Question

The voltage ???? in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance ???? is slowly increasing as the resistor heats up. Use Ohm’s law, ????=????????, to find how the resistance ???? is changing at the moment when ????= ???? Volts, ????=???? Ampere. The rate at which voltage is decreasing is ????.???? V/s, and the rate at which the current is increasing is ????.???? ampere/s. Interpret the solution in context of the problem, don’t forget to write th

Answer 1

Answer:

The change in current at  $R =456 \Omega$ is  $\frac{dI}{dt} = 7.032 * 10^{-5} A/s$

Explanation:

From the question we are told that

    The resistance is $R = 465 \Omega$

     The current is  $I = 0.09A$

    The change in voltage with respect to time is $\frac{dV}{dt} = - 0.03 V/s$

     The change in resistance with time is  $\frac{dR}{dt} = 0.03 \Omega /s$

According to ohm's law

        $V = IR$

differentiating with respect to time using chain rule

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

substituting value  at R = 456

             $-0.0327 = 0.09 * 0.03 + 456* \frac{dI}{dt}$

              $\frac{dI}{dt} = 7.032 * 10^{-5} A/s$