Question

A constant current exists in an inductor-coil connected to a battery. The coil is short-circuited and the battery is removed. Show that the charge flown through the coil after the short-circuiting is the same as that which flows in one time constant before the short-circuiting.

Answer 1

Consider an inductance L, a resitance R and a source of emf $\xi$ are connected in series.
Time constant of this LR circuit is, $\tau =\frac{L}{R}$
Let a constant current i₀ ($=\frac{\xi }{R}$) is maitened in the circuit before removal of the battery.
Charge flown in one time constant before the short-circuiting is,
$Q_{\tau }=i_0\tau$ ...(i)

Discahrge equation for LR circuit after short circuiting is given as,
$i=i_0{e}^{-\frac{t}{\tau }}$

Change flown from the inductor in small time dt after the short circuiting is given as,
$dQ=idt$

Chrage flown from the inductor after short circuting can be found by interating the above eqation within the proper limits of time,
$$\begin{gathered} Q={\int }_0^{\infty }idt \\ \Rightarrow Q={\int }_0^{\infty }{i}_0{e}^{-\frac{t}{\tau }}dt \\ \Rightarrow Q={\left[-\tau i_0{e}^{-\frac{t}{\tau }}\right]}_0^{\infty } \\ \Rightarrow Q=-\tau i_0\left[0-1\right] \\ \Rightarrow Q=\tau i_0...\left(\mathrm{ii}\right) \end{gathered}$$
Hence, proved.