In a simple electric circuit, Ohm's law states that $V=IR,$ where $V$ is the voltage in Volts, I is the current in Amperes,

and $R$ is the resistance in Ohms. Assume that, as the battery wears out, the voltage decreases at $0\mathrm{.}03$ Volts per second

and, as the resistor heats up$,$ the resistance is increasing at $0\mathrm{.}04$ Ohms per second. When the resistance is $200$ Ohms

and the current is $0\mathrm{.}01$ Amperes, at what rate is the current changing?

Amperes per second