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ludmilkaskok [199]
3 years ago
9

The voltage ???? in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance ???? is slowly inc

reasing 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
Physics
1 answer:
zimovet [89]3 years ago
6 0

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

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