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.)

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

Step-by-step explanation:

We are given the following in the question:

According to the Ohm's Law:

V = IR

Differentiating we get,

Putting values, we get,

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

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step-by-step explanation:

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