A 10 μF capacitor stores 4.5 mJ of energy - AQA - A-Level Physics - Question 22 - 2020 - Paper 2

To find the maximum current during the discharge of the capacitor, we can use the formula derived from Ohm's Law and the energy stored in the capacitor.

First, we know that the energy (E) stored in a capacitor can be expressed as: E = rac{1}{2}CV^2 where:

• E = energy in joules (J)
• C = capacitance in farads (F)
• V = voltage across the capacitor in volts (V)

Rearranging this formula to find V, we have: V = rac{2E}{C}

Substituting the known values:

• E = 4.5 mJ = 4.5 \times 10^{-3} J
• C = 10 μF = 10 \times 10^{-6} F

We calculate: $V = \frac{2 \times 4.5 \times 10^{-3}}{10 \times 10^{-6}} = 900 V$

Next, to determine the maximum current (I), we use Ohm's Law: $I = \frac{V}{R}$ where:

• I = current in amperes (A)
• R = resistance in ohms (Ω)

Given R = 25 Ω, substituting the values gives us: $I = \frac{900}{25} = 36 A$

Therefore, the maximum current during the discharge of the capacitor is 36 A.