Physics 3LB Final Quizzes|University of California-
Irvine|UPDATED 2026
Oscilloscope me base problem. You are asked to display a 1 kHz sine wave signal on the screen.
Since the period T is related to the frequency f by the formula T = 1 / f, the period of this wave is
1/(103/s)=1ms, so a reasonable se)ng for the "Time/div" dial is 1 ms/div. But suppose you
select an inappropriate se)ng for the Time/div dial by mistake. What would you see if you set
the dial on 1 μ s/div? 100 ms/div?
A (nearly) horizontal line on 1 μ s/div and many (nearly) ver cal lines of 100 ms/div
Explana on:
Time required to complete one cycle, T is 1/f = 1 ms. 100 ms/div shows 100 cycles of sine wave,
resembling closely packed ver cal lines so many ver cal lines would be shown. For 1 μ s/div we
are looking at 1/1000 of a cycle per division, so a horizontal line would be displayed.
3 mul%ple choice op%ons
Oscilloscope channel addi on problem. Read the following oscilloscope se)ngs carefully.
Suppose you display a 1 kHz, 2 V sine wave in channel A and a DC, 1V signal in channel B with
sensi vi es set on 1V/div. You select 'Add' so that the two signals are combined and you
readjust the posi on so the trace is in the middle of the screen. If you switch the signal in
channel B from DC to AC, what will happen? (Draw a picture if it helps.)
The trace jumps down 1 division
Explana on:
AC signal, the sine wave has an amplitude of 2V, so it has a max of +2V and a min of -2V, with
the addi on of a DC signal at 1V (flat line), it adds a constant offset of +1V to the sine wave
moves up by one division. Re-centering it is achieved by moving it down a division (since it
moved up), but if you switch to AC now (only displays AC), there is no +1V offset so the signal
will move down by 1 division.
3 mul%ple choice op%ons
Oscilloscope amplitude and frequency problem. Study the above graph. The "volts/div" dial is
set to 2 volts/div and the " me/div" dial is set to 5 msec/div. What is the peak-to-peak
amplitude of the displayed signal (express your answer without decimal places)? What is the
, frequency in kHz (express your answer with 3 decimal places, write the leading zero before the
decimal point)
8 Volts, 0.067 kHz
Explana on:
Peak to peak (ver cally) is 4 divisions
4 divisions * (2 volts/division) = 8 Volts
Min to min (horizontally) is 3 divisions
3 divisions (5 milliseconds/division) = 15 milliseconds (15 10^-3 seconds)
1/T = f, so 1/ 15 *10^-3 seconds = 66.67 Hz = 0.067 kHz
Oscilloscope trigger problem. The 'trigger' se)ngs determine when the trace begins. Suppose
you have displayed a sine wave with V = Vosin(ωt) on the screen, where V o
=5V and ω/2π=5kHz. Assume further that the signal triggers the scope when V=2.5V and ωt=30
degrees.
EXAMPLE:
Suppose you double the amplitude of the wave without changing any oscilloscope se)ngs.
What are V and ωt now when the scope is triggered? [Answer: the level hasn't changed so the
scope s ll triggers when V=2.5V. V = 2.5 = 10sin (ωt); ωt=sin−1(2.5/10)=14.4775121851511
degrees.]
Assume that you reduce V o back to 5V and switch the trigger slope polarity from '+' to '-
' without changing the trigger level dial. What are V (express your answer with 1 decimal place)
and ωt (express your answer without decimal places) when the scope is triggered now?
2.5 Volts, 150 degrees
Explana on:
The oscilloscope s ll triggers at V= 2.5 V, but phase changes
wt = sin^-1(5/10) = 30 degrees, looked at from downward slope (slope polarity changes makes it
trigger downward)
wt = 180-30 = 150 degrees
Ammeter and voltmeter circuit problem. Study the circuit above. Which of the following
formulas correctly relates resistance to current and voltage?
I[RA+(RvR/(R+RV))]
Irvine|UPDATED 2026
Oscilloscope me base problem. You are asked to display a 1 kHz sine wave signal on the screen.
Since the period T is related to the frequency f by the formula T = 1 / f, the period of this wave is
1/(103/s)=1ms, so a reasonable se)ng for the "Time/div" dial is 1 ms/div. But suppose you
select an inappropriate se)ng for the Time/div dial by mistake. What would you see if you set
the dial on 1 μ s/div? 100 ms/div?
A (nearly) horizontal line on 1 μ s/div and many (nearly) ver cal lines of 100 ms/div
Explana on:
Time required to complete one cycle, T is 1/f = 1 ms. 100 ms/div shows 100 cycles of sine wave,
resembling closely packed ver cal lines so many ver cal lines would be shown. For 1 μ s/div we
are looking at 1/1000 of a cycle per division, so a horizontal line would be displayed.
3 mul%ple choice op%ons
Oscilloscope channel addi on problem. Read the following oscilloscope se)ngs carefully.
Suppose you display a 1 kHz, 2 V sine wave in channel A and a DC, 1V signal in channel B with
sensi vi es set on 1V/div. You select 'Add' so that the two signals are combined and you
readjust the posi on so the trace is in the middle of the screen. If you switch the signal in
channel B from DC to AC, what will happen? (Draw a picture if it helps.)
The trace jumps down 1 division
Explana on:
AC signal, the sine wave has an amplitude of 2V, so it has a max of +2V and a min of -2V, with
the addi on of a DC signal at 1V (flat line), it adds a constant offset of +1V to the sine wave
moves up by one division. Re-centering it is achieved by moving it down a division (since it
moved up), but if you switch to AC now (only displays AC), there is no +1V offset so the signal
will move down by 1 division.
3 mul%ple choice op%ons
Oscilloscope amplitude and frequency problem. Study the above graph. The "volts/div" dial is
set to 2 volts/div and the " me/div" dial is set to 5 msec/div. What is the peak-to-peak
amplitude of the displayed signal (express your answer without decimal places)? What is the
, frequency in kHz (express your answer with 3 decimal places, write the leading zero before the
decimal point)
8 Volts, 0.067 kHz
Explana on:
Peak to peak (ver cally) is 4 divisions
4 divisions * (2 volts/division) = 8 Volts
Min to min (horizontally) is 3 divisions
3 divisions (5 milliseconds/division) = 15 milliseconds (15 10^-3 seconds)
1/T = f, so 1/ 15 *10^-3 seconds = 66.67 Hz = 0.067 kHz
Oscilloscope trigger problem. The 'trigger' se)ngs determine when the trace begins. Suppose
you have displayed a sine wave with V = Vosin(ωt) on the screen, where V o
=5V and ω/2π=5kHz. Assume further that the signal triggers the scope when V=2.5V and ωt=30
degrees.
EXAMPLE:
Suppose you double the amplitude of the wave without changing any oscilloscope se)ngs.
What are V and ωt now when the scope is triggered? [Answer: the level hasn't changed so the
scope s ll triggers when V=2.5V. V = 2.5 = 10sin (ωt); ωt=sin−1(2.5/10)=14.4775121851511
degrees.]
Assume that you reduce V o back to 5V and switch the trigger slope polarity from '+' to '-
' without changing the trigger level dial. What are V (express your answer with 1 decimal place)
and ωt (express your answer without decimal places) when the scope is triggered now?
2.5 Volts, 150 degrees
Explana on:
The oscilloscope s ll triggers at V= 2.5 V, but phase changes
wt = sin^-1(5/10) = 30 degrees, looked at from downward slope (slope polarity changes makes it
trigger downward)
wt = 180-30 = 150 degrees
Ammeter and voltmeter circuit problem. Study the circuit above. Which of the following
formulas correctly relates resistance to current and voltage?
I[RA+(RvR/(R+RV))]
