An ideal inductor (no internal resistance for the coil) or 20 mH is connected in series with an AC ammeter to an AC source whose emf is given by `e=20sqrt(2)sin (200t+pi//3)V`, where t is in seconds. Find the reading of the ammeter ?

Explanation

Given that `L = 20 mH = 20 xx 10^(-3)H`
`e=20sqrt(2)sin(200t+(pi)/(3))V`
`e=E_(0)sin(omega t+phi)`
`therefore E_(0)=20 sqrt(2), omega = 200`
`E_("rms")=(E_(0))/(sqrt(2))=(20sqrt(2))=(20sqrt(2))/(sqrt(2))=20V`
`i_("rms")=(E_("rms"))/(X_(L))=(E_("rms"))/(omega L)`
`= (20)/(200xx20xx10^(-3))=(10)/(2)=5A`