RC discharge

What you are seeing: a capacitor charged to $V_0$ discharging through a resistor $R$. Kirchhoff's loop equation gives $V_C(t) = V_0 \exp(-t/\tau)$ with $\tau = RC$. The current is just $I(t) = V_C(t)/R$, also exponentially decaying. The orange ball on the circuit diagram represents the charge $Q(t) = C V(t)$ leaking from the capacitor through the resistor.

Look for the canonical time constants: $V$ drops to $1/e \approx 37$ percent at $t = \tau$, to half at $t = \tau \ln 2 \approx 0.693 \tau$, and to 1 percent at $t \approx 4.6 \tau$. Total energy dissipated as heat in $R$ equals the initial $U_C(0) = \tfrac{1}{2} C V_0^2$.