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 0.04 volts per second and, as the resistor heats up, the resistance is increasing at 0.04 ohms per second. When the resistance is 100 ohms and the current is 0.03 amperes, at what rate is the current changing?

Answer 1

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

dV/dt = -0.04, dR/dt = 0.04, R = 100, I = 0.03, V = 3:

dI/dt = (-0.04 - 0.04*3)/100

= -0.0016 amperes per second

Answer 2

it says thats the wrong answer

Answer 3

pardon but the voltage sign in SI is U