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 = 400 Ω, I = 0.02 A, dV/dt = −0.05 V/s, and dR/dt = 0.04 Ω/s. (Round your answer to six decimal places.)

Answer 1

dI/dt= -0,000127 A/s

Explanation:

From Ohm's Law we have V = IR . deriving on both sides of equality

dV/dt = dI/dt R + I dR/dt . clearing, dI/dt = 1/R (dV/dt - I dR/dt)

replacing by the values ​​of the statement

dI/dt = 1/400Ω (-0,05 V/s - 0,02 A 0,04Ω/s) . making calculations

dI/dt = -0,0508V/s / 400Ω = -0,000127 V/sΩ. simplifying the units

dI/dt = -0,000127 A/s