A coil of self-inductance 15 H is connected to a battery of 10 V through a 5 `Omega` resistor and a switch. The switch is suddenly opened in 2 ms. Calculate the average emf induced in the coil.

Explanation

Steady state current through the coil, when the switch is closed, is
`I = (10 V)/(5 Omega)=2A`
The current through the coil when the switch is opened, `I^(.)= 0`
`"Rate of change of current" = (I^(.)-I)/(t)`
Here, t = 2 ms = 0.002 s
`(dI)/(dt)=(I^(.)-I)/(t)=(0-2)/(0.002)=-1 xx 10^(3) A // s`
Emf induced in the coil is
`epsilon=-L(dI)/(dt)`
Here, L= 15 H
`implies epsilon =-15 xx (-1 xx 10^(3))=15,000 V`