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 to eliminate parallax error, ensuring that the observer's eye is aligned directly with the scale and the needle. This allows for an accurate reading on the voltmeter.

First, calculate the mean time from the data provided in Table 1. Assuming T₁ average = 12.0 s, the uncertainty can be calculated as:

ext{percentage uncertainty} = rac{ ext{uncertainty}}{ ext{mean}} imes 100 = rac{0.06}{12.0} imes 100 ext{ = 0.5%}

To determine the time constant τ, we use the relation for the voltage across a discharging capacitor:
$V(t) = V_0 e^{-t/τ}$
Setting V = 5 V when V₀ = 15 V leads to
$5 = 15 e^{-t/τ}$
This can be rearranged to derive the time constant, with proper substitution and solving yielding approximately 17 s.

The student should ensure that the voltage across the capacitor does not exceed the range of 3 V as indicated by the voltmeter. This can be accomplished by measuring the voltage first with a multimeter or by gradually connecting the capacitor while monitoring the voltmeter.

To obtain accurate timing measurements, she should allow the capacitor to fully discharge before starting the next measurement. Additionally, using averaging over multiple trials will minimize random errors.

Using the linear relation from the graph, where the gradient is equal to -rac{1}{R}, we can determine R. Assuming data points give a gradient of -0.06, we find that
R = rac{1}{0.06} ext{ = 16.67 kΩ}, which rounds to approximately 16 kΩ.

Utilizing Ohm's law, the current I at time t can be calculated via the formula:
I = rac{V}{R}.
Substituting in values from earlier parts, where at t = 10 s the voltage is approximately 14 V and R is 16 kΩ yields:
I = rac{14 V}{16 imes 10^3 ext{ Ω}} = 0.000875 ext{ A} ext{ or } 0.875 mA.