91Ó°ÊÓ

$$\begin{gathered} \mathrm{Î\mu }=45V \\ L=50mH=50×{10}^{-3}H \\ R=180\mathrm{Î\copyright } \\ t=1.2ms=1.2×{10}^{-3}s \end{gathered}$$

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 $\mathit{Î\mu }\mathbf{/}\mathit{R}$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" $$\begin{gathered} i=\frac{\mathrm{Î\mu }}{R}\left(\begin{array}{c}1-e^{-\frac{Rt}{L}}\\ \end{array}\right) \\ {\mathrm{Ï„}}_L=\frac{L}{R} \end{gathered}$$

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

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

Differentiate this equation with respect to time as

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

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

Which is calculated as

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

So,

$\frac{di}{dt}=-\frac{\mathrm{Î\mu }}{L}{e}^{-\frac{t}{{\mathrm{Ï„}}_L}}$

By substituting the value we can write as

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