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

From Figure 7, if the peak voltage corresponds to a vertical scale where 15 V peaks correspond to 3 divisions, the voltage gain (y-voltage gain) can be calculated as:

$\text{y-voltage gain} = \frac{5.0}{\text{div}}$

Thus, the y-voltage gain is 5 V div^-1.

The trace of the output of a dc supply would be a horizontal line at a constant level, equivalent to the V_rms calculated, i.e., approximately 10.6 V on the vertical axis.

In Figure 8, the period (T) corresponds to 8 divisions on the oscilloscope. Given that the time-base setting is 5.0 x 10^-5 s per division, the total time period is:

$T = 8 \times 5.0 \times 10^{-5} = 4.0 \times 10^{-4} \text{ s}$

Calculating the frequency (f):

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

To find the time constant (τ) from Figure 10:

1. Identify the voltage change from the graph: If the voltage rises to approximately 63.2% of its maximum, this corresponds to one time constant.

2. Calculate the time corresponding to this voltage change based on the x-axis divisions from the oscilloscope trace.

3. Use the formula:

$τ = R \times C$

to deduce τ by substituting known values for R and C obtained from the experiment.

One control setting to reduce uncertainty is adjusting the time-base setting to a finer scale. This allows for more divisions to be displayed per second, providing a more detailed view of the charging and discharging curve, which aids in determining the accurate time at which the voltage reaches 63.2%.