Electromagnetic Effects
Chapter 24 MCQs
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
1
An oscilloscope displays a steady horizontal line 2.0 divisions above the centre line.
The Y-gain setting is 5.0 V/division.
What is the input voltage?
A 0.40 V d.c.
B 2.5 V d.c.
C 10 V d.c.
D 20 V a.c.
2
An oscilloscope trace is a horizontal line 3.5 divisions below the centre line.
The Y-gain setting is 2.0 V/division.
What is the input voltage?
A +1.75 V
B −1.75 V
C +7.0 V
D −7.0 V
3
A d.c. supply is connected to an oscilloscope. The trace is a horizontal line above the centre.
The supply terminals are reversed.
What happens to the trace?
A it becomes a sine wave
B it moves below the centre
C it becomes vertical
D it disappears because d.c. cannot be displayed
4
An a.c. voltage is connected to an oscilloscope. The trace is centred on the screen and has maximum displacement 4.0 divisions above the centre line.
The Y-gain is 0.50 V/division.
What is the peak voltage?
A 1.0 V
B 2.0 V
C 4.0 V
D 8.0 V
5
An a.c. waveform has peak-to-peak height 6.0 divisions.
The Y-gain is 2.0 V/division.
What is the peak voltage?
A 3.0 V
B 6.0 V
C 12 V
D 24 V
6
A sine wave on an oscilloscope has peak-to-peak height 8.0 cm. The vertical scale is 0.25 V/cm.
What is the amplitude of the voltage?
A 1.0 V
B 2.0 V
C 4.0 V
D 8.0 V
7
A waveform has peak voltage 3.0 V.
The Y-gain is changed from 1.0 V/division to 0.50 V/division.
What happens to the vertical height from centre to peak?
A halves
B doubles
C stays the same
D becomes zero
8
An oscilloscope displays an a.c. waveform. The Y-gain is increased from 2.0 V/division to 4.0 V/division.
What happens to the displayed height of the waveform?
A it doubles
B it halves
C it stays the same
D the frequency doubles
9
A waveform has one complete cycle occupying 5.0 horizontal divisions.
The time-base setting is 2.0 ms/division.
What is the period?
A 0.40 ms
B 2.5 ms
C 10 ms
D 250 ms
10
One complete cycle of an a.c. waveform occupies 4.0 divisions on an oscilloscope screen.
The time-base is 5.0 ms/division.
What is the frequency?
A 20 Hz
B 50 Hz
C 200 Hz
D 800 Hz
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
11
An oscilloscope trace shows 3 complete cycles across 12 divisions.
The time-base is 1.0 ms/division.
What is the frequency of the signal?
A 83 Hz
B 250 Hz
C 1000 Hz
D 4000 Hz
12
An a.c. signal has frequency 250 Hz.
The time-base is 1.0 ms/division.
How many horizontal divisions are occupied by one complete cycle?
A 0.25 divisions
B 2.5 divisions
C 4.0 divisions
D 250 divisions
13
A 50 Hz mains waveform is displayed on an oscilloscope.
The time-base is 5.0 ms/division.
How many divisions are occupied by one complete cycle?
A 0.25
B 2.5
C 4.0
D 10
14
A waveform has a period of 0.0020 s.
What is the frequency?
A 0.0020 Hz
B 2.0 Hz
C 50 Hz
D 500 Hz
15
An oscilloscope screen shows 2.5 complete cycles in 10 ms.
What is the frequency?
A 25 Hz
B 250 Hz
C 2500 Hz
D 25 000 Hz
16
A waveform has peak-to-peak height 4.0 divisions and period 6.0 divisions.
The Y-gain is 1.5 V/division and the time-base is 0.50 ms/division.
Which row is correct?
| peak voltage | frequency | |
|---|---|---|
| A | 3.0 V | 333 Hz |
| B | 6.0 V | 333 Hz |
| C | 3.0 V | 3000 Hz |
| D | 6.0 V | 3000 Hz |
17
An oscilloscope shows a sine wave. The distance from the top peak to the bottom trough is 5.0 divisions. The Y-gain is 4.0 V/division.
What is the peak-to-peak voltage?
A 2.5 V
B 10 V
C 20 V
D 40 V
18
An oscilloscope shows an a.c. signal with peak voltage 6.0 V.
What is the peak-to-peak voltage?
A 3.0 V
B 6.0 V
C 12 V
D 36 V
19
An a.c. voltage has peak-to-peak voltage 20 V.
What is its peak voltage?
A 5.0 V
B 10 V
C 20 V
D 40 V
20
A 12 V d.c. supply is connected to an oscilloscope set at 3.0 V/division.
Where is the trace?
A horizontal line 4.0 divisions above or below centre depending on polarity
B sine wave with amplitude 4.0 divisions
C horizontal line 3.0 divisions above centre only
D dot at the centre because d.c. has no frequency
21
A signal generator is connected to an oscilloscope. The displayed trace has a smaller horizontal spacing between peaks after the frequency setting is increased.
Which statement is correct?
A period has increased
B period has decreased
C amplitude has increased
D voltage has decreased
22
A signal generator output is shown on an oscilloscope. The vertical height increases, but the horizontal spacing between identical points is unchanged.
What has changed?
A frequency increased
B period decreased
C amplitude increased
D time-base increased
23
A waveform is too tall to fit on the oscilloscope screen.
Which adjustment should make it fit vertically?
A increase the V/division setting
B decrease the V/division setting
C increase the time-base speed only
D reverse the input terminals only
24
A waveform has too many cycles squeezed close together on the screen.
Which adjustment helps display fewer cycles across the screen?
A increase the time per division
B decrease the time per division
C increase the Y-gain sensitivity only
D move the trace vertically
25
An oscilloscope time-base is switched off. A d.c. voltage is applied.
What is seen on the screen?
A a stationary dot displaced vertically from the centre
B a horizontal line across the screen
C a sine wave
D a complete circle
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
26
An oscilloscope time-base is switched off. An a.c. voltage is applied.
What is most likely seen?
A a vertical line
B a horizontal line
C a single dot fixed at centre
D no trace because a.c. cannot affect the Y-plates
27
With no input signal and the time-base on, an oscilloscope shows:
A a horizontal line through the centre
B a vertical line through the centre
C a sine wave
D a square wave
28
A microphone is connected to an oscilloscope. A louder sound of the same pitch is made.
What happens to the trace?
A vertical amplitude increases, horizontal spacing unchanged
B vertical amplitude decreases, horizontal spacing unchanged
C horizontal spacing decreases, vertical amplitude unchanged
D horizontal spacing increases, vertical amplitude unchanged
29
A microphone is connected to an oscilloscope. A higher-pitched sound of the same loudness is made.
What happens to the trace?
A peaks become farther apart, height unchanged
B peaks become closer together, height unchanged
C height increases, peak spacing unchanged
D height decreases, peak spacing unchanged
30
A sound signal displayed on an oscilloscope has one wave occupying 2.0 ms.
What is the frequency of the sound?
A 2.0 Hz
B 50 Hz
C 500 Hz
D 2000 Hz
31
A sound wave displayed on an oscilloscope has frequency 1.0 kHz.
What is the period of the waveform?
A 0.0010 s
B 0.010 s
C 0.10 s
D 1000 s
32
A microphone detects a sound. The oscilloscope trace has peak-to-peak height 3.0 divisions.
The Y-gain is 0.20 V/division.
What is the voltage amplitude of the microphone signal?
A 0.30 V
B 0.60 V
C 1.5 V
D 6.0 V
33
A waveform is displayed with time-base 0.20 ms/division. One cycle occupies 5.0 divisions.
The time-base is changed to 0.10 ms/division. The signal itself is unchanged.
How many divisions does one cycle now occupy?
A 2.5 divisions
B 5.0 divisions
C 10 divisions
D 25 divisions
34
A waveform is displayed with Y-gain 2.0 V/division. Its peak height is 3.0 divisions.
The Y-gain is changed to 1.0 V/division. The signal itself is unchanged.
What is the new peak height?
A 1.5 divisions
B 3.0 divisions
C 6.0 divisions
D 12 divisions
35
A waveform is displayed with peak height 4.0 divisions at 0.50 V/division.
The Y-gain is changed to 1.0 V/division.
What is the new peak height?
A 1.0 division
B 2.0 divisions
C 4.0 divisions
D 8.0 divisions
36
A student wants to measure the frequency of an unknown a.c. signal using an oscilloscope.
Which pair of measurements/settings is needed?
A vertical height and Y-gain
B horizontal length of one cycle and time-base
C peak voltage and current
D resistance and voltage
37
A student wants to measure the peak voltage of an a.c. signal using an oscilloscope.
Which pair is needed?
A number of cycles and time-base
B distance from centre line to peak and Y-gain
C wavelength and speed
D period and frequency
38
A student wants to measure the peak-to-peak voltage of an a.c. signal.
Which measurement should be taken on the screen?
A horizontal distance for one cycle
B vertical distance from highest peak to lowest trough
C distance from centre to next identical point
D number of complete waves in one second directly
39
A 5.0 V d.c. supply is connected to an oscilloscope. The trace is 2.0 divisions above the centre.
What is the Y-gain setting?
A 0.40 V/division
B 2.5 V/division
C 7.0 V/division
D 10 V/division
40
An a.c. signal has peak-to-peak voltage 16 V. The trace peak-to-peak height is 8.0 divisions.
What is the Y-gain setting?
A 0.50 V/division
B 1.0 V/division
C 2.0 V/division
D 8.0 V/division
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
41
A signal has frequency 400 Hz. One complete cycle occupies 5.0 divisions on an oscilloscope screen.
What is the time-base setting?
A 0.20 ms/division
B 0.50 ms/division
C 2.0 ms/division
D 5.0 ms/division
42
An oscilloscope shows 4 complete cycles across 8.0 divisions. The time-base is 0.25 ms/division.
What is the frequency?
A 500 Hz
B 1000 Hz
C 2000 Hz
D 4000 Hz
43
A waveform has frequency 2.0 kHz. The time-base is 0.10 ms/division.
How many divisions does one cycle occupy?
A 0.20 divisions
B 2.0 divisions
C 5.0 divisions
D 20 divisions
44
A sinusoidal trace has peak-to-peak height 7.0 divisions and one cycle width 4.0 divisions.
Y-gain = 0.80 V/division.
Time-base = 0.50 ms/division.
Which row is correct?
| peak voltage | frequency | |
|---|---|---|
| A | 2.8 V | 500 Hz |
| B | 5.6 V | 500 Hz |
| C | 2.8 V | 2000 Hz |
| D | 5.6 V | 2000 Hz |
45
A d.c. voltage is applied to an oscilloscope. The time-base is on and the trace is horizontal.
Why is the trace horizontal?
A the voltage is constant with time
B the voltage changes sinusoidally
C the frequency is very high
D the Y-gain is zero
46
An a.c. voltage is displayed as a sine wave centred on the horizontal axis.
Why is the wave centred on the axis?
A the voltage alternates equally positive and negative about zero
B the voltage is always positive
C the time-base is switched off
D the frequency is zero
47
A student connects a battery to an oscilloscope and sees a line below the centre.
What can be concluded?
A the battery is connected with opposite polarity to the positive Y-input
B the battery voltage is alternating
C the battery has zero voltage
D the time-base is off
48
Two a.c. signals are shown on the same oscilloscope settings.
Signal X has twice the vertical amplitude of signal Y and half the horizontal period of signal Y.
Which statement is correct?
A X has twice the peak voltage and twice the frequency of Y
B X has twice the peak voltage and half the frequency of Y
C X has half the peak voltage and twice the frequency of Y
D X has half the peak voltage and half the frequency of Y
49
An oscilloscope trace of an a.c. signal has period 8.0 ms and peak voltage 5.0 V.
Which row gives frequency and peak-to-peak voltage?
| frequency | peak-to-peak voltage | |
|---|---|---|
| A | 125 Hz | 10 V |
| B | 125 Hz | 5.0 V |
| C | 800 Hz | 10 V |
| D | 800 Hz | 5.0 V |
50
A student says, “The oscilloscope measures frequency from the vertical height of the trace.”
Which correction is best?
A frequency is found from the horizontal period and time-base
B frequency is found from peak voltage and Y-gain only
C frequency is found from peak-to-peak voltage only
D frequency cannot be measured with an oscilloscope
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
Chapter 24 Answer Key
| Q | Ans | Q | Ans | Q | Ans | Q | Ans | Q | Ans |
|---|---|---|---|---|---|---|---|---|---|
| 1 | C | 11 | B | 21 | B | 31 | A | 41 | B |
| 2 | D | 12 | C | 22 | C | 32 | A | 42 | C |
| 3 | B | 13 | C | 23 | A | 33 | C | 43 | C |
| 4 | B | 14 | D | 24 | B | 34 | C | 44 | A |
| 5 | B | 15 | B | 25 | A | 35 | B | 45 | A |
| 6 | A | 16 | A | 26 | A | 36 | B | 46 | A |
| 7 | B | 17 | C | 27 | A | 37 | B | 47 | A |
| 8 | B | 18 | C | 28 | A | 38 | B | 48 | A |
| 9 | C | 19 | B | 29 | B | 39 | B | 49 | A |
| 10 | B | 20 | A | 30 | C | 40 | C | 50 | A |
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
Detailed Explanations
1. C
-
D.C. gives a steady horizontal line.
-
Voltage = vertical displacement × Y-gain
-
V = 2.0 × 5.0
-
V = 10 V d.c.
2. D
-
The trace is below the centre line, so the voltage is negative.
-
Voltage = 3.5 × 2.0
-
Voltage = 7.0 V
-
Input voltage = −7.0 V
3. B
-
A d.c. voltage gives a horizontal line.
-
Reversing the terminals reverses the polarity.
-
A line above the centre moves below the centre.
4. B
-
Peak voltage is measured from centre line to maximum displacement.
-
Peak height = 4.0 divisions
-
Y-gain = 0.50 V/division
-
Peak voltage = 4.0 × 0.50
-
Peak voltage = 2.0 V
5. B
-
Peak-to-peak height = 6.0 divisions
-
Peak-to-peak voltage = 6.0 × 2.0 = 12 V
-
Peak voltage = 12 / 2
-
Peak voltage = 6.0 V
6. A
-
Peak-to-peak height = 8.0 cm
-
Peak-to-peak voltage = 8.0 × 0.25 = 2.0 V
-
Amplitude = peak voltage = 2.0 / 2
-
Amplitude = 1.0 V
7. B
-
Original display:
-
peak voltage = 3.0 V
-
Y-gain = 1.0 V/division
-
height = 3 divisions
-
-
New Y-gain = 0.50 V/division
-
New height = 3.0 / 0.50 = 6 divisions
-
The height doubles.
8. B
-
Increasing Y-gain from 2.0 V/division to 4.0 V/division means each division represents more volts.
-
The same voltage now takes fewer divisions.
-
Displayed height halves.
9. C
-
One cycle = 5.0 divisions
-
Time-base = 2.0 ms/division
-
Period = 5.0 × 2.0
-
Period = 10 ms
10. B
-
One cycle = 4.0 divisions
-
Time-base = 5.0 ms/division
-
Period = 4.0 × 5.0 = 20 ms = 0.020 s
-
Frequency = 1 / period
-
Frequency = 1 / 0.020
-
Frequency = 50 Hz
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
11. B
-
3 cycles occupy 12 divisions.
-
1 cycle occupies 12 / 3 = 4 divisions.
-
Time-base = 1.0 ms/division.
-
Period = 4 × 1.0 = 4.0 ms = 0.0040 s.
-
Frequency = 1 / 0.0040
-
Frequency = 250 Hz
12. C
-
Frequency = 250 Hz.
-
Period = 1 / 250 = 0.0040 s = 4.0 ms.
-
Time-base = 1.0 ms/division.
-
Divisions per cycle = 4.0 / 1.0
-
Divisions = 4.0
13. C
-
Frequency = 50 Hz.
-
Period = 1 / 50 = 0.020 s = 20 ms.
-
Time-base = 5.0 ms/division.
-
Divisions = 20 / 5.0
-
Divisions = 4.0
14. D
-
Frequency = 1 / period
-
f = 1 / 0.0020
-
f = 500 Hz
15. B
-
2.5 cycles occur in 10 ms.
-
10 ms = 0.010 s.
-
Frequency = number of cycles / time
-
f = 2.5 / 0.010
-
f = 250 Hz
16. A
-
Peak-to-peak height = 4.0 divisions.
-
Peak height = 2.0 divisions.
-
Peak voltage = 2.0 × 1.5 = 3.0 V
-
Period = 6.0 × 0.50 ms = 3.0 ms = 0.0030 s.
-
Frequency = 1 / 0.0030 = 333 Hz
17. C
-
Peak-to-peak height = 5.0 divisions.
-
Y-gain = 4.0 V/division.
-
Peak-to-peak voltage = 5.0 × 4.0
-
Peak-to-peak voltage = 20 V
18. C
-
Peak-to-peak voltage = 2 × peak voltage
-
Peak-to-peak voltage = 2 × 6.0
-
Peak-to-peak voltage = 12 V
19. B
-
Peak voltage = peak-to-peak voltage / 2
-
Peak voltage = 20 / 2
-
Peak voltage = 10 V
20. A
-
D.C. gives a horizontal line.
-
Displacement = voltage / Y-gain
-
Displacement = 12 / 3.0 = 4.0 divisions
-
It is above or below the centre depending on polarity.
21. B
-
Peaks become closer together.
-
Horizontal spacing represents period.
-
Smaller spacing means smaller period.
-
Frequency increases and period decreases.
22. C
-
Vertical height represents amplitude/voltage.
-
Horizontal spacing represents period/frequency.
-
Height increases but spacing is unchanged.
-
Therefore amplitude increased.
23. A
-
If the waveform is too tall, reduce vertical sensitivity.
-
Increasing the V/division setting means each division represents more volts.
-
The displayed height becomes smaller and fits on the screen.
24. B
-
Too many cycles are shown because too much time is displayed across the screen.
-
To show fewer cycles, reduce the time per division.
-
Decrease the time-base value, and the waveform spreads out horizontally.
25. A
-
With time-base off, there is no horizontal sweep.
-
A d.c. voltage gives a steady vertical deflection only.
-
So a stationary dot appears above or below the centre.
26. A
-
With time-base off, the spot does not sweep sideways.
-
A.C. voltage moves the spot up and down repeatedly.
-
This appears as a vertical line.
27. A
-
With no input signal, vertical deflection is zero.
-
With time-base on, the spot sweeps horizontally.
-
The display is a horizontal line through the centre.
28. A
-
Louder sound means greater amplitude.
-
Same pitch means same frequency.
-
On the oscilloscope:
-
vertical amplitude increases
-
horizontal spacing stays the same
-
29. B
-
Higher pitch means higher frequency.
-
Same loudness means same amplitude.
-
Higher frequency means shorter period.
-
Peaks become closer together, height unchanged.
30. C
-
One wave occupies 2.0 ms.
-
Period = 2.0 ms = 0.0020 s.
-
Frequency = 1 / 0.0020
-
Frequency = 500 Hz
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
31. A
-
Frequency = 1.0 kHz = 1000 Hz.
-
Period = 1 / frequency.
-
T = 1 / 1000
-
T = 0.0010 s
32. A
-
Peak-to-peak height = 3.0 divisions.
-
Peak-to-peak voltage = 3.0 × 0.20 = 0.60 V.
-
Amplitude = peak voltage = 0.60 / 2
-
Amplitude = 0.30 V
33. C
-
Original period:
-
5.0 divisions × 0.20 ms/division = 1.0 ms
-
-
New time-base = 0.10 ms/division.
-
New divisions for one cycle:
-
1.0 / 0.10 = 10 divisions
-
34. C
-
Original peak voltage:
-
3.0 divisions × 2.0 V/division = 6.0 V
-
-
New Y-gain = 1.0 V/division.
-
New peak height:
-
6.0 / 1.0 = 6.0 divisions
-
35. B
-
Original peak voltage:
-
4.0 divisions × 0.50 V/division = 2.0 V
-
-
New Y-gain = 1.0 V/division.
-
New peak height:
-
2.0 / 1.0 = 2.0 divisions
-
36. B
-
Frequency is found from period.
-
Period is found from:
-
horizontal length of one cycle
-
time-base setting
-
-
Then:
-
frequency = 1 / period
-
37. B
-
Peak voltage is found from:
-
distance from centre line to peak
-
Y-gain setting
-
-
Peak voltage = peak height × Y-gain
38. B
-
Peak-to-peak voltage is measured from the highest peak to the lowest trough.
-
This vertical distance is then multiplied by the Y-gain.
39. B
-
Voltage = 5.0 V.
-
Deflection = 2.0 divisions.
-
Y-gain = voltage / divisions
-
Y-gain = 5.0 / 2.0
-
Y-gain = 2.5 V/division
40. C
-
Peak-to-peak voltage = 16 V.
-
Peak-to-peak height = 8.0 divisions.
-
Y-gain = 16 / 8.0
-
Y-gain = 2.0 V/division
41. B
-
Frequency = 400 Hz.
-
Period = 1 / 400 = 0.0025 s = 2.5 ms.
-
One cycle occupies 5.0 divisions.
-
Time-base = 2.5 ms / 5.0
-
Time-base = 0.50 ms/division
42. C
-
4 cycles occupy 8.0 divisions.
-
1 cycle occupies 8.0 / 4 = 2.0 divisions.
-
Time-base = 0.25 ms/division.
-
Period = 2.0 × 0.25 = 0.50 ms = 0.00050 s.
-
Frequency = 1 / 0.00050
-
Frequency = 2000 Hz
43. C
-
Frequency = 2.0 kHz = 2000 Hz.
-
Period = 1 / 2000 = 0.00050 s = 0.50 ms.
-
Time-base = 0.10 ms/division.
-
Divisions = 0.50 / 0.10
-
Divisions = 5.0
44. A
-
Peak-to-peak height = 7.0 divisions.
-
Peak height = 3.5 divisions.
-
Peak voltage = 3.5 × 0.80 = 2.8 V
-
Period = 4.0 × 0.50 ms = 2.0 ms = 0.0020 s.
-
Frequency = 1 / 0.0020 = 500 Hz
45. A
-
A d.c. voltage is constant with time.
-
Since vertical position does not change, the trace is horizontal.
-
The height shows the size and polarity of the d.c. voltage.
46. A
-
An a.c. voltage alternates positive and negative.
-
If centred on the horizontal axis, it varies equally above and below zero.
-
The centre line represents zero voltage.
47. A
-
A line below the centre means negative vertical deflection.
-
This shows the battery polarity is opposite to the positive Y-input.
48. A
-
Signal X has twice the vertical amplitude:
-
twice the peak voltage
-
-
Signal X has half the period:
-
frequency is doubled
-
-
So X has twice the peak voltage and twice the frequency.
49. A
-
Period = 8.0 ms = 0.0080 s.
-
Frequency = 1 / 0.0080 = 125 Hz
-
Peak voltage = 5.0 V.
-
Peak-to-peak voltage = 2 × 5.0 = 10 V
50. A
-
Frequency is not found from vertical height.
-
Vertical height gives voltage/amplitude.
-
Frequency is found from:
-
horizontal period
-
time-base
-
-
Then:
-
frequency = 1 / period
-
For Full Scale Course: Written and Compiled By Sir Hunain Zia (AYLOTI), World Record Holder With 154 Total Personal A Grades, 11 World Records and 7 Distinctions, Educate A Change.
Common Traps From This Chapter
| Trap | Correct Rule |
|---|---|
| D.C. on oscilloscope | horizontal line above/below centre |
| A.C. on oscilloscope | wave trace when time-base is on |
| Time-base off with d.c. | stationary displaced dot |
| Time-base off with a.c. | vertical line |
| Y-gain | volts per division |
| Peak voltage | centre line to peak × Y-gain |
| Peak-to-peak voltage | top peak to bottom trough × Y-gain |
| Peak-to-peak voltage | 2 × peak voltage |
| Period | horizontal width of one cycle × time-base |
| Frequency | 1 / period |
| Louder sound | larger vertical amplitude |
| Higher pitch | peaks closer together |
| Increasing V/division | trace height decreases |
| Decreasing V/division | trace height increases |
| Decreasing time/division | fewer cycles shown, trace spreads out |
| Increasing time/division | more cycles shown, trace compresses |
| Line below centre | negative polarity |
| Line above centre | positive polarity |
| A.C. centred on axis | alternates around zero |
| Vertical height | voltage/amplitude, not frequency |
| Horizontal spacing | period/frequency |