# 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?

Apr 22, 2018

$3 \setminus s$

#### Explanation:

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

$\frac{\mathrm{dI}}{\mathrm{dt}} = - k I$

Which is a First Order Separable Ordinary Differential Equation, so we can collect terms and "separate the variables" to get:

$\int \setminus \frac{1}{I} \setminus \mathrm{dI} = \int \setminus - k \setminus \mathrm{dt}$

Which consisting of standard integrals, so we can directly integrate to get:

$\ln | I | = - k t + C$

$\therefore | I | = {e}^{- k t + C}$

And, noting that the exponential is positive over the entire domain, we can write:

$I \left(t\right) = {e}^{- k t} {e}^{C}$

$\setminus \setminus = A {e}^{- k t}$, say, where $A = {e}^{C}$

So, the initial current, $I \left(0\right)$, flowing (at time $t = 0$) is:

$I \left(0\right) = A {e}^{0} = A$

We are given that the current drops to 10% of the initial value in the first second, so we can compute:

$I \left(1\right) = A {e}^{- k}$

And:

${I}_{1} = \frac{10}{100} \cdot I \left(0\right) \implies A {e}^{- k} = \frac{1}{10} \cdot A$

$\therefore {e}^{- k} = \frac{1}{10} \implies k = \ln \left(10\right)$

Thus we can write the solution as:

$I \left(t\right) = A {e}^{- t \ln 10}$

We want the time, $T$, such that. $I \left(T\right)$ is 0.1% of the initial current $I \left(0\right)$, so we seek $T$ satisfying:

$I \left(T\right) = \frac{0.1}{100} \cdot I \left(0\right) \implies A {e}^{- T \ln 10} = \frac{0.1}{100} \cdot A$

$\therefore {e}^{- T \ln 10} = \frac{1}{1000}$

$\therefore - T \ln 10 = \ln \left(\frac{1}{1000}\right)$

$\therefore T \ln 10 = 3 \ln 10$

$\therefore T = 3$