In this experiment, you will connect the output of a signal generator to a cathode-ray oscilloscope (CRO). You adjusts the signal generator frequency and am...

Assessment: Physics 0625 | Paper 5 Mock 01 | Practical Test Subject: Physics - 0625

Question 1 Report

In this experiment, you will connect the output of a signal generator to a cathode-ray oscilloscope (CRO). You adjusts the signal generator frequency and amplitude and records the traces displayed on the CRO screen. Fig. 4.1 shows the first trace. The Y-gain is set to 2.0 V/div and the time base is set to 5.0 ms/div. Fig. 4.2 shows a second trace from a different signal. The Y-gain for Fig. 4.2 is 5.0 V/div and the time base is 20 ms/div.

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(a) Measure the peak voltage of the waveform shown in Fig. 4.1. Show your working. [2]

(b) Measure the time period of one complete cycle of the waveform in Fig. 4.1. Show your working. [2]

(c) Use your answer to (b) to calculate the frequency of the signal. Show your working. [1]

(d) You increases the frequency of the signal generator without changing the CRO settings. Describe how the trace on the CRO screen changes. [1]

(e) Record the number of complete cycles shown on the screen in Fig. 4.2. [1]

(f) Measure the peak-to-peak voltage of the waveform in Fig. 4.2. Show your working. [1]

(g) State what you should do to the time base setting to display more complete cycles on the screen. [1]

(h) Explain why the CRO is more suitable than a moving-coil voltmeter for examining a.c. signals. [1]

Answer Details

(a) In Fig. 4.1, the waveform peaks at 3.0 divisions above the centre line (the horizontal axis through the middle of the screen). With the Y-gain set to 2.0 V/div:

\[ \text{peak height} = 3.0 \text{ divisions} \] [1]

\[ V_{\text{peak}} = 3.0 \times 2.0 = 6.0 \text{ V} \] [1]

(b) One complete cycle of the waveform in Fig. 4.1 spans from one peak to the next (or any equivalent full wavelength). Counting the grid squares, one cycle covers 4.0 divisions. With the time base set to 5.0 ms/div:

\[ \text{divisions per cycle} = 4.0 \] [1]

\[ T = 4.0 \times 5.0 = 20 \text{ ms} = 0.020 \text{ s} \] [1]

(c) Frequency is the reciprocal of the time period:

\[ f = \frac{1}{T} = \frac{1}{0.020} = 50 \text{ Hz} \] [1]

This is the standard mains frequency in many countries.

(d) If the frequency increases while the CRO settings remain unchanged, the time period of each cycle becomes shorter. Since the time base still sweeps at the same rate, each cycle occupies fewer divisions on the screen. The waves appear closer together (compressed horizontally) and more complete cycles are visible on the screen. [1]

(e) Counting the complete cycles in Fig. 4.2 (from one point where the wave crosses the centre line going upwards to the next equivalent point, repeated):

\[ \text{Number of complete cycles} = 2 \] [1]

(f) In Fig. 4.2, the waveform extends 2.0 divisions above and 2.0 divisions below the centre line, giving a peak-to-peak height of 4.0 divisions. With the Y-gain set to 5.0 V/div:

\[ V_{\text{peak-to-peak}} = 4.0 \times 5.0 = 20 \text{ V} \] [1]

The peak voltage (amplitude) is half this value: 10 V.

(g) To display more complete cycles on the screen, you should decrease the time base setting (reduce the number of ms per division). [1]

A lower time base means the beam sweeps faster across the screen, so more cycles fit into the same screen width.

(h) A CRO displays the waveform shape, showing how the voltage varies with time. This allows you to see the amplitude, frequency, and shape of the signal. A moving-coil voltmeter can only display a single steady reading (and reads zero for a.c. with zero mean, or shows r.m.s. values), so it cannot reveal the frequency or waveform shape. [1]

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