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?

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Answer 1 · 2018-04-22T19:12:13

# 3 \ s#

We have a current, #I#at time #t# in a circuit that flows according to the DE:

# (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, where #A=e^(C)#

So, the initial current, #I(0)#, flowing (at time #t=0#) is:

# 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, #T#, such that. #I(T)# is #0.1%# of the initial current #I(0)#, so we seek #T# satisfying:

# 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 #