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In the circuit shown below, the initial charge on the capacitor is $2.5 \mathrm{mC}$, with the voltage polarity as indicated. The switch is closed at time $t=0$. The current $i(t)$ at a time $t$ after the switch is closed is

- $i(t)=15 \exp (-2 \times 10^3 \mathrm{t) A}$
- $i(t)=5 \exp (-2 \times 10^3 \mathrm{t) A}$
- $i(t)=10 \exp (-2 \times 10^3 \mathrm{t) A}$
- $i(t)=-5 \exp (-2 \times 10^3 \mathrm{t) A}$