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Figure 6 shows an oscilloscope connected across resistor R which is in series with an ac supply - AQA - A-Level Physics - Question 4 - 2020 - Paper 2
Question 4
Figure 6 shows an oscilloscope connected across resistor R which is in series with an ac supply. The supply provides a sinusoidal output of peak voltage 15 V. Figur... show full transcript
Worked Solution & Example Answer:Figure 6 shows an oscilloscope connected across resistor R which is in series with an ac supply - AQA - A-Level Physics - Question 4 - 2020 - Paper 2
Step 1
04.1 Calculate the rms voltage of the supply.
Answer
To find the rms (root mean square) voltage from the peak voltage, we use the formula:
oot{2}} $$ Given that the peak voltage is 15 V: $$ V_{rms} = rac{15}{ oot{2}} = 10.6 ext{ V} $$ Thus, the rms voltage of the supply is approximately 10.6 V.Step 2
04.2 Determine the y-voltage gain of the oscilloscope used for Figure 7.
Answer
The y-voltage gain can be found from the peak voltage trace on the oscilloscope. Given that the peak voltage corresponds to 3 divisions, we have:
y ext{-voltage gain} = rac{V_{peak}}{ ext{divisions}}
Substituting the given values:
y ext{-voltage gain} = rac{15 ext{ V}}{3 ext{ div}} = 5 ext{ V div}^{-1}
Step 3
04.3 Draw the trace of the output of the dc supply on Figure 7.
Answer
The output of the dc supply would appear as a horizontal line since it outputs a constant voltage, which is equal to the rms voltage calculated in 04.1. Therefore, the trace should be a horizontal line at approximately 10.6 V on the oscilloscope.
Step 4
04.4 Calculate the frequency of the square waves.
Answer
To calculate the frequency, first note that one full cycle corresponds to the time period (T), which in this case is indicated by the number of divisions on the oscilloscope. If each division corresponds to 0.1 ms, and there are 8 divisions for one complete cycle, then:
The frequency (f) can then be derived from the period:
f = rac{1}{T} \ = rac{1}{0.8 imes 10^{-3}} = 1.25 ext{ kHz}
Step 5
04.5 Deduce the time constant for the RC circuit, explaining each step of your method.
Answer
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Observe the waveform displayed on the oscilloscope (Figure 10). The time constant (τ) can be deduced from the rate of charge and discharge of the capacitor.
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Identify the voltage levels corresponding to 63.2% of the peak voltage on the charging curve. Use the graph to find these points.
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Assuming voltage rises to 10V (63.2% of 15V), determine how long it takes to reach this voltage level. This can be read off the oscilloscope trace where the capacitor charges to 10V.
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Let's say this corresponds to about 2 divisions, each representing 0.1 ms. Therefore:
Step 6
04.6 State and explain a change to one control setting on the oscilloscope that would reduce the uncertainty in the value of the time constant.
Answer
One effective change is to reduce the time-base setting of the oscilloscope. By doing this, the waveform will be displayed over more divisions on the screen, allowing for a more precise measurement of the time taken for the voltage to rise to 63.2%. This increased granularity in the time measurement directly reduces uncertainty in the calculation of the time constant.