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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

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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. Calc... 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

Calculate the rms voltage of the supply.

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Answer

The rms voltage ( ( V_{rms} ) ) can be calculated from the peak voltage ( ( V_{peak} ) ) using the formula:

Vrms=Vpeak2=15V2=10.6VV_{rms} = \frac{V_{peak}}{\sqrt{2}} = \frac{15 V}{\sqrt{2}} = 10.6 V

Thus, the rms voltage of the supply is 10.6 V.

Step 2

Determine the y-voltage gain of the oscilloscope used for Figure 7.

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To find the y-voltage gain:

y-voltage gain=Peak Voltagediv=15V3 divisions=5V div1\text{y-voltage gain} = \frac{\text{Peak Voltage}}{\text{div}} = \frac{15 V}{3\text{ divisions}} = 5 V \text{ div}^{-1}

Therefore, the y-voltage gain of the oscilloscope is 5 V div⁻¹.

Step 3

Draw the trace of the output of the dc supply on Figure 7.

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The trace for the dc supply would be a horizontal line at a level corresponding to the average voltage of the dc supply. Given the values in this scenario, the trace would be a straight horizontal line drawn at the 10.6 V level on the oscilloscope display.

Step 4

Calculate the frequency of the square wave.

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As seen in Figure 8, one period of the square wave corresponds to 8 divisions, and the time-base is set to 5.0 × 10⁻⁵ s div⁻¹:

T=8×5.0×105s=4.0×104sT = 8 \times 5.0 \times 10^{-5} s = 4.0 \times 10^{-4} s

Where:

( T ) is the period. Therefore, the frequency ( ( f ) ) is:

f=1T=14.0×104=250Hzf = \frac{1}{T} = \frac{1}{4.0 \times 10^{-4}} = 250 Hz

Thus, the frequency of the square waves is 250 Hz.

Step 5

Deduce the time constant for the RC circuit, explaining each step of your method.

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Answer

The time constant ( ( \tau ) ) for an RC circuit can be determined from the discharge graph shown in Figure 10. The important steps involve:

  1. Measuring the voltage drop over time in the discharge phase of the capacitor.
  2. Starting from the initial voltage, find when it drops to 37% of its maximum voltage.
  3. Using the voltage at this time to apply the time constant formula:
V(t)=V0et/τV(t) = V_{0} e^{-t/\tau}

Where:

( V(t) ) is the voltage at time ( t )

Thus, to find ( \tau ):

  1. Identify the time it takes for the voltage to fall to 37%.
  2. Substitute that time into the above equation to solve for ( \tau ).

Typical calculations might yield:

τ=tln(V0V0V)\tau = \frac{t}{\ln(\frac{V_{0}}{V_{0}-V})}

Hence, we conclude the time constant based on these calculations.

Step 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.

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To reduce uncertainty, one possible adjustment is to change the time-base settings:

  1. Reducing the time-base to reflect a smaller interval per division enhances the granularity of the readings.
  2. This allows for more precise measurement of the discharging time of the capacitor.
  3. By ensuring the trace is not overly spread out, the accuracy of timing when the voltage drops to 37% can be improved, resulting in a better estimate of the time constant.

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