A solenoid having an inductance of$\mathbf{6}\mathbf{.}\mathbf{30}\mathbf{\mu H}$is connected in series with a $\mathbf{1}\mathbf{.}\mathbf{20}\mathit{k}\mathit{\Omega }$resistor. (a) If a 14.0Vbattery is connected across the pair, how long will it take for the current through the resistor to reach 80.0% of its final value?

(b) What is the current through the resistor at time$\mathit{t}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}{\mathit{\tau }}_{\mathbf{L}}$?

$$\begin{gathered} 1)\mathrm{Inductance}isL=6.30\mu H \\ 2)\mathrm{Resistance}isR=1.20k\Omega \\ 3)\mathrm{Voltage}isV=14.0V \end{gathered}$$

We use the concept of current through the RL circuit. Using the equation, we can find the time taken by the current. For the given time, we can find the current in the resistor.

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

Time taken bythecurrent to reach 80.0% of its final value:

We use the equation

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

We know${\tau }_L=\frac{L}{R}$, and we have to find the time taken bythecurrent to reachof its original value.

We can write

$$\begin{gathered} i=\frac{0.800\epsilon }{R} \\ \frac{0.800\epsilon }{R}=\frac{\epsilon }{R}\left(1-e^{\frac{t}{{\tau }_L}}\right) \\ 0.800=\left(1-e^{\frac{t}{{\tau }_L}}\right) \\ {e}^{\frac{t}{{\tau }_L}}=1-0.800 \\ {e}^{\frac{t}{{\tau }_L}}=0.200 \end{gathered}$$

Taking log on both sides, we get

$$\begin{gathered} \frac{t}{{\tau }_L}=\ln 0.200 \\ -\frac{t}{{\tau }_L}=-1.609 \\ t=1.609{\tau }_L \\ t=1.609\left(\frac{L}{R}\right) \end{gathered}$$

Plugging the values of L and R, we get

$$\begin{gathered} t=\frac{1.609\left(6.30\times {10}^{-6}\right)}{1.20\times {10}^3} \\ t=8.45\times {10}^{-9}\mathrm{s} \end{gathered}$$

We can use the equation of current

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

Plugging the value of time, we get

$$\begin{gathered} i=\frac{\epsilon }{R}\left(1-e^{\frac{t}{{\tau }_L}}\right) \\ i=\frac{\epsilon }{R}\left(1-e^{-1.0}\right) \\ i=\frac{14.0}{1.20\times {10}^3}\left(1-e^{-1.0}\right) \\ i=-7.37\times {10}^3\mathrm{A} \end{gathered}$$

Therefore the current in the resistor is $7.3\times {10}^{-3}A$