Question

At time$\mathbf{t}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{a}\mathbf{}\mathbf{\epsilon }\mathbf{=}\mathbf{45}\mathbf{}\mathbf{V}$potential difference is suddenly applied to the leads of a coil with inductance $\mathbf{L}\mathbf{=50}\mathbf{mH}$and resistance $\mathit{R}\mathbf{=}\mathbf{180}$. At what rate is the current through the coil increasing at $\mathit{t}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{2}\mathit{m}\mathit{s}$?

Short answer

The rate of current through the coil increasing at $t=1.2\mathrm{ms}$is$\frac{di}{dt}=12\mathrm{A}/\mathrm{s}$

Step-by-step solution

Given

$$\begin{gathered} \epsilon =45V \\ L=50mH=50\times {10}^{-3}H \\ R=180\Omega \\ t=1.2ms=1.2\times {10}^{-3}s \end{gathered}$$

Understanding the concept

If constant emf is introduced into a single loop circuit containing a resistance R and inductance L, the current rises to an equilibrium value of $\mathit{\epsilon }\mathbf{/}\mathit{R}$the current is given by equation 30-40. We need to differentiate that equation with respect to time to find the rate of increase in current

Formula:

role="math" localid="1661425495110" $$\begin{gathered} i=\frac{\epsilon }{R}\left(\begin{array}{c}1-e^{-\frac{Rt}{L}}\\ \end{array}\right) \\ {\tau }_L=\frac{L}{R} \end{gathered}$$

Calculate the rate of current through the coil increasing at t = 1.2 ms

The constantamfin the loop circuit containing a resistance R and inductance L, the current rises to an equilibrium value of $\epsilon /R$ the current is given by

$i=\frac{\epsilon }{R}\left(\begin{array}{c}1-e^{-\frac{Rt}{L}}\\ \end{array}\right)$

Differentiate this equation with respect to time as

$$\begin{gathered} \frac{di}{dt}=\frac{\epsilon }{R}\frac{d}{dt}{e}^{-\frac{Rt}{L}} \\ \frac{di}{dt}=\frac{\epsilon }{R}\left(-\frac{R}{L}\right)e^{-\frac{Rt}{L}} \end{gathered}$$

Where$\frac{L}{R}={\tau }_L$is the inductive time constant.

Which is calculated as

$$\begin{gathered} {\tau }_L=\frac{L}{R} \\ {\tau }_L=\frac{50\times {10}^{-3}}{180} \\ {\tau }_L=2.8\times {10}^{-4}s \end{gathered}$$

So,

$\frac{di}{dt}=-\frac{\epsilon }{L}{e}^{-\frac{t}{{\tau }_L}}$

By substituting the value we can write as

$$\begin{gathered} \frac{di}{dt}=\frac{45}{50\times {10}^{-3}}\times e^{-4.29} \\ \frac{di}{dt}=900\times 0.0137 \\ \frac{di}{dt}=12A/s \end{gathered}$$