The voltage $V$ in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance $R$ is slowly increasing as the resistor heats up. Use Ohm's law, $V = IR$, to find how the current $I$ is changing (in A/s) at the moment when $R = 327 \Omega$, $I = 0.08 A$, $\frac{dV}{dt} = -0.04 V/s$, and $\frac{dR}{dt} = 0.05 \Omega/s$. (Round your answer to six decimal places.)