Consider a circuit as shown with a battery, $$ V_b $$
and two identical capacitors
$$ C_1, C_2 $$
(with C_2 being the one that's off to the side). Both have the same capacitance C. The resistor has resistance R.
Before adding the new capacitor C_2, the current in the ammeter is, say, I. The reed switch flips back and forth such that the capacitor charges fully and discharges fully, (only one capacitor at this moment). Thus when the switch is to the right, the capacitor has voltage V and current is V/R = I.
Now when I add the second capacitor, $$ C_2 $$ can I treat them as a single capacitor with capacitance 2C? If I do, then the current should decrease because:
$$ V = Q/2C $$ $$ I = V/R $$
Even if I consider them to be separate capacitors, Each capacitor will store charge
$$ Q $$
and thus total charge will be, $$2Q$$
Thus, voltage should be
$$ 2Q/2C = V $$
and current should stay the same. Why is this incorrect?
Note that I understand that current should double because you have twice the charge being stored and at each discharge cycle, the rate of charge flow should double and hence current should double as well.