A `100 Omega` resistance is connected in series with a 4H inductor. The voltage across the resistor is, `V_(R) = (2.0V) sin (10^(3) rads^(-1))`.
(i) Find the expression of circular current .
(ii) Find the inductive reactance.
(iii) 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}`.