Question

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 0.02 volts per second and, as the resistor heats up, the resistance is increasing at 0.01 ohms per second. When the resistance is 200 ohms and the current is 0.02 amperes, at what rate is the current changing?

Answer 1

dV/dt = I*dR/dt + R*dI/dt

Answer 2

ok but how do i set up the equation to find I, dR/dt,R, and dl/dt?

Answer 3

Your problem gives you a lot of these things;
e.g., dV/dt = 0.02 V/second.

Answer 4

i am looking for the change in I... how do get that?

Answer 5

to find the rate of change of I with respect to time, you simply solve dV/dt = I*dR/dt + R*dI/dt for I, which is easy to do since you've been given all the values

Answer 6

Let V' = dV/dt etc., then we have
V' = IR' + RI'
=> I' = (V' - IR')/R

Now you have in the problem already all of the terms on the right hand side and hence you can solve for I' = dI/dt