explanation: An open in the resistor will cause an infinite resistance. No current can flow in the circuit. Since no current flows, there is no voltage drop across the switch or the lamp, and the entire battery voltage is read on the voltmeter.
90
8074.
(Refer to Figure 21.) Which symbol represents a variable resistor?
A— 2.
B — 1.
C— 3.
A— 2.
Explanation: Symbol 2 is a variable resistor (rheostat). This is not the most widely used symbol as it can be used only for a rheostat (a variable resistor with only two connections) and not for a potentiometer. Symbol 1 is a tapped resistor. Symbol 3 is a tapped resistor—tapped at two places.
91
8075.
In a P-N-P transistor application, the solid-state device is turned on when the
A— base is negative with respect to the emitter.
B— base is positive with respect to the emitter.
C— emitter is negative with respect to the base.
A— base is negative with respect to the emitter.
A P-N-P transistor conducts between the emitter and collector (is turned on) when a small amount of current flows into the base. This current flows when the emitter-base junction is forward biased. It is forward biased when the base is negative with respect to the emitter.
92
8076.
In an N-P-N transistor application, the solid-state device is turned on when the
A— emitter is positive with respect to the base.
B— base is negative with respect to the emitter.
C— base is positive with respect to the emitter.
C— base is positive with respect to the emitter.
An N-P-N transistor conducts between the emitter and collector (is turned on) when a small amount of current f lows from the base. This current flows when the emitter base junction is forward biased. It is forward biased when the base is positive with respect to the emitter.
93
8077. Typical application for zener diodes is as
A— full-wave rectifiers.
B— half-wave rectifiers.
C— voltage regulators.
C— voltage regulators.
A zener diode is a special type of semiconductor diode that is designed to operate with current flowing through it in its reverse direction. When a specific amount of inverse voltage is applied between the cathode and anode, the diode breaks down and conducts in its reverse direction. This principle is used as the voltage-sensing element in a voltage regulator.
94
8078.
(Refer to Figure 22.) Which illustration is correct concerning bias application and current (positive charge) flow?
A— 1.
B— 2.
C— 3.
A— 1.
In 1, the base of this N-P-N transistor is positive with respect to the emitter (the emitter-base junction is forward biased). Base-emitter current flows and therefore collector emitter current flows as is shown by the current-flow arrow. In 2, the base and emitter of this N-P-N transistor have the same polarity and no emitter-base current flows. There is no flow between the emitter and the collector. In 3, the base and emitter of this P-N-P transistor have the same polarity, and no emitter-base current flows. There is no flow between the emitter and the collector.
95
8079.
Forward biasing of a solid-state device will cause the device to
A— conduct via zener breakdown.
B— conduct.
C— turn off.
B— conduct.
When a solid-state device such as a diode is forward biased, the N-material is negative with respect to the P- material. A forward-biased device will conduct.
96
8080.
(Refer to Figure 23.) If an open occurs at R 1 , the light
A— cannot be turned on.
B— will not be affected.
C— cannot be turned off.
C— cannot be turned off.
If there is an open circuit in R 1 , the base and the collector will have the same voltage. The base current will be maximum, and the light cannot be turned off.
97
8081.
(Refer to Figure 23.) If R 2 sticks in the up position, the light will
A— be on full bright.
B— be very dim.
C— not illuminate.
A— be on full bright.
An N-P-N transistor circuit such as this conducts the maximum amount of current between the emitter and collector (the lamp is burning brightest) when the maximum amount of current is flowing from the base. This occurs when the base is most positive with respect to the emitter. This would be the case if R 2 were stuck in the up position
98
8082.
(Refer to Figure 24.) Which statement concerning the depicted logic gate is true?
A— Any input being 1 will produce a 0 output.
B— Any input being 1 will produce a 1 output.
C— All inputs must be 1 to produce a 1 output.
B— Any input being 1 will produce a 1 output.
Explanation: This is an OR gate. Any time there is a 1 on any of the inputs, there will be a 1 on the output.
99
8083.
(Refer to Figure 25.) In a functional and operating circuit, the depicted logic gate’s output will be 0
A— only when all inputs are 0.
B— when all inputs are 1.
C— when one or more inputs are 0.
C— when one or more inputs are 0.
Explanation: This is an AND gate. Any time one or more of the inputs do not have a 1 on them (have a 0 on them), the output will be 0. For the output to be 1, all of the inputs must be 1.
100
8084.
(Refer to Figure 26.) Which of the logic gate output conditions is correct with respect to the given inputs?
A— 1.
B— 2.
C— 3.
B— 2.
Explanation: These are EXCLUSIVE OR (XOR) gates. There will be a 1 on the output when one, and only one, input has a 1 on it. View 1 is incorrect. It has a 1 on both inputs so the output should be 0. View 2 is correct. Only one of the inputs has a 1 on it and the output is a 1. View 3 is incorrect. It has a 1 on one of its inputs, but the output is shown to be 0.
101
8084-1.
Which of the following logic gates will provide an active high out only when all inputs are different?
A— XNOR.
B— NAND.
C— XOR.
C— XOR.
The output of the XOR gate will only be at an active high when one and only one of the two inputs is high. If both inputs are low, or both inputs are high, then the output of the gate will be low.
102
8085.
A lead-acid battery with 12 cells connected in series (no-load voltage = 2.1 volts per cell) furnishes 10 amperes to a load of 2-ohms resistance. The internal resistance of the battery in this instance is
A— 0.52 ohm.
B— 2.52 ohms.
C— 5.0 ohms.
A— 0.52 ohm.
Explanation: This battery has a no-load voltage of 12 × 2.1 = 25.2 volts. When it supplies 10 amperes to a 2-ohm load, the loaded voltage is 10 × 2 = 20 volts. The voltage dropped across its internal resistance is 25.2 – 20.0 = 5.2 volts. The internal resistance of the battery is found by dividing the voltage dropped across it by the load current. 5.2 ÷ 10 = 0.52 ohms.
103
8086.
If electrolyte from a lead-acid battery is spilled in the battery compartment, which procedure should be followed?
A— Apply boric acid solution to the affected area followed by a water rinse.
B— Rinse the affected area thoroughly with clean water.
C— Apply sodium bicarbonate solution to the affected area followed by a water rinse.
C— Apply sodium bicarbonate solution to the affected area followed by a water rinse.
Explanation: The electrolyte in a lead-acid battery is a solution of sulfuric acid and water. If any is spilled in the battery compartment, it should be neutralized with a solution of sodium bicarbonate (baking soda) and water, and the area rinsed with fresh water.
104
8087.
Which statement regarding the hydrometer reading of a lead-acid storage battery electrolyte is true?
A— The hydrometer reading does not require a temperature correction if the electrolyte temperature is 80°F.
B— A specific gravity correction should be added to the hydrometer reading if the electrolyte temperature is below 59°F.
C— The hydrometer reading will give a true indication of the capacity of the battery regardless of the electrolyte temperature.
A— The hydrometer reading does not require a temperature correction if the electrolyte temperature is 80°F.
Explanation: When testing the specific gravity of the electrolyte of a lead-acid battery, the temperature must be taken into consideration. No correction is necessary when the electrolyte temperature is between 70°F and 90°F. A correction of 0.4 should be added to the specific gravity reading for each 10° above 80°F, and 0.4 should be subtracted from the specific gravity reading for each 10° below 80°F.
105
8088.
A fully charged lead-acid battery will not freeze until extremely low temperatures are reached because
A— the acid is in the plates, thereby increasing the specific gravity of the solution.
B— most of the acid is in the solution.
C— increased internal resistance generates sufficient heat to prevent freezing.
B— most of the acid is in the solution.
Explanation: When a lead-acid battery is charged, the sulfate radicals from both plates join with hydrogen atoms from the water in the electrolyte and form sulfuric acid. Sulfuric acid has a much lower freezing point than water, and the electrolyte in a fully charged battery has a freezing point much lower than that of the electrolyte in a discharged battery.
106
8089.
What determines the amount of current which will flow through a battery while it is being charged by a constant voltage source?
A— The total plate area of the battery.
B— The state-of-charge of the battery.
C— The ampere-hour capacity of the battery.
B— The state-of-charge of the battery.
Explanation: When a battery is charged by the constant-voltage method, a voltage somewhat higher than the open-circuit voltage of the battery is placed across the battery terminals. When the battery is in a low state of charge, its voltage is low and the constant-voltage charger will put a large amount of current into it. As the charge continues, the battery voltage rises and the current going into the battery decreases. When the battery is fully charged, only enough current f lows into it to compensate for the power lost in its internal resistance.
107
8090.
Which of the following statements is/are generally true regarding the charging of several aircraft batteries together?
1) Batteries of different voltages (but similar capacities) can be connected in series with each other across the charger, and charged using the constant current method.
2) Batteries of different ampere-hour capacity and same voltage can be connected in parallel with each other across the charger, and charged using the constant voltage method.
3) Batteries of the same voltage and same ampere-hour capacity must be connected in series with each other across the charger, and charged using the constant current method.
A— 3.
B— 2 and
C— 1 and 2.
C— 1 and 2.
Batteries having different voltages, but similar capacity can be connected in series and charged by the constant current method. Batteries having the same voltage but different ampere-hour capacity can be connected in parallel and charged at the same time, if they are charged by the constant-voltage method.
108
8091.
The method used to rapidly charge a nickel-cadmium battery utilizes
A— constant current and constant voltage.
B— constant current and varying voltage.
C— constant voltage and varying current.
C— constant voltage and varying current.
Explanation: Nickel-cadmium batteries can be charged rapidly by using constant voltage and varying current.
109
8092.
The purpose of providing a space underneath the plates in a lead acid batter’s cell container is to
A— prevent sediment buildup from contacting the plates and causing a short circuit.
B— allow for convection of the electrolyte in order to provide for cooling of the plates.
C— ensure that the electrolyte quantity ratio to the number of plates and plate area is adequate.
A— prevent sediment buildup from contacting the plates and causing a short circuit.
Explanation: There is a space left below the plates in a lead acid battery cell to allow for the accumulation of sediment, thus preventing the sediment from coming in contact with the plates and causing a short circuit.
110
8093.
Which condition is an indication of improperly torqued cell link connections of a nickel-cadmium battery?
A— Light spewing at the cell caps.
B— Toxic and corrosive deposits of potassium carbonate crystals.
C— Heat or burn marks on the hardware.
C— Heat or burn marks on the hardware.
Explanation: Nickel-cadmium batteries are made differently from lead-acid batteries in that the individual cells are removable and are connected together with metal straps secured to the tops of the cells with nuts. If the cell-link connections are not properly torqued, they will cause a high-resistance path for the current. They will become overheated and will show burn marks.
111
8094.
The presence of any small amount of potassium carbonate deposits on the top of nickel-cadmium battery cells in service is an indication of
A— normal operation.
B— excessive gassing.
C— plate sulfation.
A— normal operation.
Explanation: When a nickel-cadmium battery is fully charged, the battery becomes hot and the electrolyte bubbles. This causes some of the electrolyte to spew out of the top of the cell through the cell vent. When the water evaporates from the spewed-out electrolyte, it leaves a deposit of potassium carbonate, a white powder.
112
8095. What is the likely result, if any, of servicing and charging nickel-cadmium and lead-acid batteries together in the same service area?
A— The electrolytes in each battery are the same, so there are no significant differences.
B— The nickel-cadmium would not be affected, but the lead-acid battery could become contaminated.
C— Contamination of both types of batteries would occur.
C— Contamination of both types of batteries would occur.
Explanation: The chemistry of lead-acid and nickel-cadmium batteries is incompatible. Electrolyte from one type of battery will contaminate the other. For this reason, battery shops keep the two types of batteries separated. Separate charging systems, cleaning facilities, and installation tools should be used for each type of battery.
113
8095-1.
Which of the following best describes the operating principal in a nickel-cadmium battery installed in an aircraft?
A— At full charge, the electrolyte will be at its lowest level and should be filled.
B— To completely charge a nickel-cadmium battery, some gassing must take place; thus, some water will be used.
C— When positive plates slowly give up oxygen, which is regained by the negative plates, the battery is charging.
B— To completely charge a nickel-cadmium battery, some gassing must take place; thus, some water will be used.
Explanation: Toward the end of the charging cycle, the cells emit gas. This will also occur if the cells are overcharged. This gas is caused by decomposition of the water in the electrolyte into hydrogen at the negative plates and oxygen at the positive plates. To completely charge a nickel-cadmium battery, some gassing, however however slight, must take place; thus, some water will be used.
114
8096.
The electrolyte of a nickel-cadmium battery is the lowest when the battery is
A— being charged.
B— in a discharged condition.
C— under a heavy load condition.
B— in a discharged condition.
Explanation: As a nickel-cadmium battery becomes discharged, some of the electrolyte is absorbed by the plates when the battery is fully discharged, its electrolyte level is the lowest.
115
8096-1.
The electrolyte of a nickel cadmium battery is highest when the battery is
A— in a fully charged condition.
B— in a discharged condition.
C— under a no-load condition.
A— in a fully charged condition.
Explanation: As a nickel-cadmium battery is discharged, the plates absorb some of the electrolyte and the electrolyte level is lowered. As the battery is charged, the plates give up some of the electrolyte. The level of the electrolyte is highest when the battery is in a fully charged condition.
116
8097.
The end-of-charge voltage of a 19-cell nickel-cadmium battery, measured while still on charge,
A— must be 1.2 to 1.3 volts per cell.
B— must be 1.4 volts per cell.
C— depends upon its temperature and the method used for charging.
C— depends upon its temperature and the method used for charging.
Explanation: The end-of-charge voltage, measured while the cell is still on charge, depends upon its temperature and the method used for charging it. A cell of a 19-cell battery being charged at room temperature at 28.5-volt constant voltage would read about 1.5 volts at the end of charge; at a 27.5-volt constant- voltage condition, the cell would read about 1.45 volts. Under constant-current charging conditions, this value would depend upon temperature and charge current; at the end of seven hours charging at the five-hour rate, the voltage of a cell would be about 1.58 volts, at 75°F.
117
8098.
During discharge, nickel-cadmium batteries will show a lower liquid level than when at full charge because
A— a chemical action causes the electrolyte to evaporate through the vents.
B— current leakage from individual cells causes a rise in temperature.
C— the electrolyte becomes absorbed into the plates.
C— the electrolyte becomes absorbed into the plates.
Explanation: If a nickel-cadmium battery is stored for a long period of time, some of the electrolyte in the cells will be absorbed into the plates, and the level will drop. The electrolyte level will rise when the battery is given a freshening charge before it is put into service.
118
8099.
How can the state-of-charge of a nickel-cadmium battery be determined?
A— By measuring the specific gravity of the electrolyte.
B— By a measured discharge.
C— By the level of the electrolyte.
B— By a measured discharge.
Explanation: The specific gravity of the electrolyte of a nickel-cadmium battery does not change as the state-of-charge changes. The voltage of a nickel-cadmium battery remains relatively constant as its state-of-charge changes. The only way to know for sure the amount of charge a nickel-cadmium battery has in it (its state-of-charge) is to completely discharge it and then to put back into it a known number of ampere-hours of charge.
119
8100.
What may result if water is added to a nickel-cad- mium battery when it is not fully charged?
A— Excessive electrolyte dilution.
B— Excessive spewing is likely to occur during the charging cycle.
C— No adverse effects since water may be added anytime.
B— Excessive spewing is likely to occur during the charging cycle.
Explanation: The level of the electrolyte in a nickel-cadmium cell changes as the cell is discharged and charged. The level is lowest when the cell is discharged and highest at the end of the charging cycle. If water is added to a cell when some of the electrolyte has been absorbed into the plates, the level will be too high when the cell is fully charged. Some of this excess liquid will likely spew out of the cell when it is near the end of its charge cycle.
120
8101.
In nickel-cadmium batteries, a rise in cell temperature
A— causes an increase in internal resistance.
B— causes a decrease in internal resistance.
C— increases cell voltage.
B— causes a decrease in internal resistance.
Explanation: One of the desirable features of a nickel-cadmium battery is its low internal resistance which gives it the ability to discharge at a high rate and to accept a high rate of charge. The voltage and internal resistance of a nickel-cadmium cell vary inversely as the temperature. When the cell temperature increases, the voltage and internal resistance both decrease. This allows the battery to accept an excessive amount of charging current which produces more heat and can cause a thermal runaway.
121
8101-1.
Which of the following best describes the contributing factors to thermal runaway in a nickel-cadmium battery installed in an aircraft?
A— High internal resistance intensified by high cell temperatures and a high current discharge/charge rate in a constant potential (voltage) charging system.
B— Low internal resistance intensified by high cell temperatures and a high voltage discharge/charge rate in a constant current charging system.
C— Low internal resistance intensified by high cell temperatures and a high current discharge/charge rate in a constant potential (voltage) charging system.
C— Low internal resistance intensified by high cell temperatures and a high current discharge/charge rate in a constant potential (voltage) charging system.
Explanation: One of the problems with a nickel-cadmium battery is the danger of a thermal runaway. When the center cells of the battery become overheated, their resistance decreases and allows more current to flow. This increased current causes additional heating and further decreased resistance. This condition can continue until the battery is destroyed and a fire hazard is created. Thermal runaway occurs during a high current discharge/charge rate in a constant potential charging system.
122
8102.
When a charging current is applied to a nickel cadmium battery, the cells emit gas
A— toward the end of the charging cycle.
B— throughout the charging cycle.
C— especially if the electrolyte level is high.
A— toward the end of the charging cycle.
Explanation: Gassing occurs in the cells of a nickel-cadmium battery at the end of the charging cycle when all of the oxygen has been removed from the negative plate. This gassing is caused by the decomposition of the electrolyte.
