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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 = 337Ω, I = 0.01A, dV/dt = -0.07V/s, and dR/dt = 0.06Ω/s. (Round your answer to six decimal places.)

Respuesta :

Answer:

The current is decreasing at a rate 0.000209 ampere per second.

Step-by-step explanation:

We are given the following in the question:

[tex]R = 337\Omega\\I = 0.01 A\\\\\dfrac{dV}{dt} = -0.07\text{ V/s}\\\\\dfrac{dR}{dt} = 0.06\text{ ohm/s}[/tex]

According to the Ohm's Law:

V = IR

Differentiating we get,

[tex]\dfrac{dV}{dt} = I\dfrac{dR}{dt} + R\dfrac{dI}{dt}[/tex]

Putting values, we get,

[tex]-0.07 = (0.01)(0.06)+ (337)\dfrac{dI}{dt}\\\\(337)\dfrac{dI}{dt} = -0.07-(0.01)(0.06) =-0.0706\\\\\dfrac{dI}{dt} = \dfrac{-0.0706}{337} = -0.000209\text{ A/s}[/tex]

Thus, the current is decreasing at a rate 0.000209 ampere per second.