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. `0.8Å`
B. `4/7Å`
C. `1.0Å`
D. `(4)/(sqrt7)Å`

Answer

A

Explanation

`E=E_(0)cos omegat" "…(i)`
Given `E=4 cos 1000t" "…(ii)`
From Eqs. (i) and (ii), we get
Peak value of emf, `E_(0)=4V`
Angular frequency,` omega=1000 Hz`
Now peak value of 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`