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 when reading the voltmeter. It ensures that the observer's line of sight is aligned correctly with the indicator needle, helping to provide an accurate reading by eliminating reflection errors.

To calculate the percentage uncertainty in T1/2, first find the mean value of T1/2. After calculating the uncertainty in individual readings, use the equation:

$\text{Percentage Uncertainty} = \left( \frac{\text{Uncertainty}}{\text{Mean} \ T_{1/2}} \right) \times 100$

For example, if the mean is 12.04 s and the uncertainty is found to be 0.07 s, the calculation will yield:

$\text{Percentage Uncertainty} = \left( \frac{0.07}{12.04} \right) \times 100 \approx 0.58\%$.

To determine the time constant, the equation for the discharging capacitor is used:

$V(t) = V_0 e^{-t/RC}$

where R is the resistance and C is the capacitance. Timing measurements lead to an estimated time constant τ = R x C. By substituting appropriate values from the experiment, one may find that τ approximates to 17 s.

• The student should ensure that the voltage across the capacitor does not exceed 3 V when she connects it to the X and Y sockets, to avoid exceeding the voltmeter's full-scale reading.

• She should discharge the capacitor fully before starting her timing procedure to ensure accurate and consistent readings for the time constant across multiple trials.

To compute the resistance from the graph in Figure 8, determine the gradient from the line plotted as:

$V = IR$

where I is the current. By choosing appropriate values of V and corresponding readings from T1/2, one can deduce that:

Using the time constant and the voltmeter reading at t = 10 s, apply Ohm's law combined with the exponential decay formula to find the current. If the voltage is noted as a certain value, say Vt, the current can be determined as:

$I = \frac{V_t}{R}$.

For example, if Vt is observed to be 10 V and R is 16 kΩ, then:

$I = \frac{10}{16000} = 0.000625 \, A (625 \, µA).$