A coil is connected to an AC source with peak emf, 8 V and frequency `(30)/(pi)`Hz. The coil has resistance of `8Omega`. If the average power dissipated by the coil is 0.4 W, then the inductance of the coil is

Options

A. 0.8 H
B. 2.0 H
C. 1.4 H
D. 0.4 H

Answer

D

Explanation

Given peak, emf, `V_(0)=8V`
Frequency of Ac source, `f=(30)/(pi)` Hz and resistance,
`R=8Omega`
`V_(rms)=(8)/(sqrt2)V" "("":'V_(rms)=(V_(0))/(sqrt2))`
`:.V_(rms)xxI_(rms)xxcosphi=0.4`
`V_(rms)xx(V_(rms))/(Z)xx(R)/(Z)=0.4`
`("":'` Power factor, `cosphi=(R)/(Z)`)
`(V_(rms)^(2))/(Z^(2))xxR=0.4`
`implies(((8)/(sqrt2))^(2)xx8)/(X_(L)^(2)+R^(2))=0.4`
`X_(L)^(2)+R^(2)=640`
`X_(L)^(2)+64=640`
`omega^(2)L^(2)=640=64" "(":'X_(L)=omegaL)`
`L^(2)=(576)/(omega^(2))`
`impliesL=(24)/(omega)=(24)/(2pixx(30)/(pi))=(24)/(60)=0.4H`
The inductance of the coil is 0.4 H.