4/30/2012
General Class Element 3 Course
P
Presentation
t ti
¾ ELEMENT 3 SUB‐ELEMENTS
SUB ELEMENTS
G1 – Commission’s Rules
G2 – Operating Procedures
G3 – Radio Wave Propagation
G4 – Amateur Radio Practices
G5 – Electrical Principles
G6 – Circuit Components
G7 – Practical Circuits
G8 – Signals and Emissions
G9 – Antennas
A t
G0 – Electrical and RF Safety
General Licensing Class
Subelement G5
Electrical Principles
3 Exam Questions, 3 Groups
2
Electrical Principles
ElectricElectrical Principlesrinciples
¾ Impedance Z, is the opposition to the flow of
current in an AC circuit.
1
2πFC
Resonance occurs in a circuit when XL is equal to XC.
xL = 2πFL
(G5A01)
¾ Reactance is opposition
pp
to the flow of alternatingg
current caused by capacitance or inductance.
2πFL =
(G5A02)
1
2πFC
XC=
XL=2
2πFL
ElectriElectrical Principlesinciples
p
p
1
2 πFC
This is XL=XC
What we do to the left side of the equation, we must do to the right side, and what
we do to the numerator we must do to the denominator, to maintain equality
Therefore…..
When XL equals XC, it creates a special frequency called ‘resonant frequency’
XC =
1
F2= (2πL)(2πC)
F2=
1
(2π)2(LC)
Mulitplied
p
both sides by
y F and
divided both sides by 2πL
Multiplied
p
denominator
Electrical Electrical PrinciplesPrinciples
¾ Reactance causes opposition to the flow of alternating
current in an inductor.
¾ Reactance causes opposition to the flow of alternating
current in a capacitor.
¾ As the frequency of the applied AC increases,
increases the
reactance of an inductor increases.
(G5A03)
F2=
1
((2π))2 LC
From previous slide
(G5A04)
(G5A05)
F=
1
2π √LC
See XL formula
Take square root of both
sides of equation
This is the resonant frequency formula
formula.
¾ As the frequency of the applied AC increases, the
reactance of a capacitor decreases.
(G5A06)
See XC formula
¾ When the impedance of an electrical load is equal to the
internal impedance of the power source, the source can
deliver maximum power to the load
load.
(G5A07)
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Electrical Electrical PrinciplesPrinciples
Electrical Electrical PrinciplesPrinciples
¾ Impedance matching is important so the source can
power to the load.
deliver maximum p
¾ Ohm is the unit used to measure reactance.
(G5A08)
(G5A09)
¾ One reason to use an impedance matching
transformer is to maximize the transfer of
power.
(G5A12)
¾ Ohm is the unit used to measure impedance.
impedance
¾ One method of impedance matching between two
AC circuits is to insert an LC network between the
two circuits.
(
(G5A10)
)
¾ Devices that can be used for impedance
matching at radio frequencies
(G5A13)
(G5A11)
ƒ A transformer
ƒ A Pi‐network
Pi network
ƒ A length of transmission line
All of these choices are correct
Electrical Electrical PrinciplesPrinciples
Electrical Principles
¾A two‐times increase or decrease in power results
in a change of approximately 3 db.
(G5B01)
(G5
0 )
Definition of a Decibel
200 watts of electrical power are
used
d if 400 VDC is
i supplied
li d tto an
800-ohm load. (G5B03)
• See P on chart
¾The total current entering a parallel circuit equals
the sum of the currents through each branch.
(G5B02)
•
•
•
•
P=E²/R
P=(400) /800
P=(400)²/800
P=160,000 / 800
P= 200 Watts
IT = I1 + I2 + I3
Electrical Principles
¾ 2.4 watts of electrical power are used
by a 12‐VDC
12 VDC light bulb that draws 0.2
0 2 amperes.
amperes
(G5B04)
• P= E * I
• P=(12) * 0.2
• P= 2.4 Watts
See P on chart
¾ Approximately 61 milliwatts are
being dissipated when a current of 7.0
milliamperes flows through 1
1.25
25
kilohms.
(G5B05)
•
•
•
•
•
P= I² R
See P on chart
P =(0.007)² * 1250
P = 0.000049 * 1250
P=0.0613 watts
0.061 Watts = 61.3 Milliwatts
ElectricalElectrical Principles Principles
¾The output PEP from a transmitter is 100
watts if an oscilloscope measures 200 volts
peak‐to‐peak across a 50‐ohm dummy load
connected to the transmitter output.
(G5B06)
¾ PEP =[[ (200
(
/ 2)) x .707]] ² / R
¾ PEP= [70.7] ² / 50
¾ PEP= 4,998 / 50
¾ PEP= 99.97 Watts
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Electrical Principles
Electrical Principles
¾ The RMS value of an AC signal is the voltage that
causes the same power dissipation as a DC voltage
off the
h same value.
l
(G5B07)
¾ 339.4 volts is the peak‐to‐peak
voltage of a sine wave that has
an RMS voltage of 120 volts.
(G5B08)
•
•
•
•
– If you combined two or more sine wave voltages, the RMS voltage would be the square
root of the average of the sum of the squares of each voltage waveform.
Peak to Peak = 2 (1.41
(1 41 * RMS)
PP= 2(1.41 * 120)
PP= 2(169.68)
PP = 339.36 Volts
¾ 12 volts is the RMS voltage of a
sine wave with a value of 17 volts
peak.
(G5B09)
• RMS = Peak * 0.707
• RMS = 17 * 0.707
• RMS = 12 Volts
ElectricalElectrical Principles Principles
¾ The percentage of power loss that would result from a
transmission line loss of 1 dB would be approx
approx. 20
20.5
5%
%.
¾ The ratio of peak envelope power to average power for an
unmodulated carrier is 1.00.
¾ 245 volts
lt would
ld b
be th
the voltage
lt
across a 50
50‐ohm
h d
dummy lload
d
dissipating 1200 watts.
(
(G5B10)
)
(G5B11)
(G5B12)
•
See E on chart
•
•
•
•
E =√ (P*R)
E = √ (1200*50)
E = √ 60,000
E = 244.9 Volts RMS
Electrical Principles
¾ 1060 watts is the output PEP of an unmodulated carrier if
an average reading
di wattmeter connected
d to the
h
transmitter output indicates 1060 watts.
¾ 625 watts is the output PEP from a transmitter if an
oscilloscope measures 500 volts peak‐to‐peak across a 50‐
ohm resistor connected to the transmitter output.
(G5B13)
(G5B14)
ElectricalElectrical Principles Principles
¾ Mutual inductance causes a voltage to appear
across the secondary winding of a transformer when
an AC voltage source is connected across its primary
winding.
(G5C01)
–
PEP =[[ (500 / 2) x .707]
707] ² / R
–
PEP= [ 250 * .707] ² / 50
–
PEP= [176.75]
[176 75] ² / 50
–
PEP= 31,240. 56 / 50
–
PEP = 624.81 Watts
ElectricalElectrical Principles Principles
¾ The source of energy
is normally
connected
t d to
t the
th
primary winding in a
transformer
transformer.
(G5C02)
Mutual Inductance examples
– The simplest transformer has two
windings: a primary winding and a
secondary winding.
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Electrical Electrical PrinciplesPrinciples
¾ A resistor in series should be added to an existing resistor in a
circuit to increase circuit resistance
resistance.
¾ The total resistance of three 100‐ohm resistors in parallel is
33.3 ohms.
(G5C03)
Electrical Electrical PrinciplesPrinciples
¾ The turns ratio of a transformer used to match an audio
amplifier having a 600‐ohm output impedance to a speaker
having a 4‐ohm
4 ohm impedance is 12.2
12 2 to 1
1.
(G5C07)
This is a ‘turns ratio’ problem.
(G5C04)
–
–
–
–
For identical resistors in parallel simply divide the resistance of one resistor by the number of resistors to
find the total network resistance.
R = resistor value / number of resistors
R = 100 / 3
Or the long way.
R = 33.333 Ohms
NP = turns on the primary
NP
NS = turns on the secondary NS
¾ 150 ohms is the value of each resistor which, when three of
them are connected in parallel, produce 50 ohms of
resistance, and the same three resistors in series produce 450
ohms.
h
¾ The resistance of a carbon resistor will change depending on
the resistor's temperature coefficient rating if the ambient
temperature is increased.
d
=
=
ZP
ZP = p
primary
y impedance
p
ZS
ZS = secondary impedance
600
4
(G5C05)
=
=
(G6A06)
Electrical Electrical PrinciplesPrinciples
¾The equivalent
q
capacitance
p
of two 5000 p
picofarad
capacitors and one 750 picofarad capacitor
parallel is 10750 p
picofarads.
connected in p
(G5C08)
• Capacitors in parallel simply add together,
therefore the total capacity would be:
• 5000 pf + 5000pf + 750 pf
• 10750 pf
150
12.2
This is a ‘turns ratio’ problem.
Electrical Electrical PrinciplesPrinciples
¾ The capacitance
p
of three 100 microfarad capacitors
p
connected in series 33.3 microfarads.
(G5C09)
– For identical capacitors in series simply divide the
capacitance of one capacitor by the number of Capacitors
Capacitors.
(Only for equal values.)
– C=capacitance value / number of capacitors
– C = 100 / 3
– C = 33.333 microfarads
Capacitors in parallel formula.
Capacitors in parallel.
ElectricalElectrical Principles Principles
¾ The inductance of three 10 millihenry inductors
connected in parallel is 3.3 millihenrys.
(G5C10)
– For identical inductors in parallel simply divide the inductance of one inductor by the
number of inductors.
inductors
– L=Inductor value / number of inductors
– L = 10 / 3
Or the long way.
– L = 3.333 millihenrys
y
¾ The inductance of a 20 millihenry inductor in series
with a 50 millihenry inductor is 70 millihenrys
Electrical Electrical PrinciplesPrinciples
¾ The capacitance of a 20 microfarad capacitor in
series with a 50 microfarad capacitor is 14.3
microfarads.
(G5C12)
– CT= 1// [(1/C
[( / 1) + ((1/C
/ 2)]
– CT = 1/ [(1/20) + (1/50)]
– CT = 1/ [(.050)+(1/.020)]
(G5C11)
– Inductors in series simply add.
– Therfore L = 20 + 50
– L = 70 millihenrys.
Just like resistors in series.
– CT = (1/.07)
– CT = 14.285 microfarads
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Electrical Electrical PrinciplesPrinciples
¾ A capacitor in parallel should be added to a capacitor in a
circuit to increase the circuit capacitance.
¾ An
A inductor
i d
in
i series
i should
h ld b
be added
dd d to an iinductor
d
iin a
circuit to increase the circuit inductance.
¾ 5.9 ohms is the total resistance of a 10 ohm,, a 20 ohm,,
and a 50 ohm resistor in parallel.
(G5C13)
G5A01
What is impedance?
A. The electric charge
g stored byy a capacitor
p
(G5C14)
B. The inverse of resistance
(G5C15)
–
–
–
–
–
RT= 1/ [(1/R1) + (1/R2) + (1/R3)]
RT= 1/ [(1/10) + (1/20) + (1/50)]
RT = 1/ [(0.1) + (0.05) + (0.02)]
RT =1/ .17
RT = 5.88 ohms
C. The opposition to the flow of current in an AC circuit
D. The force of repulsion between two similar electric fields
• Remember that the total resistance in a parallel circuit will always be
less than the smallest resistor in the parallel network.
G5A02
What is reactance?
G5A03
Which of the following causes opposition to the
flow of alternating current in an inductor?
A. Opposition
pp
to the flow of direct current caused byy resistance
A Conductance
A.
B. Opposition to the flow of alternating current caused by
capacitance or inductance
B. Reluctance
C. Admittance
C A property of ideal resistors in AC circuits
C.
D. Reactance
D. A large spark produced at switch contacts when an inductor is
de‐energized
G5A04
Which of the following causes opposition to the
flow of alternating current in a capacitor?
G5A05
How does an inductor react to AC?
A Conductance
A.
A. As the frequency of the applied AC increases, the reactance
decreases
B. Reluctance
B. As the amplitude
p
of the applied
pp
AC increases, the reactance
increases
C. Reactance
C. As the amplitude
p
of the applied
pp
AC increases,, the reactance
decreases
D. Admittance
D As the frequency of the applied AC increases,
D.
increases the reactance
increases
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G5A07
G5A06
How does a capacitor react to AC?
What happens when the impedance of an
electrical load is equal to the internal
i
impedance
d
off the
h power source??
A. As the frequency of the applied AC increases, the reactance
decreases
A The source delivers minimum power to the load
A.
B As the frequency of the applied AC increases,
B.
increases the reactance
increases
B. The electrical load is shorted
C. No current can flow through the circuit
C. As the amplitude of the applied AC increases, the reactance
increases
D. The source can deliver maximum power to the load
D. As the amplitude of the applied AC increases, the reactance
decreases
Whyy is impedance
p
matchingg
important?
G5A08
A So the source can deliver maximum power to the load
A.
B. So the load will draw minimum p
power from the source
What unit is used to measure
reactance?
G5A09
A. Farad
B. Ohm
C. Ampere
C. To ensure that there is less resistance than reactance in the
circuit
D. Siemens
D To ensure that the resistance and reactance in the circuit are
D.
equal
What unit is used to measure
impedance?
G5A10
A
A.
Volt
l
G5A11
Which of the following describes one method
iimpedance
d
matching
hi b
between two AC
C circuits?
i i ?
of
A Insert an LC network between the two circuits
A.
B Ohm
B.
B. Reduce the p
power output
p of the first circuit
C. Ampere
C. Increase the power output of the first circuit
D. Watt
D. Insert a circulator between the two circuits
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What is one reason to use an
i
impedance
d
matching
hi transformer?
f
?
G5A12
G5A13
Which of the following devices can be used for
iimpedance
d
matching
t hi att radio
di ffrequencies?
i ?
A To minimize transmitter power output
A.
A A transformer
A.
B. To maximize the transfer of power
B. A Pi‐network
C. To reduce power supply ripple
C. A length of transmission line
D. To minimize radiation resistance
D. All of these choices are correct
A two‐times increase or decrease in
power results in a change
p
g of how manyy dB?
G5B02
A Approximately 2 dB
A.
A. It equals
q
the average
g of each branch current
B. Approximately 3 dB
B. It decreases as more parallel branches are added to the
circuit
G5B01
How does the total current relate to the individual
parallel circuit?
currents in each branch of a p
C. Approximately 6 dB
C It equals the sum of the currents through each branch
C.
D. Approximately 12 dB
D. It is the sum of the reciprocal of each individual voltage drop
G5B03
How many watts of electrical power are used if
400 VDC is supplied to an 800
800‐ohm
ohm load?
A 0.5
A.
0 5 watts
B. 200 watts
G5B04
How many watts of electrical power are
used
by a 12‐VDC light bulb that draws 0.2
amperes?
A 2.4
A.
2 4 watts
B. 24 watts
C. 6 watts
C. 400 watts
D. 60 watts
D. 3200 watts
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G5B06
G5B05
How many watts are being dissipated when
a current of 7.0 milliamperes flows through
kilohms?
1.25
What is the output PEP from a transmitter if an oscilloscope
measures 200 volts peak‐to‐peak across a 50‐ohm dummy
load
connected to the transmitter output?
A Approximately
A.
A
i t l 61 milliwatts
illi tt
A 1.4
A.
1 4 watts
B Approximately 61 watts
B.
B. 100 watts
C. 353.5 watts
C. Approximately 11 milliwatts
D. 400 watts
D. Approximately 11 watts
G5B08
G5B07
Which value of an AC signal results in the
same power dissipation as a DC voltage of
same value?
the
A. The peak‐to‐peak value
A 84.8
A.
84 8 volts
B The peak value
B.
B. 169.7 volts
C. The RMS value
C. 240.0 volts
D. The reciprocal of the RMS value
D. 339.4 volts
G5B09
What is the RMS voltage of
sine wave
with a value of
17 volts peak?
What is the peak‐to‐peak voltage of a sine
wave that has an RMS voltage of 120 volts?
What percentage of power loss
would
result from a
transmission line loss of 1 dB?
G5B10
A 8.5
A.
8 5 volts
l
A 10.9
A.
10 9 %
B 12 volts
B.
B. 12.2 %
C. 24 volts
C. 20.5 %
D. 34 volts
D. 25.9 %
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G5B11
What is the ratio of peak envelope
power to average power for an
unmodulated carrier?
A.
G5B12
What would be the RMS voltage across a
50‐ohm
50
ohm dummy load dissipating 1200 watts?
A 173 volts
A.
0.707
B. 1.00
B. 245 volts
C. 1.414
C. 346 volts
D. 2.00
D. 692 volts
G5B14
G5B13
What is the output PEP of an unmodulated
carrier if an average reading wattmeter connected
transmitter
i
output iindicates
di
1060 watts??
to the
What is the output PEP from a transmitter if an oscilloscope
measures 500 volts peak‐to‐peak across a 50‐ohm resistor
connected to the transmitter output?
A 530 watts
A.
A 8.75
A.
8 75 watts
B. 1060 watts
B. 625 watts
C. 1500 watts
C. 2500 watts
D. 2120 watts
D. 5000 watts
G5C01
What causes a voltage to appear across the secondary
winding of a transformer when an AC voltage source is
connected across its primary winding?
G5C02
Which part of a transformer is normally
connected to the incoming source of energy?
A Capacitive coupling
A.
A. The secondaryy
B. Displacement
p
current coupling
p g
B. The primary
C. Mutual inductance
C. The core
D. Mutual capacitance
D. The plates
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G5C03
Which of the following components
should be added to an existing resistor
t increase
to
i
th
the resistance?
it
?
What is the total resistance of
three
h
100 h resistors
100‐ohm
i
iin parallel?
ll l?
G5C04
A A resistor in parallel
A.
A 0.30
A.
0 30 ohms
B. A resistor in series
B. 0.33 ohms
C. A capacitor in series
C. 33.3 ohms
D. A capacitor in parallel
D. 300 ohms
G5C05
If three equal value resistors in parallel produce 50 ohms
of
resistance, and the same three resistors in series
produce 450 ohms,
what
h iis the
h value
l off each
h resistor?
i
?
G5C06
What is the RMS voltage across a 500‐turn
secondary winding in a transformer if the 2250‐
primary is connected to 120 VAC?
A 1500 ohms
A.
A 2370 volts
A.
B. 90 ohms
B. 540 volts
C. 150 ohms
C. 26.7 volts
D. 175 ohms
D. 5.9 volts
G5C07
What is the turns ratio of a transformer used to match an
audio amplifier having a 600‐ohm output impedance to a
speaker having
a4
4‐ohm
ohm impedance?
turn
G5C08
What is the equivalent capacitance of two 5000
picofarad capacitors and one 750 picofarad
capacitor connected in parallel?
A 12.2
A.
12 2 to 1
A 576.9
A.
576 9 picofarads
B. 24.4 to 1
B. 1733 picofarads
C. 150 to 1
C. 3583 picofarads
D. 300 to 1
D. 10750 picofarads
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What is the capacitance of three 100
microfarad capacitors connected in series?
G5C09
What is the inductance of three 10
millihenry
illih
i d
inductors
connected
d iin parallel?
ll l?
G5C10
A 0.30
A.
0 30 microfarads
A 0.30
A.
0 30 Henrys
B. 0.33 microfarads
B. 3.3 Henrys
C. 33.3 microfarads
C. 3.3 millihenrys
D. 300 microfarads
D. 30 millihenrys
G5C11
What is the inductance of a 20 millihenry
i d t iin series
inductor
i with
ith a 50 millihenry
illih
iinductor?
d t ?
G5C12
What is the capacitance of a 20 microfarad
capacitor
i iin series
i with
i h a 50
0 microfarad
i f d capacitor?
i ?
A 0.07
A.
0 07 millihenrys
A. 0
0.07
0 microfarads
c o a ads
B. 14.3 millihenrys
y
B. 14.3 microfarads
C. 70 millihenrys
D. 1000 millihenrys
G5C13
Which of the following components should be
added
dd d to a capacitor
i to iincrease the
h capacitance?
i
?
C. 70 microfarads
D. 1000 microfarads
G5C14
Which of the following components should be
to an inductor to increase the inductance?
A An inductor in series
A.
A A capacitor
A.
it iin series
i
B. A resistor in series
B A resistor in parallel
B.
C. A capacitor in parallel
C. An inductor in parallel
D. A capacitor in series
D. An inductor in series
added
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G5C15
What is the total resistance of a 10 ohm,
a 20 ohm,
h and
d a 50 ohm
h resistor
i t iin parallel?
ll l?
A 5.9
A.
5 9 ohms
B. 0.17 ohms
C. 10000 ohms
D. 80 ohms
12