Answer
(a) $I_L = 0.80~A$
(b) $I_L = 0.80~mA$
Work Step by Step
(a) We can find the peak current:
$I_L = \frac{V_L}{X_L}$
$I_L = \frac{V_L}{\omega ~L}$
$I_L = \frac{V_L}{2\pi ~f ~L}$
$I_L = \frac{10.0~V}{(2\pi) (100~Hz) ~(0.020~H)}$
$I_L = 0.80~A$
(b) We can find the peak current:
$I_L = \frac{V_L}{X_L}$
$I_L = \frac{V_L}{\omega ~L}$
$I_L = \frac{V_L}{2\pi ~f ~L}$
$I_L = \frac{10.0~V}{(2\pi) (100\times 10^3~Hz) ~(0.020~H)}$
$I_L = 0.80\times 10^{-3}~A$
$I_L = 0.80~mA$
