Question

the voltage, V in a circuit that satisfies the law V=IR is slowly dropping at the rate of 0.01V/s as the battery wears out. at the same time, the resistance, R is increasing at the rate of 0.5ohm/s as the resistor heats up. at what rate is the current, I changing if at this instant R=600ohms and I=0.04A?

Answer 1

\[\frac{dI}{dt}=\frac{dI}{dV}\frac{dV}{dt}+\frac{dI}{dR}\frac{dR}{dt}?\]