Inductive Source?

Using the Laplace transform, we have

$I(s) = \dfrac{10 \omega}{ ( s^2 + \omega^2)(\dfrac{s}{\omega} + 1) } = \dfrac{ 10 \omega^2}{ (s^2 + \omega^2)(s + \omega) }$

By partial fractions, this becomes

$I(s) = \dfrac{A}{(s + \omega) } + \dfrac{B s + C}{s^2 + \omega^2}$

Multiplying left and right by $(s + \omega) (s^2 + \omega^2)$ yields,

$10 \omega^2 = A (s^2 + \omega^2) + (B s + C) (s + \omega)$

Equating the coefficients of the powers of $s$ on both sides of the equation, yields,

$A + B = 0$

$B \omega + C = 0$

$A \omega^2 + C \omega = 10 \omega^2$

And this linear system solves to $A = 5 , B = -5 , C = 5 \omega$ , so that

$I(s) = \dfrac{5}{s + \omega } + \dfrac{ -5 s + 5 \omega }{s^2 + \omega^2 }$

Taking the inverse Laplace transform, we get,

$\bold {i(t) = 5 e^{-\omega t } - 5 \cos \omega t + 5 \sin \omega t}$

Now the instantaneous power through the resistor is

$p(t) = v(t) i(t) = R i^2(t) = i^2(t)$

And the power supplied by the source is

$p_S(t) = V_S(t) i(t) = (10 \sin \omega t) i(t)$

So that the power supplied by the inductor is

$p_L(t) = p(t) - p_S(t)$

Plugging in $\omega = 120 \pi$ and $t=0.008$ , gives $p = 34.01503686$ and $p_S =7.309736426$ , and hence, the required fraction is

$\text{Fraction of powers} = \dfrac{ p - p_S }{ p } = 0.785$