In the picture below, we are charging up the capacitor, by connecting it (and the resistor to a battery of voltage $V_0$, at time $t = 0$). In terms of what is happening physicially, how would you interpret the given equation $$V_0 = \frac{Q}{C} + IR?$$

Given that the potential difference accross the battery is $V_0$, and it charges the capacitor (which then has it's own potential difference $\frac{Q}{C}$), which produces an electric field along the conductors (wires) in a different direction to that of the battery, would it be correct to say that $V_0 - \frac{Q}{C}$ is then the net potential difference accross the battery which drives the current $I$. Hence using ohms law we have $V_0 - \frac{Q}{C} = IR$? Is this what is happening in this equation, or is it at least close?

Thanks.