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 = 475 Ω, I = 0.06 A, dV/dt = −0.04 V/s, and dR/dt = 0.03Ω/s.

Answer 1

Answer:

-0.000088A/s

Step-by-step explanation:

We are given that

V=IR

$R=475\Omega$

$I=0.06A$

$\frac{dV}{dt}=-0.04V/s$

$\frac{dR}{dt}=0.03\Omega/s$

Differentiate w.r.t t

$\frac{dV}{dt}=\frac{dI}{dt}R+I\frac{dR}{dt}$

Using the formula

$(uv)'=u'v+uv'$

Substitute the values

$-0.04=\frac{dI}{dt}\times 475+0.06\times 0.03$

$-0.04=475\frac{dI}{dt}+0.0018$

$475\frac{dI}{dt}=-0.04-0.0018$

$\frac{dI}{dt}=\frac{-0.04-0.0018}{475}=-0.000088A/s$