   Chapter 14.5, Problem 40E

Chapter
Section
Textbook Problem

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 = 0.08 A, dV/dt = −0.01 V/s, and dR/dt = 0.03 Ω/s.

To determine

To find: The rate of change of I when R=400Ω,I=0.08A,dVdt=0.01V/s and dRdt=0.03Ω/s .

Explanation

Given:

The Ohm’s Law is V=IR where V is voltage, I is current and R is resistance.

The values of R=400Ω,I=0.08A,dVdt=0.01V/s and dRdt=0.03Ω/s .

Chain Rule:

“Suppose that z=f(x,y) is a differentiable function of x and y , where x=g(t)andy=h(t) are both differentiable functions of t then, z is differentiable function of t and dzdt=zxdxdt+zydydt ”.

Calculation:

The Ohm’s Law V = IR can be written as, I=VR .

By using chain rule, obtain the value of dIdt .

dIdt=IVdVdt+IRdRdt=1RdVdt+(VR2)dRdt

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