'A' Level Questions - The Oscilloscope
Q5. The circuit below shows a sinusoidal ac source connected to two resistors, R1 and R2, which form a potential divider. Oscilloscope 1 is connected across the source and oscilloscope 2 is connected across R2.
(a) The diagram below shows the trace obtained on the screen of oscilloscope 1. The time base of the oscilloscope is set at 10 ms per division and the voltage sensitivity at 15 V per division.
T = 40 ms
V0 = 45 V
For the ac source, calculate
(i) the frequency,
T = 40 ms 
f = 1/T = 1/(40 x 10-3) = 25 Hz
(ii) the rms voltage.
Vrms = V0 / 
=45/
= 32V
(4 marks)
(b) The resistors have the following values: R1 = 450 
and R2 = 90 . Calculate:
(i) the rms current in the circuit,
V = IR
I = V/R
= 32/(450 + 90)
= 32/540
= 0.059 A or 59 mA
(ii) the rms voltage across R2.
V = IR
= 0.059 x 90
= 5.3 V
(2 marks)
(c) Oscilloscope 2 is used to check the calculated value of the voltage across R2. The screen of oscilloscope 2 is identical to that of oscilloscope 1 and both are set to the same time base.
Oscilloscope 2 has the following range for voltage sensitivity: 1 V per div., 5 V per div., 10 V per div. and 15 V per div.
State which voltage sensitivity would give the most suitable trace. Explain the reasons for your choice.
Vpeak = 5.3 x =7.5
Peak to peak will be 15V - that will be best filling the height of the screen.
5 V/div is the best choice
reason: others would give too large or too small a trace
(3 marks)
(Total 9 marks)
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