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 = 389 Ω, I = 0.03 A, dV/dt = −0.06 V/s, and dR/dt = 0.07 Ω/s. (Round your answer to six decimal places.)

Accepted Solution

Answer:[tex]\frac{dI}{dt}=-0.00016 A/s[/tex]Step-by-step explanation:We are given that By ohm's law [tex]V=IR[/tex]R=389 ohm I=0.03 A[tex]\frac{dV}{dt}=-0.06V/s[/tex][tex]\frac{dR}{dt}=0.07ohm/s[/tex]We have to find rate of change of current I means [tex]\frac{dI}{dt}[/tex]Differentiate the equation w.r.t t[tex]\frac{dV}{dt}=\frac{dI}{dt}R+I\frac{dR}{dt}[/tex]Substitute the values then we get [tex]-0.06=\frac{dI}{dt}\times 389+0.03\times 0.07[/tex][tex]-0.06=389\frac{dI}{dt}+0.0021[/tex][tex]-0.06-0.0021=389\frac{dI}{dt}[/tex][tex]-0.0621=389\frac{dI}{dt}[/tex][tex]\frac{dI}{dt}=\frac{-0.0621}{389}=-0.000160 A/s[/tex]Hence, the current I is changing at the rate=-0.00016A/s