If a larger value of resistance were used with the same value of capacitance in the above circuit, then a smaller current would flow, therefore it would take longer for the capacitor to charge up ...

The gradient of this graph is equal to the capacitance of the capacitor. \(C= \frac{Q}{V}\) And the area under the graph is the energy stored by the capacitor.

The more current you draw, the more you are going to droop the capacitors when they are storing energy instead of charging and that’s going to affect the output, too. So how bad is it?