A `100 Omega` resistasnce is connected in series with a `4 H` inductor. The voltage across the resistor is `V_R=(2.0V)sin(10^3 rad//s)t`:
(a) Find the expessinocircuit current
(b) Find the inductive reactance
(c) derive an expression for the voltage across the inductor,

Explanation

(a)`i=V_R/(R)=((2.0V)sin(10^3rad//s)t)/(100)`
`=(2.0xx10^-2A)sin(10^3rad//s)t`
(b) `X_L=omegaL=(10^3rad//s )(4H)`
`=4.0xx10^(3) Omega`
(c) The amplitude of voltage across inductor,
`V_0=i_0X_L=(2.0 xx 10^-2A)(4.0xx10^3Omega)`
`=80V`
In an `AC`, voltage across the inductor leads the current by `90^@` or `pi//2` rad. Hence,
`V_L=V_0sin(omegat+pi//2)`
`=(80V)sin{(10^3rad//s)t+pi/2rad}`.