A2 October 2020 Paper 2 Q5
5 A capacitor is an electrical component which stores charge. The value of the charge stored by the capacitor, in suitable units, is denoted by \(Q\). The capacitor is placed in an electrical circuit.
At any time \(t\) seconds, where \(t \geqslant 0\), \(Q\) can be modelled by the differential equation
\[\frac{\mathrm{d}^2Q}{\mathrm{d}t^2} - 2\frac{\mathrm{d}Q}{\mathrm{d}t} - 15Q = 0.\]Initially the charge is 100 units and it is given that \(Q\) tends to a finite limit as \(t\) tends to infinity.
(a) Determine the charge on the capacitor when \(t = 0.5\). [6]
(b) Determine the finite limit of \(Q\) as \(t\) tends to infinity. [1]
| Scheme | Marks | AO |
|---|---|---|
| AE: \(m^2 - 2m - 15 = 0 \Rightarrow m = 5\) or \(-3\) So GS is \(Q = A\mathrm{e}^{-3t} + B\mathrm{e}^{5t}\) | M1 A1 | 1.1 1.1 |
| \(Q\) tends to finite limit as \(t \to \infty \Rightarrow B = 0\) | B1 | 2.2a |
| \(t = 0,\ Q = 100 \Rightarrow A = 100\) | M1 | 3.4 |
| So \(Q = 100\mathrm{e}^{-3t}\) | A1 | 1.1 |
| \(t = 0.5 \Rightarrow Q = 100\mathrm{e}^{-1.5} = 22.3\) | A1 | 3.4 |
| [6] |
Notes
B1: Or \(\mathrm{d}Q/\mathrm{d}t\) tends to zero as \(t \to \infty\). www
M1: Using initial condition to find \(A\) (or \(A + B\))
A1: soi
| Scheme | Marks | AO |
|---|---|---|
| (As \(t \to \infty\) \(\mathrm{e}^{-3t} \to 0\) so \(Q\) tends to) 0. | B1 | 3.4 |
| [1] |
Notes
B1: Only if from \(Q = k\mathrm{e}^{-at},\ a \gt 0\)
It must be clear that the limit is 0; “\(Q\) is approximately 0” would not be sufficient for B1.