N 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 0.04 Ohms per second. When the resistance is 300 Ohms and the current is 0.02 Amperes, at what rate is the current changing?

Given Information:  

Rate of change of Voltage = dV/dt = -0.02 V/s

Rate of change of Resistance = dR/dt = 0.04 Ω /s

Resistance = R = 300 Ω

Current = I = 0.02 A

Required Information:  

Rate of change of current = dI/dt = ?

dI/dt = -6.93x10⁻⁵ A/s

Explanation:  

The Ohm's law is given by

V = IR  eq. 1

Where V is the voltage across Resistance R and I is the current flowing through the Resistance R

Taking derivative of eq. 1 with respect to time t yields

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

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

dI/dt = (-I*dR/dt + dV/dt)/R  eq. 2

Now substitute the given values into eq. 2

dI/dt = (-0.02*0.04 + (-0.02))/300

dI/dt = (-0.0008 - 0.02)/300

dI/dt = -0.0208/300

dI/dt = -6.93x10⁻⁵ A/s

Therefore, the current is decreasing at the rate of 6.93x10⁻⁵ A/s