The circuit shown in in the figure, with $R=\frac{1}{3} \Omega$, $\mathrm{L}=\frac{1}{4} \mathrm{H}, \mathrm{C}=3 \mathrm{~F}$ has input voltage $v(t)=\sin 2 t$. The resulting current $i(t)$ is
- $5 \sin (2t + 53.1^{\circ})$
- $5 \sin (2t - 53.1^{\circ})$
- $25 \sin (2t + 53.1^{\circ})$
- $25 \sin (2t - 53.1^{\circ})$