Question

How long would it take, following the removal of the battery, for the potential difference across the resistor in an RL circuit (with L = 2.00H, R = 3.00) to decay to 10.0% of its initial value?

Short answer

The time required for the current to decay to 10% of its initial value is t = 1.54s

Step-by-step solution

Given

$$\begin{gathered} L=2.00H \\ R=3.00 \\ i=0.1i_0 \end{gathered}$$

Understanding the concept

When the source of constant emf is removed, the current decay from a valueaccording to the equation of decay of current given by 30.45. So we can use that equation to find the time.

Formula:

$$\begin{gathered} i=i_0{e}^{-t/{\tau }_L} \\ {\tau }_L=\frac{L}{R} \end{gathered}$$

Calculate the time required for the current to decay to 10% of its initial value.

We have, the equation of decay of current as,

$$\begin{gathered} i=i_0{e}^{\frac{-t}{{\tau }_L}} \\ \frac{i}{i_0}=e^{\frac{-t}{{\tau }_L}} \\ \frac{i_0}{i}=e^{\frac{t}{{\tau }_L}} \end{gathered}$$

Taking logofboththe sides

$$\begin{gathered} \ln \left(\frac{i_0}{i}\right)=\ln \left(e^{\frac{-t}{{\tau }_L}}\right) \\ \ln \left(\frac{i_0}{i}\right)=\frac{t}{{\tau }_L} \end{gathered}$$

Where ${\tau }_L=\frac{L}{R}\mathrm{and}i=0.1i_0$

$$\begin{gathered} \ln \ln \left(\frac{i_0}{0.1i_0}\right)=\frac{tR}{L} \\ t=\frac{L}{R}\ln \ln \left(10\right) \\ t=\frac{2.00}{3.00}\ln \ln \left(10\right) \\ t=1.54s \end{gathered}$$