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 = 389 Ω, I = 0.03 A, dV/dt = −0.06 V/s, and dR/dt = 0.07 Ω/s. (Round your answer to six decimal places.)

Answer 1

`(dI)/(dt)=-0.00016 A/s`

Explanation:

We are given that

By ohm's law

`V=IR`

R=389 ohm

I=0.03 A

`(dV)/(dt)=-0.06V/s`

`(dR)/(dt)=0.07ohm/s`

We have to find rate of change of current I means
`(dI)/(dt)`

Differentiate the equation w.r.t t

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

Substitute the values then we get

`-0.06=(dI)/(dt)* 389+0.03* 0.07`

`-0.06=389(dI)/(dt)+0.0021`

`-0.06-0.0021=389(dI)/(dt)`

`-0.0621=389(dI)/(dt)`

`(dI)/(dt)=(-0.0621)/(389)=-0.000160 A/s`

Hence, the current I is changing at the rate=-0.00016A/s