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

V = IR
dV/dt = -0.01 (V/s)
dR/dt = 0.5 (Ohms/s)
R = 600 (Ohms)
I = 0.04 (As)

Find dI/dt

Answer 2

This is a pretty straightforward related rates problem,

\[\frac{dV}{dt} = I*\frac{dR}{dt}+R*\frac{dI}{dt}\]

Answer 3

Thanks a bunch, I'll give it a shot (y)