003 - Measurements Flashcards
A-003-001-001
What is the easiest amplitude dimension to measure by viewing a pure sine wave on an oscilloscope?
(a) Peak-to-peak voltage
(b) Peak voltage
(c) RMS voltage
(d) Average voltage
A-003-001-001
What is the easiest amplitude dimension to measure by viewing a pure sine wave on an oscilloscope?
(a) Peak-to-peak voltage
A-003-001-002
What is the RMS value of a 340 volt peak-to-peak pure sine wave?
(a) 120 volts
(b) 170 volts
(c) 240 volts
(d) 300 volts
A-003-001-002
What is the RMS value of a 340 volt peak-to-peak pure sine wave?
(a) 120 volts
A-003-001-003
What is the equivalent to the RMS value of an AC voltage?
(a) The AC voltage causing the same heating of a given resistor as a DC voltage of the same value
(b) The AC voltage found by taking the square root of the peak AC voltage
(c) The DC voltage causing the same heating of a given resistor as the peak AC voltage
(d) The AC voltage found by taking the square root of the average AC value
A-003-001-003
What is the equivalent to the RMS value of an AC voltage?
(a) The AC voltage causing the same heating of a given resistor as a DC voltage of the same value
A-003-001-004
If the peak value of a 100 Hz sinusoidal waveform is 20 volts, the RMS value is:
(a) 28.28 volts
(b) 7.07 volts
(c) 16.38 volts
(d) 14.14 volts
A-003-001-004
If the peak value of a 100 Hz sinusoidal waveform is 20 volts, the RMS value is:
(d) 14.14 volts
A-003-001-005
In applying Ohm’s law to AC circuits, current and voltage values are:
(a) peak values times 0.707
(b) average values
(c) average values times 1.414
(d) none of the proposed answers
A-003-001-005
In applying Ohm’s law to AC circuits, current and voltage values are:
(a) peak values times 0.707
A-003-001-006
The effective value of a sine wave of voltage or current is:
(a) 50% of the maximum value
(b) 70.7% of the maximum value
(c) 100% of the maximum value
(d) 63.6% of the maximum value
A-003-001-006
The effective value of a sine wave of voltage or current is:
(b) 70.7% of the maximum value
A-003-001-007
AC voltmeter scales are usually calibrated to read:
(a) peak voltage
(b) instantaneous voltage
(c) average voltage
(d) RMS voltage
A-003-001-007
AC voltmeter scales are usually calibrated to read:
(d) RMS voltage
A-003-001-008
An AC voltmeter is calibrated to read the:
(a) peak-to-peak value
(b) average value
(c) peak value
(d) effective value
A-003-001-008
An AC voltmeter is calibrated to read the:
(d) effective value
A-003-001-009
Which AC voltage value will produce the same amount of heat as a DC voltage, when applied to the same resistance?
(a) The average value
(b) The RMS value
(c) The peak value
(d) The peak-to-peak value
A-003-001-009
Which AC voltage value will produce the same amount of heat as a DC voltage, when applied to the same resistance?
(b) The RMS value
A-003-001-010
What is the peak-to-peak voltage of a sine wave that has an RMS voltage of 120 volts?
(a) 339.5 volts
(b) 84.8 volts
(c) 169.7 volts
(d) 204.8 volts
A-003-001-010
What is the peak-to-peak voltage of a sine wave that has an RMS voltage of 120 volts?
(a) 339.5 volts
A-003-001-011
A sine wave of 17 volts peak is equivalent to how many volts RMS?
(a) 24 volts
(b) 12 volts
(c) 34 volts
(d) 8.5 volts
A-003-001-011
A sine wave of 17 volts peak is equivalent to how many volts RMS?
(b) 12 volts
A-003-002-001
The power supplied to the antenna transmission line by a transmitter during an RF cycle at the highest crest of the modulation envelope is known as:
(a) peak-envelope power
(b) mean power
(c) carrier power
(d) full power
A-003-002-001
The power supplied to the antenna transmission line by a transmitter during an RF cycle at the highest crest of the modulation envelope is known as:
(a) peak-envelope power
A-003-002-002
To compute one of the following, multiply the peak-envelope voltage by 0.707 to obtain the RMS value, square the result and divide by the load resistance. Which is the correct answer?
(a) PIV
(b) ERP
(c) PEP
(d) power factor
A-003-002-002
To compute one of the following, multiply the peak-envelope voltage by 0.707 to obtain the RMS value, square the result and divide by the load resistance. Which is the correct answer?
(c) PEP
A-003-002-003
Peak-Envelope Power (PEP) for SSB transmission is:
(a) peak-voltage multiplied by peak current
(b) equal to the RMS power
(c) Peak-Envelope Voltage (PEV) multiplied by 0.707, squared and divided by the load resistance
(d) a hypothetical measurement
A-003-002-003
Peak-Envelope Power (PEP) for SSB transmission is:
(c) Peak-Envelope Voltage (PEV) multiplied by 0.707, squared and divided by the load resistance
A-003-002-004
The formula to be used to calculate the power output of a transmitter into a resistor load using a voltmeter is:
(a) P = EI/R
(b) P = EI cos 0
(c) P = IR
(d) P = (E exponent 2) /R
A-003-002-004
The formula to be used to calculate the power output of a transmitter into a resistor load using a voltmeter is:
(d) P = (E exponent 2) /R
A-003-002-005
How is the output Peak-Envelope Power of a transmitter calculated if an oscilloscope is used to measure the Peak-Envelope Voltage across a dummy resistive load (where PEP = Peak-Envelope Power, PEV = Peak-Envelope Voltage, Vp = peak-voltage, RL = load resistance)?
(a) PEP = [(Vp)(Vp)] / (RL)
(b) PEP = [(0.707 PEV)(0.707 PEV)] / RL
(c) PEP = (Vp)(Vp)(RL)
(d) PEP = [(1.414 PEV)(1.414 PEV)] / RL
A-003-002-005
How is the output Peak-Envelope Power of a transmitter calculated if an oscilloscope is used to measure the Peak-Envelope Voltage across a dummy resistive load (where PEP = Peak-Envelope Power, PEV = Peak-Envelope Voltage, Vp = peak-voltage, RL = load resistance)?
(b) PEP = [(0.707 PEV)(0.707 PEV)] / RL
A-003-002-006
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) 400 watts
(b) 1000 watts
(c) 100 watts
(d) 200 watts
A-003-002-006
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?
(c) 100 watts
A-003-002-007
What is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50-ohm dummy load connected to the transmitter output?
(a) 1250 watts
(b) 2500 watts
(c) 500 watts
(d) 625 watts
A-003-002-007
What is the output PEP from a transmitter if an oscilloscope measures 500 volts peak-to-peak across a 50-ohm dummy load connected to the transmitter output?
(d) 625 watts
A-003-002-008
What is the output PEP of an unmodulated carrier transmitter if a wattmeter connected to the transmitter output indicates an average reading of 1060 watts?
(a) 2120 watts
(b) 1500 watts
(c) 1060 watts
(d) 530 watts
A-003-002-008
What is the output PEP of an unmodulated carrier transmitter if a wattmeter connected to the transmitter output indicates an average reading of 1060 watts?
(c) 1060 watts
A-003-002-009
What is the output PEP from a transmitter, if an oscilloscope measures 400 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output?
(a) 200 watts
(b) 400 watts
(c) 600 watts
(d) 1000 watts
A-003-002-009
What is the output PEP from a transmitter, if an oscilloscope measures 400 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output?
(b) 400 watts
A-003-002-010
What is the output PEP from a transmitter, if an oscilloscope measures 800 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output?
(a) 800 watts
(b) 6400 watts
(c) 3200 watts
(d) 1600 watts
A-003-002-010
What is the output PEP from a transmitter, if an oscilloscope measures 800 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output?
(d) 1600 watts
A-003-002-011
An oscilloscope measures 500 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output during unmodulated carrier conditions. What would an average-reading power meter indicate under the same transmitter conditions?
(a) 427.5 watts
(b) 884 watts
(c) 442 watts
(d) 625 watts
A-003-002-011
An oscilloscope measures 500 volts peak-to-peak across a 50 ohm dummy load connected to the transmitter output during unmodulated carrier conditions. What would an average-reading power meter indicate under the same transmitter conditions?
(d) 625 watts
A-003-003-001
What is a dip meter?
(a) An SWR meter
(b) A marker generator
(c) A variable frequency oscillator with metered feedback current
(d) A field-strength meter
A-003-003-001
What is a dip meter?
(c) A variable frequency oscillator with metered feedback current
A-003-003-002
What does a dip meter do?
(a) It gives an indication of the resonant frequency of a circuit
(b) It measures transmitter output power accurately
(c) It measures field strength accurately
(d) It measures frequency accurately
A-003-003-002
What does a dip meter do?
(a) It gives an indication of the resonant frequency of a circuit
A-003-003-003
What two ways could a dip meter be used in an amateur station?
(a) To measure resonant frequencies of antenna traps and to measure a tuned circuit resonant frequency
(b) To measure antenna resonance and impedance
(c) To measure antenna resonance and percentage modulation
(d) To measure resonant frequency of antenna traps and percentage modulation
A-003-003-003
What two ways could a dip meter be used in an amateur station?
(a) To measure resonant frequencies of antenna traps and to measure a tuned circuit resonant frequency
A-003-003-004
A dip meter supplies the radio frequency energy which enables you to check:
(a) the calibration of an absorption-type wavemeter
(b) the impedance mismatch in a circuit
(c) the adjustment of an inductor
(d) the resonant frequency of a circuit
A-003-003-004
A dip meter supplies the radio frequency energy which enables you to check:
(d) the resonant frequency of a circuit