# General Exam-Electrical Principles

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1. 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
• (C)
• Impedance is the opposition to the flow of current in an alternating current (ac) circuit made up of capacitive reactance, inductive reactance, and resistance .
2. G5A02 What is reactance?
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
• (B)
• The opposition to flow of ac current caused by inductance and capacitance is referred to as reactance. Reactance is one component of impedance, along with resistance.
3. G5A03 Which of the following causes opposition to the flow of alternating current in an inductor?
A. Conductance
B. Reluctance
D. Reactance
• (D)
• The opposition to flow of ac current caused by inductance and capacitance is referred to as reactance. The opposition to flow of current in an alternating current (ac) circuit caused by an inductor is referred to as inductive reactance.
4. G5A04 Which of the following causes opposition to the flow of alternating current in a capacitor?
A. Conductance
B. Reluctance
C. Reactance
• (C)
• The opposition to flow of ac current caused by inductance and capacitance is referred to as reactance. The opposition to flow of current in an alternating current (ac) circuit caused by a capacitor is referred to as capacitive reactance.
5. G5A05 How does an inductor react to AC?
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 increases
(D)The opposition to flow of current caused by the coil in an alternating current (ac) circuit is referred to as inductive reactance. Inductive reactance increases as the ac frequency increases.
6. G5A06 How does a capacitor react to AC?
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
D. As the amplitude of the applied AC increases, the reactance decreases
• (A)
• The opposition to flow of current caused by a capacitor in an alternating current (ac) circuit is referred to as capacitive reactance. This reactance decreases as the ac frequency increases.
7. G5A07 What happens when the impedance of an electrical load is equal to the internal impedance of the power source?
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)
• A power source delivers maximum power to a load when the impedance of the load is equal to (matched to) the impedance of the source. When the impedances are not matched, the power source can not transfer as much power to the load. This is true for both dc power sources (such as batteries or power supplies) and ac power sources (such as transmitters).
8. G5A08 Why is impedance matching important?
A. So the source can deliver maximum power to the load
B. So the load will draw minimum power from the source
C. To ensure that there is less resistance than reactance in the circuit
D. To ensure that the resistance and reactance in the circuit are equal
• (A)
• A power source delivers maximum power to a load when the impedance of the load is equal to (matched to) the impedance of the source. When the impedances are not matched, the power source can not transfer as much power to the load. This is true for both dc power sources (such as batteries or power supplies) and ac power sources (such as transmitters).
9. G5A09 What unit is used to measure reactance?
B. Ohm
C. Ampere
D. Siemens
• (B)
• The ohm is the unit used to measure any opposition to the flow of current. In an ac circuit, this opposition is referred to as impedance which includes both reactance and resistance.
10. G5A10 What unit is used to measure impedance?
A. Volt
B. Ohm
C. Ampere
D. Watt
• (B)
• The ohm is the unit used to measure any opposition to the flow of current. In an ac circuit, this opposition is referred to as impedance which includes both reactance and resistance.
11. G5A11 Which of the following describes one method of impedance matching between two AC circuits?
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
• (A)
• An LC network such as a pi-network uses the exchange of stored energy between the inductor and capacitors to transform the ratio of voltage and current (impedance) between the input and output. Common examples of impedance matching LC networks are the L-, T-, and pi-network, named for the resemblance to a letter of the arrangement of their components on a schematic.
12. G5A12 What is one reason to use an impedance matching transformer?
A. To minimize transmitter power output
B. To maximize the transfer of power
C. To reduce power supply ripple
• (B)
• An impedance matching transformer changes the ratio of voltage and current between the load and source. Since impedance is the ratio of voltage and current, the transformer can also match different impedances. Matching impedances also maximizes power transfer between the source and load.
13. G5A13 Which of the following devices can be used for impedance matching at radio frequencies?
A. A transformer
B. A Pi-network
C. A length of transmission line
D. All of these choices are correct
• (D)
• All of these can alter the ratio of voltage and current in a circuit, effectively changing the impedance as well. A pi-network is an LC-circuit that uses the exchange of stored energy between the inductor and capacitors to transform the ratio of voltage and current (impedance) between the input and output. Special lengths of transmission lines (called “stubs”) set up patterns of reflections in the feed line that cancel the reflections from a mismatched load, making the load impedance appear as if it was the same as that of the feed line.
14. G5B01 A two-times increase or decrease in power results in a change of how many dB?
A. Approximately 2 dB
B. Approximately 3 dB
C. Approximately 6 dB
D. Approximately 12 dB
• (B)
15. G5B02 How does the total current relate to the individual currents in each branch of a parallel circuit?
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
• (C)In a circuit with several parallel branches, the total current flowing into the junction of the branches is equal to the sum of the current through each branch. This is Kirchoff’s Current Law.
16. G5B03 How many watts of electrical power are used if 400 VDC is supplied to an 800-ohm load?
A. 0.5 watts
B. 200 watts
C. 400 watts
D. 3200 watts
• (B)Since P = I x E and I = E/R, the power in a circuit can also be expressed as:P = (E x E) / R = (400 x 400) / 800 = 200 W
17. G5B04 How many watts of electrical power are used by a 12-VDC light bulb that draws 0.2 amperes?
A. 2.4 watts
B. 24 watts
C. 6 watts
D. 60 watts
• (A)
• Use the Power Circle drawing to find the equation to calculate power. The power in a circuit is equal to the voltage times the current: P = I x E = 0.2 x 12 = 2.4 watts
18. G5B05 How many watts are 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
• (A)
• Use the Ohm’s Law Circle and Power Circle drawings to find the equations to calculate power. Since P = I × E and E = R × I, the power in a circuit can also be expressed as P = I × I × R, so P = 0.007 × 0.007 × 1250 = 0.06125 W = 61.25 mW. Remember that 7 milliamperes is equal to 0.007 ampere, 1.25 kilohms is equal to 1250 ohms and 0.06125 watt is equal to approximately 61 milliwatts.
19. 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. 1.4 watts
B. 100 watts
C. 353.5 watts
D. 400 watts
• (B)
• PEP is the output power measured at the peak of the RF cycle at the peak of the signal's envelope and is equal to:

• Penv voltage = Peak-to-peak voltage / 2
• PEP = (Penv voltage x 0.707)2 / R
• PEP = (100 x 0.707)2 / 50 = 100 W
20. G5B07 Which value of an AC signal results in the same power dissipation as a DC voltage of the same value?
A. The peak-to-peak value
B. The peak value
C. The RMS value
D. The reciprocal of the RMS value
• (C)
• RMS, or root mean square, voltage values convert a constantly-varying ac voltage to the equivalent of a constant dc voltage. The RMS value of an ac voltage is the value that would deliver the same amount of power to a resistance as a dc voltage of the same value.
21. G5B08 What is the peak-to-peak voltage of a sine wave that has an RMS voltage of 120 volts?
A. 84.8 volts
B. 169.7 volts
C. 240.0 volts
D. 339.4 volts
• (D)If you know the RMS voltage and want to know the peak value, multiply the RMS by the square root of 2 (which is 1.414). If you want to know peak-to-peak, double the result. In this case, 120 × 1.414 × 2 = 339.4 V.
22. G5B09 What is the RMS voltage of a sine wave with a value of 17 volts peak?
A. 8.5 volts
B. 12 volts
C. 24 volts
D. 34 volts
• (B)If you know the peak voltage, you can find the RMS value by multiplying the peak voltage by 0.707 (which is the same as dividing by the square root of 2). 17 × 0.707 = 12 V.
23. G5B10 What percentage of power loss would result from a transmission line loss of 1 dB?
A. 10.9%
B. 12.2%
C. 20.5%
D. 25.9%
• (C)
• % Power = 100% x log-1 (dB / 10)100% x log-1 (-1 / 10) = 79.43% gets to the antenna. The loss is ~20.5%
24. G5B11 What is the ratio of peak envelope power to average power for an unmodulated carrier?
A. .707
B. 1.00
C. 1.414
D. 2.00
• (B)
• It’s 1.0 because for an unmodulated carrier all RF cycles have the same voltage, meaning that the envelope’s average and peak values are the same.
25. G5B12 What would be the RMS voltage across a 50-ohm dummy load dissipating 1200 watts?
A. 173 volts
B. 245 volts
C. 346 volts
D. 692 volts
• (B)
• It’s 245 V because P = E2 / R, so E = √(1200 / 50). This is the RMS voltage across the 50-W load.
26. G5B13 What is the output PEP of an unmodulated carrier if an average reading wattmeter connected to the transmitter output indicates 1060 watts?
A. 530 watts
B. 1060 watts
C. 1500 watts
D. 2120 watts
• (B)
• The PEP and average power of an unmodulated carrier are the same. For an unmodulated carrier all RF cycles have the same voltage, meaning that the envelope’s average and peak values are the same.
27. 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. 8.75 watts
B. 625 watts
C. 2500 watts
D. 5000 watts
• (B)
• PEP = ( ERMS)2 = ( 250 X 0.707 )2 / 50 = 625 Watts
28. 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?
A. Capacitive coupling
B. Displacement current coupling
C. Mutual inductance
D. Mutual capacitance
• (C)
• A transformer consists of two coils (windings) sharing a common core so that the flux from one winding is shared by both windings An ac signal in the input coil, called the primary winding produces an AC signal in the output coil, called the secondary winding. The core material might be layers of steel, a powdered iron mixture, some other magnetic material, or even air. The coupling between the primary and secondary windings is called mutual inductance. When a current flows through the primary winding it creates a magnetic held in the core. That magnetic field changes polarity and strength as the primary ac voltage changes. The changing magnetic field in the common core is shared by the secondary winding, inducing a voltage across the turns of the secondary winding and creating a current in the secondary circuit.
29. G5C02 Which part of a transformer is normally connected to the incoming source of energy?
A. The secondary
B. The primary
C. The core
D. The plates
• (B)
• The primary winding is generally considered to be the input. However energy can flow in either direction, primary-to-secondary or secondary-to-primary.
30. 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
• (B)
31. G5C04 What is the total resistance of three 100-ohm resistors in parallel?
A. .30 ohms
B. .33 ohms
C. 33.3 ohms
D. 300 ohms
• (C)
32. 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?
A. 1500 ohms
B. 90 ohms
C. 150 ohms
D. 175 ohms
33. G5C06 What is the RMS voltage across a 500-turn secondary winding in a transformer if the 2250-turn primary is connected to 120 VAC?
A. 2370 volts
B. 540 volts
C. 26.7 volts
D. 5.9 volts
• (C)
34. 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 a 4-ohm impedance?
A. 12.2 to 1
B. 24.4 to 1
C. 150 to 1
D. 300 to 1
• (A)
35. G5C08 What is the equivalent capacitance of two 5000 picofarad capacitors and one 750 picofarad capacitor connected in parallel?
• (D)
36. G5C09 What is the capacitance of three 100 microfarad capacitors connected in series?
• (C)
37. G5C10 What is the inductance of three 10 millihenry inductors connected in parallel?
A. .30 Henrys
B. 3.3 Henrys
C. 3.3 millihenrys
D. 30 millihenrys
• (C)
38. G5C11What is the inductance of a 20 millihenry inductor in series with a 50 millihenry inductor?
A. .07 millihenrys
B. 14.3 millihenrys
C. 70 millihenrys
D. 1000 millihenrys
• (C)
39. G5C12 What is the capacitance of a 20 microfarad capacitor in series with a 50 microfarad capacitor?
• (B)
40. G5C13 Which of the following components should be added to a capacitor to increase the capacitance?
A. An inductor in series
B. A resistor in series
C. A capacitor in parallel
D. A capacitor in series
• (C)
41. G5C14 Which of the following components should be added to an inductor to increase the inductance?
A. A capacitor in series
B. A resistor in parallel
C. An inductor in parallel
D. An inductor in series
• (D)