Question

A battery is connected to a series RL circuit at time t = 0 . At what multiple of${\mathit{\tau }}_{\mathbf{L}}$will the current be 0.100% less than its equilibrium value?

Short answer

$t=6.90{\tau }_L$

Step-by-step solution

Given

A battery is connected to a series RL circuit at time t = 0

Understanding the concept

We use the concept of current through RL circuit. Using the equation, we can find the time at the given value of current.

Formulae:

$i=\frac{V}{R}\left(1-e^{-\frac{t}{{\tau }_L}}\right)$

Calculate at what multiple of τL will the current be 0.100%  less than its equilibrium value

We know ${\tau }_L=\frac{L}{R}$,and 0.100% less means 0.999% , so we can write

$i=0.999\left(\frac{V}{R}\right)$

Using this intheequation of current, we get

$$\begin{gathered} 0.999\left(\frac{V}{R}\right)=\frac{V}{R}\left(1-e^{-\frac{t}{{\tau }_L}}\right) \\ 0.999=1-e^{-\frac{t}{{\tau }_L}} \\ {e}^{-\frac{t}{{\tau }_L}}=1-0.999 \\ {e}^{-\frac{t}{{\tau }_L}}=1-0.0010 \end{gathered}$$

Taking log on both sides,

$$\begin{gathered} -\frac{t}{{\tau }_L}=-6.90 \\ t=6.90{\tau }_L \end{gathered}$$