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1 General Class Element 3 Course Presentation ti ELEMENT 3 SUB ELEMENTS General Licensing Class Subelement G5 3 Exam Questions, 3 Groups G1 Commission s Rules G2 Operating Procedures G3 Radio Wave Propagation G4 Amateur Radio Practices G5 G6 Circuit Components G7 Practical Circuits G8 Signals and Emissions G9 Antennas G0 Electrical and RF Safety 2 Impedance Z, is the opposition to the flow of current in an AC circuit. (G5A01) Reactance is opposition to the flow of alternating current caused by capacitance or inductance. (G5A02) X 1 L =2πFL X C = 2πFC When X L equals X C, it creates a special frequency called resonant frequency Electricrinciples 1 x L = 2πFL X C = 2πFC Resonance occurs in a circuit when X L is equal to X C. Therefore.. 1 2ππ FL = 2πFC This is X L =X C 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 F 2 = (2πL)(2πC) F 2 = 1 Mulitplied both sides by F and divided both sides by 2πL 1 Multiplied denominator (2π) 2 (LC) Electriinciples p F 2 1 From previous slide = (2π) 2 LC F= 1 2π LC 1 Take square root of both sides of equation This is the resonant frequency formula. Reactance causes opposition to the flow of alternating current in an inductor. (G5A03) Reactance causes opposition to the flow of alternating current in a capacitor. (G5A04) As the frequency of the applied AC increases, the reactance of an inductor increases. (G5A05) See X L formula As the frequency of the applied AC increases, the reactance of a capacitor decreases. (G5A06) See X C formula 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. (G5A07) 1

2 Impedance matching is important so the source can deliver maximum power to the load. (G5A08) Ohm is the unit used to measure reactance. (G5A09) Ohm is the unit used to measure impedance. (G5A10) One method of impedance matching between two AC circuits is to insert an LC network between the two circuits. (G5A11) One reason to use an impedance matching transformer is to maximize the transfer of power. (G5A12) Devices that can be used for impedance matching at radio frequencies (G5A13) A transformer APinetwork Pi network A length of transmission line All of these choices are correct A two times increase or decrease in power results in a change of approximately 3 db. (G5B01) 200 watts of electrical power are used if 400 VDC is supplied to an 800-ohm load. (G5B03) Definition of a Decibel The total current entering a parallel circuit equals the sum of the currents through each branch. (G5B02) I T =I 1 +I 2 +I 3 See P on chart P=E²/R P=(400)²/800 P=160,000 / 800 P= 200 Watts 2.4 watts of electrical power are used by a 12 VDC light bulb that draws 0.2 amperes. (G5B04) P= E * I P=(12) * 0.2 P= 2.4 Watts See P on chart Approximately 61 milliwatts are being dissipated when a current of 7.0 milliamperes flows through 1.25 kilohms. (G5B05) P= I² R See P on chart P =(0.007)² * 1250 P = * 1250 P= watts Watts = 61.3 Milliwatts Electrical Principles The 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) PEP =[ [(200 / 2) x.707] ² / R PEP= [70.7] ² / 50 PEP= 4,998 / 50 PEP= Watts 2

3 The RMS value of an AC signal is the voltage that causes the same power dissipation as a DC voltage of the same value. (G5B07) volts is the peak to peak voltage of a sine wave that has an RMS voltage of 120 volts. (G5B08) Peak to Peak = 2 (1.41 * RMS) PP= 2(1.41 * 120) PP= 2(169.68) PP = Volts 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. 12 volts is the RMS voltage of a sine wave with a value of 17 volts peak. (G5B09) RMS = Peak * RMS = 17 * RMS = 12 Volts Electrical Principles The percentage of power loss that would result from a transmission line loss of 1 db would be approx %. (G5B10) The ratio of peak envelope power to average power for an unmodulated carrier is (G5B11) 245 volts would be the voltage across a 50 ohm dummy load dissipating 1200 watts. (G5B12) See E on chart E= (P*R) E = (1200*50) E = 60,000 E = Volts RMS 1060 watts is the output PEP of an unmodulated carrier if an average reading wattmeter connected to the transmitter output indicates 1060 watts. (G5B13) 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. (G5B14) PEP =[ (500 / 2) x.707] ² / R PEP= [ 250 *.707] ² / 50 PEP= [176.75] 75] ² /50 PEP= 31, / 50 PEP = Watts Electrical Principles 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) Electrical Principles The source of energy is normally connected dto the primary winding in a transformer. (G5C02) Mutual Inductance examples The simplest transformer has two windings: a primary winding and a secondary winding. 3

4 A resistor in series should be added to an existing resistor in a circuit to increase circuit resistance. (G5C03) The total resistance of three 100 ohm resistors in parallel is 33.3 ohms. (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 = Ohms 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. (G5C05) The resistance of a carbon resistor will change depending on the resistor's temperature coefficient rating if the ambient temperature is increased. (G6A06) 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 impedance is 12.2to1 1. (G5C07) N P = turns on the primary N S = turns on the secondary N P N S = = Z P Z S This is a turns ratio problem = 150 Z P = primary impedance Z S = secondary impedance = 12.2 This is a turns ratio problem. The equivalent capacitance of two 5000 picofarad capacitors and one 750 picofarad capacitor connected in parallel is picofarads. (G5C08) Capacitors in parallel simply add together, therefore the total capacity would be: 5000 pf pf pf pf The capacitance of three 100 microfarad capacitors connected in series 33.3 microfarads. (G5C09) For identical capacitors in series simply divide the capacitance of one capacitor by the number of Capacitors. C=capacitance value / number of capacitors C = 100 / 3 C = microfarads (Only for equal values.) Capacitors in parallel formula. Capacitors in parallel. Electrical Principles 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. L=Inductor value / number of inductors L = 10 / 3 L = millihenrys Or the long way. The inductance of a 20 millihenry inductor in series with a 50 millihenry inductor is 70 millihenrys (G5C11) Just like resistors in series. Inductors in series simply add. Therfore L = L = 70 millihenrys. The capacitance of a 20 microfarad capacitor in series with a 50 microfarad capacitor is 14.3 microfarads. (G5C12) C T = 1/ /[(1/C 1 ) + (1/C 2 )] C T = 1/ [(1/20) + (1/50)] C T = 1/ [(.050)+(1/.020)] C T = (1/.07) C T = microfarads 4

5 A capacitor in parallel should be added to a capacitor in a circuit to increase the circuit capacitance. (G5C13) An inductor in series should ldbe added dto an inductor in a circuit to increase the circuit inductance. (G5C14) 5.9 ohms is the total resistance of a 10 ohm, a 20 ohm, and a 50 ohm resistor in parallel. (G5C15) R T = 1/ [(1/R 1 ) + (1/R 2 ) + (1/R 3 )] R T = 1/ [(1/10) + (1/20) + (1/50)] R T = 1/ [(0.1) + (0.05) + (0.02)] R T =1/.17 R T = 5.88 ohms Remember that the total resistance in a parallel circuit will always be less than the smallest resistor in the parallel network. G5A01 What is impedance? A. The electric charge stored by a capacitor B. The inverse of resistance C. The opposition to the flow of current in an AC circuit D. The force of repulsion between two similar electric fields G5A02 What is reactance? G5A03 Which of the following causes opposition to the flow of alternating current in an inductor? A. Opposition to the flow of direct current caused by resistance B. Opposition to the flow of alternating current caused by capacitance or inductance C. A property of ideal resistors in AC circuits D. A large spark produced at switch contacts when an inductor is de energized A. Conductance B. Reluctance C. Admittance D. Reactance 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 B. Reluctance C. Reactance D. Admittance A. As the frequency of the applied AC increases, the reactance decreases B. As the amplitude of the applied AC increases, the reactance increases C. As the amplitude of the applied AC increases, the reactance decreases D As the frequency of the applied AC increases the reactance D. As the frequency of the applied AC increases, the reactance increases 5

6 G5A06 How does a capacitor react to AC? G5A07 What happens when the impedance of an electrical load is equal to the internal impedance of the power source? A. As the frequency of the applied AC increases, the reactance decreases B. As the frequency of the applied AC increases, the reactance increases C. As the amplitude of the applied AC increases, the reactance increases A. The source delivers minimum power to the load B. The electrical load is shorted C. No current can flow through the circuit D. The source can deliver maximum power to the load D. As the amplitude of the applied AC increases, the reactance decreases Why is impedance matching G5A08 important? A. So the source can deliver maximum power to the load B. So the load will draw minimum power from the source G5A09 A. Farad B. Ohm What unit is used to measure reactance? C. To ensure that there is less resistance than reactance in the circuit C. Ampere D. Siemens D. To ensure that the resistance and reactance in the circuit are equal G5A10 A. Volt B. Ohm C. Ampere D. Watt What unit is used to measure impedance? G5A11 Which of the following describes one method of impedance matching between two AC circuits? i A. Insert an LC network between the two circuits B. Reduce the power output of the first circuit C. Increase the power output of the first circuit D. Insert a circulator between the two circuits 6

7 What is one reason to use an impedance matching transformer? G5A12 G5A13 Which of the following devices can be used for impedance matching at radio frequencies? A. To minimize transmitter power output B. To maximize the transfer of power C. To reduce power supply ripple D. To minimize radiation resistance A. Atransformer B. A Pi network C. A length of transmission line D. All of these choices are correct G5B01 A two times increase or decrease in power results in a change of how many db? G5B02 How does the total current relate to the individual currents in each branch of a parallel circuit? A. Approximately 2 db B. Approximately 3 db C. Approximately 6 db D. Approximately 12 db A. It equals the average of each branch current B. It decreases as more parallel branches are added to the circuit C. It equals the sum of the currents through each branch 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 ohm load? G5B04 How many watts of electrical power are used by a 12 VDC light bulb that draws 0.2 amperes? A. 05watts 0.5 B. 200 watts C. 400 watts D watts A. 24watts 2.4 B. 24 watts C. 6 watts D. 60 watts 7

8 G5B05 How many watts are being dissipated when a current of 7.0 milliamperes flows through 1.25 kilohms? A. Approximately 61 milliwatts B. Approximately 61 watts C. Approximately 11 milliwatts D. Approximately 11 watts G5B06 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. 14watts 1.4 B. 100 watts C watts D. 400 watts G5B07 Which value of an AC signal results in the same power dissipation as a DC voltage of same value? the G5B08 What is the peak to peak voltage of a sine wave that has an RMS voltage of 120 volts? A. The peak to peak value A volts B. The peak value B volts C. The RMS value C volts D. The reciprocal of the RMS value D volts G5B09 What is the RMS voltage of sine wave with a value of 17 volts peak? A volts B. 12 volts C. 24 volts D. 34 volts G5B10 What percentage of power loss would result from a transmission line loss of 1 db? A % B % C % D % 8

9 G5B11 What is the ratio of peak envelope power to average power for an unmodulated carrier? G5B12 What would be the RMS voltage across a 50 ohm dummy load dissipating 1200 watts? A B C D A. 173 volts B. 245 volts C. 346 volts D. 692 volts G5B13 What is the output PEP of an unmodulated carrier if an average reading wattmeter connected transmitter output indicates 1060 watts? to the G5B14 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. 875watts 8.75 B watts B. 625 watts C watts C watts D watts D 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 B. Displacement current coupling C. Mutual inductance D. Mutual capacitance A. The secondary B. The primary C. The core D. The plates 9

10 G5C03 Which of the following components should be added to an existing resistor to increase the resistance? A. A resistor in parallel B. A resistor in series C. A capacitor in series D. A capacitor in parallel What is the total resistance of three 100 ohm resistors in parallel? l? G5C04 A. 030ohms 0.30 B ohms C ohms 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 is the value of each resistor? G5C06 What is the RMS voltage across a 500 turn secondary winding in a transformer if the 2250 primary is connected to 120 VAC? turn A ohms A volts B. 90 ohms B. 540 volts C. 150 ohms C 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 a4ohm 4 ohm impedance? A to 1 B to 1 C. 150 to 1 D. 300 to 1 G5C08 What is the equivalent capacitance of two 5000 picofarad capacitors and one 750 picofarad capacitor connected in parallel? A picofarads B picofarads C picofarads D picofarads 10

11 G5C09 What is the capacitance of three 100 microfarad capacitors connected in series? A. 030microfarads 0.30 B microfarads C microfarads D. 300 microfarads G5C10 What is the inductance of three 10 millihenry inductors connected in parallel? l? A. 030Henrys 0.30 B. 3.3 Henrys C. 3.3 millihenrys D. 30 millihenrys G5C11 What is the inductance of a 20 millihenry inductor in series with a 50 millihenry inductor? G5C12 What is the capacitance of a 20 microfarad capacitor in series with a 50 microfarad capacitor? A millihenrys B millihenrys C. 70 millihenrys D millihenrys A microfarads coaads B microfarads C. 70 microfarads D microfarads G5C13 Which of the following components should be added d to a capacitor to increase the capacitance? G5C14 Which of the following components should be added to an inductor to increase the inductance? A. An inductor in series B. A resistor in series C. A capacitor in parallel D. A capacitor in series A. A capacitor in series B. A resistor in parallel C. An inductor in parallel D. An inductor in series 11

12 G5C15 What is the total resistance of a 10 ohm, a 20 ohm, and a 50 ohm resistor in parallel? l? A. 59ohms 5.9 B ohms C ohms D. 80 ohms 12