A coil of self inductance 12 mH and resistance `0.8Omega` , a switch and a battery of internal resistance `0.2Omega` are connected in series. After the switch is closed, the time taken for current to attain 80%of the saturation value is `N xx 10^(-2)` sec. What is the value of N? (Take In 5 = 1.6)

Answer

`1.92`

Explanation

In a L - R circuit, the current at any time t is given as
`I= I_0[1 – exp(-t//tau)]`
since, `I = (80)/100I_0`
`implies 80/100I_0=I_0[1-"exp"(-t//tau)]`
`implies 0.8 = 1- exp ((-t)/tau)`
`implies exp((-t)/(tau))=0.2`
Taking natural logarithms on both sides, we get
`(-t)/tau=ln(1/5)`
`=t = L/Rln(5) .....[ :. tau=L/R]`
`=(12xx10^(-3))/(0.2+0.8)xx1.6....[ :. R.=R+r]`
`= 19.2 xx10^(-3)s`
`:. t = 1.92 xx10^(-2) s`