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

The change in current at
`R =456 \Omega` is
`(dI)/(dt) = 7.032 * 10^(-5) A/s`

Step-by-step 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
`(dV)/(dt) = - 0.03 V/s`

The change in resistance with time is
`(dR)/(dt) = 0.03 \Omega /s`

According to ohm's law

`V = IR`

differentiating with respect to time using chain rule

`(dV)/(dt) = I (dR)/(dt) + R * (dI)/(dt)`

substituting value at R = 456

`-0.0327 = 0.09 * 0.03 + 456* (dI)/(dt)`

`(dI)/(dt) = 7.032 * 10^(-5) A/s`