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1. The circuit shown in the figure, with R = 1/3 Ω, L = 1/4H, C = 3F has input voltage v(t) = sin2t. The resulting current i(t) is
(A): 5 sin (2t + 53.1°)
(B): 5 sin (2t - 53.1°)
(C): 25 sin (2t + 53.1°)
(D): 25 sin (2t - 53.1°)
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