When a current of 10 ampere is flowing through a resistance of 20 ohm and inductance of 10 henry, the battery is switched off. Find (i) current after 0.4 sec. (ii) the time the current takes to fall to 60% of its initial value.

Explanation

Here, `I_(0) = 10 A, R = 20 Omega, L = 10 H`
(i) `I = ?, t = 0.4 sec`.
During decay,
`I = I_(0) e^(-(R//L)t) = 10 e^(-20 xx 0.4//10) = 10 e^(-0.8)`
`log_(10) I = log_(10) 10 - 0.8 log_(10) e`
`= 1 - 0.8 xx log_(10) 2.718 = 1 - 0.8 xx 0.343`
`log_(10) I = 1 - 0.3474 = 4.49 A`
(ii) `t = ? I = (60)/(100) I_(0)` From `I = I_(0) e^(-R//L)`
`(I)/(I_(0)) = e^((-R//L)t) , (60)/(100) = e^((-20)/(10)t) = e^(-2t)`
`log_(10) 6 - log_(10) 10 = -2 t log_(10) e`
`0.7782 - 1.0000 = -2 t xx 0.4343`
or `-0.2218 = - t xx 0.8686`
`t = (0.2218)/(0.8686) = 0.255 s`