An analogue voltmeter has a resistance that is much less than that of a modern digital voltmeter - AQA - A-Level Physics - Question 3 - 2021 - Paper 3

The mirror is used when reading the meter to eliminate parallax errors. It allows the observer to see the reflection of the needle directly in line with the scale, ensuring that readings are taken at right angles to the scale, which provides a more accurate reading.

To determine the percentage uncertainty in T1s, we first find the mean of the recorded times: T_{1s, mean} = rac{12.00 + 11.94 + 12.06 + 12.04 + 12.16}{5} = 12.04 ext{ s}

Next, we calculate the individual uncertainties (Using maximum and minimum values). The uncertainty for this set is: $ext{Uncertainty} = 12.16 - 11.94 = 0.22 ext{ s}$

The percentage uncertainty is given by: ext{Percentage Uncertainty} = rac{ ext{Uncertainty}}{T_{mean}} imes 100 = rac{0.22}{12.04} imes 100 ext{ %} ext{ which is approximately } 1.83 ext{ %.}

To derive the time constant, we use the relationship for the discharge of a capacitor: $V(t) = V_0 e^{-t / au}$ where $V(t)$ is the voltage across the capacitor at time t, $V_0$ is the initial voltage, and $au$ is the time constant.

Given that the voltmeter reading decreases from 10 V to 5 V, we can set up the equation to solve for τ. The time taken to discharge from 10 V to 5 V can be used to calculate the time constant, which should equal approximately 17 seconds.

• The student should ensure that the pd across the capacitor does not exceed 3 V by checking the voltage before making the connection to avoid exceeding the full-scale reading on the voltmeter.

• To develop her procedure, the student should measure the discharge over multiple trials and calculate an average time constant from the results, ensuring to record the voltmeter's maximum and minimum readings accurately.

To find the resistance of the voltmeter, we first need to plot the ln(V'/V) versus t. From the graph in Figure 8, we can determine the gradient as: ext{Gradient} = -rac{1}{R} To find R, we can rearrange this to: R = -rac{1}{ ext{Gradient}} Substituting in values from the graph, we estimate R to be about 16 kΩ.

Using Ohm's law, the current can be calculated by: I = rac{V}{R} At t = 10 s, if V is found to be approximately 12 V from the earlier discharge formula, and knowing that $R$ was determined to be 16 kΩ, we can substitute: I = rac{12}{16000} = 0.00075 A = 0.75 ext{ mA}.