Question

The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasing as the resistor heats up. Use Ohm's Law, V = IR, to find how the current I is changing at the moment when R = 465?, I = 0.09A, dV/dt =-0.03V/s, and dR/dt = 0.03?/s. (Round your answer to six decimal places.)

Answer 1

The current I is changing at -0.000012.

Explanation:

Given : The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasing as the resistor heats up. Use Ohm's Law, V = IR

To find : How the current I is changing at the moment.

Solution :

We have given,

R = 465\ ohm, I = 0.09A,
`(dV)/(dt)=-0.03\ V/s` and
`(dR)/(dt)= 0.03\ ohm/s`

Using Ohm's law, we find V

`V=IR`

`V=0.09* 465`

`V=41.85\ V`

Re-write ohm's law in terms of I,

`I=(V)/(R)`

Derivate w.r.t to t,

`(dI)/(dt)=-(V)/(R^2)((dR)/(dt))+(I)/(R)((dV)/(dt))`

Substitute the values given,

`(dI)/(dt)=-(41.85)/((465)^2)(0.03)+(0.09)/(465)(-0.03)`

`(dI)/(dt)=-0.000005806-0.000005806`

`(dI)/(dt)=-0.000011612`

`(dI)/(dt)\approx -0.000012`

The current I is changing at -0.000012.