91Ó°ÊÓ

Figure 30-15 is the RL circuit.

We use the concept of emf induced in the inductor. Using the equation, we can find the ratio of emf induced to the battery emf for the given values of the time. Also we can find the value of time constant.

${\mathit{Î\mu }}_{\mathbf{L}}\mathbf{=}\mathbf{}\mathit{Î\mu }\mathbf{×}{\mathit{e}}^{\mathbf{-}\mathbf{}\frac{\mathbf{t}}{{\mathbf{Ï„}}_{\mathbf{L}}}}$

The ratio of inductor’s self-induced emf to the battery’s emf just after t = 0 :

We can applytheloop rule of voltage:

$\begin{array}{rcl}\mathrm{Î\mu }-{\mathrm{Î\mu }}_L-iR& =& 0\\ \mathrm{Î\mu }-{\mathrm{Î\mu }}_L& =& iR\\ & & \end{array}$
At t =0 the current through the R is also zero, so we get

$\begin{array}{rcl}\mathrm{Î\mu }-{\mathrm{Î\mu }}_L& =& 0\\ \mathrm{Î\mu }& =& {\mathrm{Î\mu }}_L\\ \frac{{\mathrm{Î\mu }}_L}{\mathrm{Î\mu }}& =& 1\\ & & \end{array}$

So, the ratio of $\frac{{\mathrm{Î\mu }}_L}{\mathrm{Î\mu }}$is 1.

The ratio of inductor’s self-induced emf to the battery’s emf just after $t=2.00{\mathrm{τ}}_L$:

Using the equation, we can write

$\begin{array}{rcl}{\mathrm{Î\mu }}_L& =& \mathrm{Î\mu }×e^{-\frac{t}{{\mathrm{Ï„}}_L}}\\ {\mathrm{Î\mu }}_L& =& \mathrm{Î\mu }×e^{-\frac{2.00{\mathrm{Ï„}}_L}{{\mathrm{Ï„}}_L}}\\ {\mathrm{Î\mu }}_L& =& \mathrm{Î\mu }×e^{-2.00}\\ {\mathrm{Î\mu }}_L& =& 0.1353\mathrm{Î\mu }\\ \frac{{\mathrm{Î\mu }}_L}{\mathrm{Î\mu }}& =& 0.1353\end{array}$

So, the ratio $\frac{{\mathrm{Î\mu }}_L}{\mathrm{Î\mu }}$of is 0.1353.

We have

$\begin{array}{rcl}{\mathrm{Î\mu }}_L& =& \mathrm{Î\mu }×e^{-\frac{t}{{\mathrm{Ï„}}_L}}\\ \frac{{\mathrm{Î\mu }}_L}{\mathrm{Î\mu }}& =& e^{-\frac{t}{{\mathrm{Ï„}}_L}}\\ & & \end{array}$

Taking log on both sides, we get

$\begin{array}{rcl}-\frac{t}{{\mathrm{Ï„}}_L}& =& \ln \left(\frac{{\mathrm{Î\mu }}_L}{\mathrm{Î\mu }}\right)\\ -\frac{t}{{\mathrm{Ï„}}_L}& =& \ln \left(0.500\right)\\ -\frac{t}{{\mathrm{Ï„}}_L}& =& -0.693\\ \frac{t}{{\mathrm{Ï„}}_L}& =& 0.693\\ & & \end{array}$

So, the ratio of $\frac{t}{{\mathrm{Ï„}}_L}$is 0.693.