A coil of inductance `8.4mH` and resistance `6Omega` is connected to a `12V` battery. The current in the coil is 1.0A` in the time (approx .)

Options

A. 500 sec
B. 20 sec
C. 35 milli sec
D. 1 milli sec

Answer

D

Explanation

Peak current in the circuits `i_0=12/6=2A`
Current decreases from 2A to 1A 1.e., becomes half in time `t=0.693 L/R = 0.693 xx(8.4xx10^(-3))/6 =1` milli sec
Aliter : `i_0 =12/6 =2A`
`i=i_0(1-e-(Rt)/L)`
`implies 1=2(1-e^((-6t)/(8.4 xx10^(-3))))implies 1/2=1-e^((-6t)/(8.4 xx10^(-3)))`
`implies (6t)/(8xx10^(-3))=ln2 implies t = (8.4 xx10^(-3)xx0.693)/(6)`
`implies t = 0.97 m sec ~~ 1 m` sec.