'A' Level Questions - The Oscilloscope
Q1. The diagram shows a trace on the screen of an oscilloscope. The Y-sensitivity of the oscilloscope is set at 5.0 V per division and the time base is set at 0.50 ms per division.
(a) For the trace, determine
(i) the maximum positive value of potential difference,
2 divisions above the zero line - at 5.0 V per division is 10 V 
(ii) the maximum negative value of potential difference,
1 division below the zero line - at 5.0 V per division is 5.0 V
(iii) the frequency of the signal.
Period T is 3 divisions - at 0.50 ms per division gives 1.5 x 10-3s
f = 1/T
= 1/(1.5 x 10-3) = 670 Hz
(4 marks)
(b) The trace shows the variation in the potential difference across a 100 
resistor. Calculate the energy dissipated in the resistor
(i) for the first 1.00 ms,
voltage = -5.0V
Energy dissipated = ItV
But I = V/R
so energy dissipated = V2t/R = 25 x 0.001/100
= 2.5 x 10-4 J = 0.25 mJ
(ii) between 1.00 ms and 1.50 ms,
voltage = 10 V
energy dissipated = V2t/R = 100 x 0.0005/100
= 5.0 x 10-4 J = 0.50 mJ
(iii) in one cycle,
The sum of (i) and (ii) = 0.75 mJ
(iv) in one second.
In one second there are 667 cycles so:
in one second 0.75 x 667 = 0.50 J of energy are dissipated.
(5 marks max)
(Total 9 marks)
About this site
