Chapter 31 - Electromagnetic Oscillations and Alternating Current - Problems - Page 939: 69b

$I = 2.76~A$

We can find the current amplitude: $I = \frac{\mathscr{E}}{\sqrt{R^2+~(X_L-X_C)^2}}$ $I = \frac{\mathscr{E}}{\sqrt{R^2+~(\omega_d~L-1/\omega_d~C)^2}}$ $I = \frac{\mathscr{E}}{\sqrt{R^2+~(2\pi~f~L-1/2\pi~f~C)^2}}$ $I = \frac{120~V}{\sqrt{(40.0~\Omega)^2+[(2\pi)~(3000~Hz)~(850\times 10^{-6}~H)-1/(2\pi)~(3000~Hz)~(1.60\times 10^{-6}~F)]^2}}$ $I = 2.76~A$