Question 1
A cathode-ray
oscilloscope (c.r.o.) is connected to an alternating voltage. The following
trace is produced on the screen.
The oscilloscope
time-base setting is 0.5 ms cm-1 and
the Y-plate sensitivity is 2 V cm-1.
Which statement about
the alternating voltage is correct?
A The
amplitude is 3.5 cm.
B The
frequency is 0.5 kHz.
C The
period is 1 ms.
D The wavelength is 4 cm.
Reference: Past Exam Paper – June 2014 Paper 12 Q3
Solution 1:
Answer: B.
Amplitude is the maximum voltage. It
can be obtained by considering the vertical position of the trace from the
central line on the screen.
Y-plate sensitivity = 2V cm-1
Amplitude = 3.5 cm = 3.5 × 2 = 7 V
[A is incorrect]
Frequency = 1 / period = 1 / T
A value of time (not ‘wavelength’)
can be obtained horizontally. [D is incorrect]
Time-base setting = 0.5 ms cm-1
Period T = 4 cm = 4 × 0.5 = 2 ms [C
is incorrect]
Frequency = 1/T = 1 / (2×10-3)
= 0.5 kHz
How to download this book
ReplyDeleteuse it online. it's not a book
DeleteIn Q1, the wavelength is also correct i.e 4cm Then why Option ?
ReplyDeleteIn Q1
ReplyDeleteWhy D is incorrect. The wavelength is 4cm.
The wavelength is NOT 4cm.
DeleteTHe horizontal axis on the CRO gives the time (on the distance).
Wavelength is in metre while time is in second.
Also, it does not even give the period directly. On a cro we need to use the time-base setting as shown above.
Thanks so much for this site!
ReplyDeleteWhy cant the answer be A the amplitude is 3.5 cm
ReplyDeletethe y-sensitivity is 2V / cm
Deletethat is,
1 cm -> 2V
3.5 cm -> 3.5 x 2 = 7 V
the amplitude is 7V
why multiply period with 10^-3
ReplyDeletethe time-base setting is in millisecond/cm (ms/cm). we need to convert to SI unit (which is second)
Delete