In a LR circuit of 3 mH dinductance and `4 Omega` resistance, emf `E=4cos 1000t` volt is applied. The amplitude of current is

Options

A. `0.8 A`
B. `4/7 A`
C. `1A`
D. `(4)/(sqrt(7))A`

Answer

A

Explanation

`epsi=epsi_(0)cosomegat` . . (i)
given, `epsi=4cos1000t` . . . (ii)
From eqs. (i) and (ii), we get
Peak value of emf, `E_(2)=4V`
Angular frequency, `omega=1000Hz`
Now peak value of current current is
`i_(0)=(E_(0))/(Z)=(E_(0))/(sqrt(R^(2)+X_(L)^(2)))=(E_(0))/(sqrt(R^(2)+omega^(2)L^(2)))`
putting `E_(0)=4V,R=4Omega,omega=1000Hz,L=3mH`
`=3xx10^(-3)H`
we get `i_(0)=0.8A`