When a transistor radio is switched off, the current falls away according to the differential equation (dI)/dt=-kI where k Is a constant . If the current drops to 10% in the first second ,how long will it take to drop to 0.1% of its original value?
1 Answer
3 \ s
Explanation:
We have a current,
(dI)/(dt) = -kI
Which is a First Order Separable Ordinary Differential Equation, so we can collect terms and "separate the variables" to get:
int \ 1/I \ dI = int \ -k \ dt
Which consisting of standard integrals, so we can directly integrate to get:
ln |I| = -kt + C
:. |I| = e^(-kt + C)
And, noting that the exponential is positive over the entire domain, we can write:
I(t) = e^(-kt)e^(C)
\ \ = Ae^(-kt) , say, whereA=e^(C)
So, the initial current,
I(0) = Ae^(0) = A
We are given that the current drops to 10% of the initial value in the first second, so we can compute:
I(1) = Ae^(-k)
And:
I_1 = 10/100 * I(0) => Ae^(-k) = 1/10 * A
:. e^(-k) = 1/10 => k = ln(10)
Thus we can write the solution as:
I(t) = Ae^(-tln10)
We want the time,
I(T) = 0.1/100 * I(0) => Ae^(-Tln10) = 0.1/100 * A
:. e^(-Tln10) = 1/1000
:. -Tln10 = ln (1/1000)
:. Tln10 = 3ln10
:. T= 3