Additional AS Level Oscilloscope
questions to try
1. |
Why should you maximize
the volts/division scale when determining the amplitude of a wave
form?
The larger the trace the smaller the error in measuring its height. If you can measure the trace height to within 2mm then the voltage error you record will be smaller if their are less volts per division on the screen. |
2. |
What effect does changing the volts/division scale from 1 volt
per division to 5 volts per division have on the display of the incoming
wave on the screen? (Does the wave amplitude increase, decrease or
remain the same?)
The wave amplitude would decrease. |
3. |
What effect does changing the volts/division scale from 1 volt
per division to 5 volts per division have on the incoming wave itself?
No change at all |
4. |
One cycle of a waveform occupies five divisions of an oscilloscope
screen. The timebase dial is set to 1 ms/div. Calculate the frequency.
T = 5ms
f = 1/T
= 1/(5 x 10-3)
= 200 Hz |
5. |
One cycle of a waveform occupies two divisions of an oscilloscope
screen. The timebase dial is set to 5 ms/div.
Calculate the frequency.
T = 10ms
f = 1/T
= 1/(10 x 10-3)
= 100 Hz |
6. |
A waveform occupies six divisions of an oscilloscope screen when
the timebase is switched off. The voltage gain is set to 0.5V/div.
Calculate the peak to peak current being tested if the resistance
of the circuit is known to be 10 ohm.
Peak to peak voltage = 6 x 0.5 = 3V
V= IR
I = V/R = 3/10 = 0.3 A |
7. |
A waveform occupies two divisions of an oscilloscope screen when
the timebase is switched off. The voltage gain is set to 10 mV/div.
Calculate the peak current being tested if the resistance of the circuit
is known to be 1 kΩ
Peak to Peak Voltage = 20 mV
Peak voltage = 10 mV
Vo = IoR
Io = V/R = 10 x 10-3/103 = 1.0 x 10-6A
Io= 1.0 μA |
8. |
A waveform occupies six divisions of an oscilloscope screen when
the timebase is switched off. The voltage gain is set to 0.25V/div.
Calculate the rms current being tested if the resistance of the circuit
is known to be 19 ohm.
Peak to peak voltage = 6 x 0.25 = 1.5V
Vo = IoR
Io = V/R = 1.5/19 A
Now, IRMS = Io/√2
= 1.5/(19 x√2)
= 0.056A
|
9. |
A waveform occupies three divisions of an oscilloscope screen when
the timebase is switched off. The voltage gain is set to 20 mV/div.
Calculate the rms current being tested if the resistance of the circuit
is known to be 1.4 kΩ.
Peak to peak voltage = 60 mV
VRMS = Vo/√2 = 60/√2
VRMS = IRMSR
IRMS = VRMS /R
= (60/√2) /(1.4 x 103)
= 0.030A |
10. |
Sketch an oscilloscope screen showing a 200 Hz frequency, 4.0 V rms
trace. Include all calculations you have done to work out what you
should draw, and don't forget to indicate clearly what the timebase
and voltage gain settings are!
f = 200 Hz
T = 1/f = 1/200 = 5.0 x 10-3s = 5.0 ms
V RMS = 4.0 V
Vo = √2V RMS
V (Peak to Peak) = 2√2V RMS = 2√2 x 4 = 11.3 V
|
LOJ
January 2002 - revised 2008 and 2021