A student uses the circuit shown to determine the capacitance of a capacitor - Scottish Highers Physics - Question 13 - 2023

To determine the absolute uncertainty in the capacitance, we first need to find the percentage uncertainty in voltage and charge:

1. Calculate the percentage uncertainty in voltage:

$$

2. Calculate the absolute uncertainty in capacitance using the formula:

$\Delta C = C \times \sqrt{\left(\frac{\Delta Q}{Q}\right)^2 + \left(\frac{\Delta V}{V}\right)^2}$

Where:

• (\Delta Q = 0.1 \times 10^{-3} C)
• (\Delta V = 0.1 V)

Using the capacitance calculated previously (C ≈ 24.0 µF):

$\Delta C \approx 24.0 \times 10^{-6} \times \sqrt{\left(\frac{0.1 \times 10^{-3}}{136.8 \times 10^{-3}}\right)^2 + \left(\frac{0.1}{5.7}\right)^2} \approx 4 \times 10^{-10} F$

Hence, the absolute uncertainty in the capacitance is approximately 4.0 nF.

The time taken (t) for the capacitor to charge fully can be estimated using the relationship:

$t = 5RC$

Where:

• R = 15 kΩ = 15 \times 10^{3} Ω
• C was calculated as 24.0 × 10^{-6} F

Substituting the values:

$t = 5 \times 15 \times 10^{3} \times 24.0 \times 10^{-6} \approx 1.8 \times 10^{3} s$

Thus, the estimated time taken for the capacitor to charge fully is approximately 1800 seconds.