# RC circuit with lossy capacitor

This circuit

simulate this circuit – Schematic created using CircuitLab

can be regarded as an RC series circuit with a lossy capacitor. The time expression for $V_C(t)$ with $R_2 \to \infty$ is available here and it is $V_C(t) = V_0 (1 - e^{-t/\tau})$.

I would like to obtain the same expression, but in this situation.

$V_0$ is the DC voltage generator; the switch is closed for $t \geq 0$ and $V_C (t = 0) = 0$ (capacitor initially discharged).

I can write

$$\frac{V_0 - V_C (t)}{R_1} = I(t)$$

$$\frac{V_0}{R_1} - \frac{1}{R_1 C} \frac{dQ(t)}{dt} = I(t)$$

which is the current across $R_1$ and so the total corrent entering $C // R_2$. $V_C (t)$ is variable during the capacitor charge. The fact is that here $I(t)$ is not simply $dQ(t) / dt$, because not all the charge exiting from $R_1$ goes through the capacitor: part of it flows across $R_2$ and this amount of "leaked" charge changes (raises) with time. So, how can this be taken into account?

Are there any hints to obtain a differential equation for the charge or the current of the capacitor?

$I_{R1}(t) = I_C(t) + I_{R2}(t)$.