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 = 465?, I = 0.09A, dV/dt =-0.03V/s, and dR/dt = 0.03?/s. (Round your answer to six decimal places.)

Answer 1

Answer:

The current I is changing at -0.000012.

Step-by-step explanation:

Given : 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.

Solution :

We have given,

R = 465\ ohm, I = 0.09A, $\frac{dV}{dt}=-0.03\ V/s$ and $\frac{dR}{dt}= 0.03\ ohm/s$

Using Ohm's law, we find V

$V=IR$

$V=0.09\times 465$

$V=41.85\ V$

Re-write ohm's law in terms of I,

$I=\frac{V}{R}$

Derivate w.r.t to t,

$\frac{dI}{dt}=-\frac{V}{R^2}(\frac{dR}{dt})+\frac{I}{R}(\frac{dV}{dt})$

Substitute the values given,

$\frac{dI}{dt}=-\frac{41.85}{(465)^2}(0.03)+\frac{0.09}{465}(-0.03)$

$\frac{dI}{dt}=-0.000005806-0.000005806$

$\frac{dI}{dt}=-0.000011612$

$\frac{dI}{dt}\approx -0.000012$

The current I is changing at -0.000012.