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 = 475 Ω, I = 0.06 A, dV/dt = −0.04 V/s, and dR/dt = 0.03Ω/s.

Answer 1

-0.000088A/s

Explanation:

We are given that

V=IR

`R=475\Omega`

`I=0.06A`

`(dV)/(dt)=-0.04V/s`

`(dR)/(dt)=0.03\Omega/s`

Differentiate w.r.t t

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

Using the formula

`(uv)'=u'v+uv'`

Substitute the values

`-0.04=(dI)/(dt)* 475+0.06* 0.03`

`-0.04=475(dI)/(dt)+0.0018`

`475(dI)/(dt)=-0.04-0.0018`

`(dI)/(dt)=(-0.04-0.0018)/(475)=-0.000088A/s`