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The voltage Vin 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 = 456 n, I = 0.05 A, dV/dt = -0.09 V/s, and dR/dt = 0.06 N/s. (Round your answer to six decimal places.) A/s

The voltage Vin 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 = 456 n, I = 0.05 A, dV/dt = -0.09 V/s, and dR/dt = 0.06 N/s. (Round your answer to six decimal places.) A/s

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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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 = 0.05 A, dV/dt = -0.09 V/s, and dR/dt = 0.06 N/s. (Round your answer to six decimal places.) 456 N, I = A/s
Transcribed Image Text: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 = 0.05 A, dV/dt = -0.09 V/s, and dR/dt = 0.06 N/s. (Round your answer to six decimal places.) 456 N, I = A/s
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