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=L/R`

let a constant current `(i_0=xi/R)` is maitened in the circuit before removal of the battery.

Charge flown in one time constant before the short-circuiting is,

\[Q_\tau  =  i_0 \tau..........(1)\]

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,

\[Q = \int_0^\infty idt\]

\[ \Rightarrow Q = \int_0^\infty i_0 e^{- \frac{t}{\tau}} dt\]

\[ \Rightarrow Q = \left[ - \tau i_0 e^{- \frac{t}{\tau}} \right]_0^\infty \]

\[ \Rightarrow Q = - \tau i_0 \left[ 0 - 1 \right]\]

\[ \Rightarrow Q = \tau i_0 ............. (2)\]

Hence, proved.