In the circuit shown, $V_{C}$ is $0$ volts at $t=0 \; \mathrm{sec}$. For $t>0$, the capacitor current $i_{c}(t)$, where $t$ is in seconds, is given by
- $0.50 \exp (-25 t) \; \mathrm{mA}$
- $0.25 \exp (-25 t) \; \mathrm{mA}$
- $0.50 \exp (-12.5 t) \; \mathrm{mA}$
- $0.25 \exp (-6.25 t) \; \mathrm{mA}$