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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 V p-p of the waveform shown in Figure 8.
Answer
To find the peak-to-peak voltage (V p-p), measure the vertical distance between the maximum and minimum points of the waveform in Figure 8. This distance indicates the peak voltage positive and negative swings. If the peak voltage is found to be approximately 6.30 V, as stated in the marking scheme, we would write:
Step 2
Determine the frequency f of the waveform shown in Figure 8.
Answer
To calculate the frequency (f), we first need the time period (T) from Figure 8. Count the number of divisions for one complete cycle and multiply by the time scale setting (e.g. if each division is 0.2 ms and you measure 5 divisions):
The frequency is the reciprocal of the period:
f = rac{1}{T} = rac{1}{1.0 ext{ ms}} = 1000 ext{ Hz} = 250 ext{ Hz}
Step 3
Step 4
A student suggests that setting the time-base to 0.2 ms division might reduce uncertainty in the determination of the time constant. State and explain any possible advantage/disadvantage in making this suggested adjustment.
Answer
The advantage of using a smaller time-base division is that it increases the temporal resolution of the observed waveform. This allows for more accurate readings of the decay of voltage across the capacitor, leading to a better calculation of the time constant. However, a disadvantage may be that if the time scale is too small, the waveform can appear compressed, which makes it difficult to distinguish between closely timed features. The setting must balance clarity with resolution.
Step 5
Draw on Figure 11 the waveform you expect the student to see.
Answer
The expected waveform on Figure 11 should reflect the effect of connecting a parallel resistor, which would cause the voltage across C to discharge more rapidly than with a single resistor. As shown in the dashed line in Figure 10, the discharge curve will flatten out more quickly leading to a steeper slope initially than the original waveform.
Step 6
Complete Figure 12a to draw the voltage across R during the time interval.
Answer
The voltage across R will show a rectangle-like waveform in Figure 12a, similar to the output from the signal generator in Figure 12a. Ensure to reflect that during the low state of the signal, the voltage across R should also be low, and during the high state, it will reflect the same amplitude as the generator output, corresponding the appropriate timing.
Step 7
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
To fully display the waveform across R, the student may need to adjust the Y-axis (voltage) scale to accommodate the peak voltage expected across R. Reducing time-base to a suitable division (possibly to 0.1 ms) may help in showing the rapid changes in the waveform better, while ensuring that the vertical scale covers the entire range of voltages observed.