In the given circuit `S_(1)" and " S_(2)`
are switches. `S_(2)` is closed for
a long time and `S_(1)` remains
Open. Now `S_(1)` is also closed. Here
`R=10 Omega,L=1` mH and
`epsilon=3 V."Let" (di)/(dt)` is the
magnitude of rate of change of
current (in"A/s"). Just after `S_(1)`
is Closed. The value for the
inductor L is
`(di)/(dt)=x xx 10^(3)"A/s".` Then find x.

Answer

2

Explanation

When `S_(1)` is closed current in
inductor remains , i= `(e)/(2R)`
`therefore (e-V_(1))/(R)+(e-V_(1))/(2R)=(e)/(2R)implies V_(1)=(2 epsilon)/(3)`

The potential difference across the
inductor, `(V_(1))=(2 epsilon)/(3)=L(di)/(dt)`
`implies(di)/(dt)=+(2 epsilon)/(3L)=(2 xx 3)/(3 xx 10^(-3))=2000`
`A//s=x xx 10^(3) A//s implies x=2`