The time constant of an inductance coil is `2 xx 10^(-3) s`. When a `90 Omega` resistance is joined in series, the same constant becomes `0.5 xx 10^(-3) s`. The inductance and resistance of the coil are

Options

A. `30mH,30Omega`
B. `30mH,60Omega`
C. `60mH,30Omega`
D. `30mH,60Omega`

Answer

C

Explanation

`tau=(L)/(R)=2xx10^(-3)`
`tau'=(L)/(R+90)=0.5xx10^(-3)`
`(i)//(ii) implies (R+90)/(R)=4`
`R=30Omega`
`L=2xx10^(-3)xx30=60mH`