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.)

Question

Lilian West

Mathematics

1 February, 15:36

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.)

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Lutz · Answer 1

dI/dt = - 0,000127 A/s

Step-by-step 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