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

The y-voltage gain (G_y) can be calculated using:

$G_{y} = \frac{V_{peak}}{y \text{ div}}$

From Figure 7, the peak voltage corresponds to 3 divisions:

$G_{y} = \frac{15 \, \text{V}}{3 \text{ div}} = 5 \, \text{V div}^{-1}$

The trace for the dc supply on the oscilloscope would be a horizontal line at the level determined by the rms value, which is approximately 10.6 V. The line should be flat and drawn across the oscilloscope grid, showing a constant voltage.

From Figure 8, we observe that the square wave spans 8 divisions for one complete period (T). Given that the time settings are set to 5.0 x 10^-5 s div^-1:

$T = 8 imes (5.0 \times 10^{-5} \, \text{s div}^{-1}) = 4.0 \times 10^{-4} \, \text{s}$

Frequency (f) is given by:

$f = \frac{1}{T} = \frac{1}{4.0 \times 10^{-4}} = 250 \, \text{Hz}$

To determine the time constant (\tau), we analyze the discharge curve as shown in Figure 10. The time constant is defined as:

$\tau = R \cdot C$

We can estimate where the voltage falls to about 37% of its initial value (V_0). This is derived from the graph showing that the voltage reaches approximately 5 V after a specified time: we can measure this time from the graph to calculate \tau.

By reading from the graph, we can find the time corresponding to this voltage drop.

One effective way to reduce uncertainty is to change the time-base setting on the oscilloscope. By reducing the time per division, the waveform is displayed with more divisions across the graph, allowing for better resolution. This increased clarity helps in accurately determining the time taken for the voltage to change, thus improving the precision of the time constant measurement.