Chapter 31 - Electromagnetic Oscillations and Alternating Current - Problems - Page 939: 68b

$I = 0.288~A$

We can find the current amplitude: $I = \frac{\mathscr{E}}{\sqrt{R^2+~(X_L-X_C)^2}}$ $I = \frac{\mathscr{E}}{\sqrt{R^2+~(\omega_d~L-1/\omega_d~C)^2}}$ $I = \frac{\mathscr{E}}{\sqrt{R^2+~(2\pi~f~L-1/2\pi~f~C)^2}}$ $I = \frac{170~V}{\sqrt{(200~\Omega)^2+[(2\pi)~(2000~Hz)~(60.0\times 10^{-3}~H)-1/(2\pi)~(2000~Hz)~(0.400\times 10^{-6}~F)]^2}}$ $I = 0.288~A$