State what is meant by a capacitance of 370 µF - AQA - A-Level Physics - Question 3 - 2018 - Paper 2

Using the time constant formula: $au = R imes C$ Given, $au = 1.0 ext{ s}$ and $C = 370 imes 10^{-6} ext{ F}$.

Rearranging gives: R = rac{ au}{C} = rac{1.0}{370 imes 10^{-6}} = 2.70 imes 10^{3} ext{ ohms}.

At $t = 2.0 ext{ s}$, the potential difference can be calculated using the charging formula: $V = V_0(1 - e^{-t/ au})$ Where $V_0 = 9.8 ext{ V}$ and $t = 2.0 ext{ s}$. Substituting gives: $$V = 9.8(1 - e^{-2.0/1.0}) \ ext{Therefore, } V ext{ is approximately } 8.65 ext{ V}.$

The graph of potential difference versus time will start from (0,0), rising quickly initially and then gradually leveling off as it approaches 9.8 V. This results in a curve that has a steep gradient at the beginning, decreasing as it becomes horizontal near the emf value.

The time taken for the charging current to fall to half its initial value can be calculated with the relation: $I(t) = I_0 e^{-t/ au}$. Setting I(t) = rac{I_0}{2}, we get: rac{I_0}{2} = I_0 e^{-t/ au} \ ext{Thus, } e^{-t/ au} = rac{1}{2} \ ext{Taking natural log on both sides: } -rac{t}{ au} = ext{ln}rac{1}{2} \ t = - au ext{ln}rac{1}{2} \ t ext{ is approximately } 0.693 ext{ s.}

Using the formula for charge: $Q(t) = C imes V(1 - e^{-t/ au})$ where $Q = 3.0 ext{ mC} = 3.0 imes 10^{-3} ext{ C}$ and $C = 370 imes 10^{-6} ext{ F}$. Setting $3.0 imes 10^{-3} = 370 imes 10^{-6} imes 9.8 (1 - e^{-t/1.0})$, we can solve for $t$ to find that $t$ is approximately 0.35 s.