The current \(I\) (in amperes) in a certain electric circuit at time \(t\) (in hours) satisfies the equation \[I = k(1 + 3\mathrm{e}^{-\frac{3t}{4}}),\] where \(k\) is a positive constant.
The initial current for this electric circuit is \(8\) amperes.
The finite area, \(A\), between the curve in part (b), the \(t\)-axis and the lines \(t = 0\) and \(t = 4\) is given by \[A = p + \frac{q}{\mathrm{e}^3},\] where \(p\) and \(q\) are integers.
Mr Sim introduces a new technology that makes the current drops slower. The current \(J\) (in amperes) in the upgraded electric circuit at time \(t\) (in hours) satisfies the equation \[J = 2(1 + 3\mathrm{e}^{-bt}),\] where \(b\) is a positive constant.
Create a free account to practise similar questions.
Need help? Join our JC Math tuition classes.
Learn more