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When switch S is open (off) the oscilloscope displays the waveform shown in Figure 7 - AQA - A-Level Physics - Question 2 - 2017 - Paper 3
Question 2
When switch S is open (off) the oscilloscope displays the waveform shown in Figure 7. When S is closed (on) the oscilloscope displays the waveform shown in Figure 8... show full transcript
Worked Solution & Example Answer:When switch S is open (off) the oscilloscope displays the waveform shown in Figure 7 - AQA - A-Level Physics - Question 2 - 2017 - Paper 3
Step 1
Determine the peak-to-peak voltage Vp-p of the waveform shown in Figure 8.
Answer
To find the peak-to-peak voltage (Vp-p), measure the distance from the highest peak to the lowest trough on the waveform displayed in Figure 8. This value is determined directly from the grid divisions on the oscilloscope. If, for example, the peak value is at 6.3 V and the trough is at 0 V, then the peak-to-peak voltage can be calculated as:
Thus, Vp-p is approximately 6.3 V.
Step 2
Determine the frequency, f of the waveform shown in Figure 8.
Answer
Frequency (f) is calculated using the period (T) of the waveform. First, identify the number of divisions for one complete cycle on the oscilloscope. Suppose the waveform spans 4 divisions and each division represents 0.5 ms:
Now calculate frequency as:
f = rac{1}{T} = rac{1}{2 imes 10^{-3}} = 500 Hz
Step 3
Step 4
State and explain any possible advantage/disadvantage in making this suggested adjustment.
Answer
Adjusting the time-base to 0.2 ms/division can provide greater detail in observing fast transient effects improving measurement accuracy. However, a disadvantage could be that the waveform might compress too much, making it difficult to differentiate between peaks and troughs, or resulting in reduced clarity in oscillations.
Step 5
Draw on Figure 11 the waveform you expect the student to see.
Answer
The expected waveform will display both the charging and discharging phases across the capacitor, with characteristic exponential curves. Please ensure the sketch accurately reflects the expected amplitude and transition behavior relative to Figure 10.
Step 6
Explain the change in the waveform.
Answer
When connecting an identical resistor in parallel with R, the resistance decreases, which influences the time constant of the circuit. This results in quicker charging and discharging times, leading to steeper rises and falls in the waveform. The overall shape of the waveform may look flatter and reach the amplitude more quickly.
Step 7
Complete Figure 12a to draw the voltage across R during the time interval.
Answer
Complete the figure by sketching the voltage across resistor R, showing a square waveform pattern that aligns with the timing of the voltage changes in previous figures. Highlight that voltage across R will mirror the square wave input, albeit with appropriate amplitude based on the circuit configuration.
Step 8
State and explain what changes, if any, the student needs to make to the settings of the oscilloscope so the waveform across R is fully displayed.
Answer
The student should adjust the vertical sensitivity (V/div) to better capture the expected amplitude of the waveform across R. If the waveform appears clipped, a lower setting on the vertical scale might be necessary to ensure full visibility of fluctuations.