Respuesta :
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, [tex]\frac{dV}{dt}=-0.03\ V/s[/tex] and [tex]\frac{dR}{dt}= 0.03\ ohm/s[/tex]
Using Ohm's law, we find V
[tex]V=IR[/tex]
[tex]V=0.09\times 465[/tex]
[tex]V=41.85\ V[/tex]
Re-write ohm's law in terms of I,
[tex]I=\frac{V}{R}[/tex]
Derivate w.r.t to t,
[tex]\frac{dI}{dt}=-\frac{V}{R^2}(\frac{dR}{dt})+\frac{I}{R}(\frac{dV}{dt})[/tex]
Substitute the values given,
[tex]\frac{dI}{dt}=-\frac{41.85}{(465)^2}(0.03)+\frac{0.09}{465}(-0.03)[/tex]
[tex]\frac{dI}{dt}=-0.000005806-0.000005806[/tex]
[tex]\frac{dI}{dt}=-0.000011612[/tex]
[tex]\frac{dI}{dt}\approx -0.000012[/tex]
The current I is changing at -0.000012.
