By Ohm's Law, if I have a well working circuit and I swap out the battery for a new battery which has double the Voltage drop, but I maintain the circuit's Resistance, what will happen to the Current?

current would double

Explanation:

According To Ohm's Law, the current in a conducting wire depends on the voltage drop across the two ends of the wire. It can be expressed as:

I ∝ V

⇒ V = I R

where, R is the resistance and proportionality constant.

In the given case, the resistance of the circuit remains same while the voltage drop has increased to twice the initial value by changing the battery. This will the cause the current to double as well from its previous value.