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

Time Constant is Ratio of Inductance & Resistance

Explanation:

Let us have an inductance L,  

• Resistance R & source of emf $\varepsilon$ are connected in series.

• LR circuit's Time constant, $\tau = \frac {L}{R}$

Here, constant current $i_0 = (\frac {\varepsilon}{R})$

which will be maintained in full circuit the battery gets removed.

• Now, With one time constant, the charge that is flown, before it gets short-circuiting,

$Q_\tau = i_0\tau$ """(1)

• We get the LR circuit's Discharge equation but after a short circuit,

$i = i_0 e^{-t/\tau}$

• In small time dt, there is charge being drawn from the inductor,  after the short circuit,

dQ = idt

• Integrating the above equation (within limits of time) gives us the charge flown from inductor after short circuit-  

Q = $\int_{0}^{\infty } idt$

⇒ Q = $\int_{0}^{\infty } i_0 e^{-t/\tau} dt$

⇒ Q = $[-\tau i_0e^{-t/\tau}]_{0}^{\infty}$

⇒ Q = $-\tau i_0[0 - 1]$

⇒ Q = $\tau i_0$

Hence Proved !