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 = 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.)

Answer 1

Answer:

$\frac{dI}{dt}=-0.00016 A/s$

Step-by-step explanation:

We are given that

By ohm's law

$V=IR$

R=389 ohm

I=0.03 A

$\frac{dV}{dt}=-0.06V/s$

$\frac{dR}{dt}=0.07ohm/s$

We have to find rate of change of current I means $\frac{dI}{dt}$

Differentiate the equation w.r.t t

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

Substitute the values then we get

$-0.06=\frac{dI}{dt}\times 389+0.03\times 0.07$

$-0.06=389\frac{dI}{dt}+0.0021$

$-0.06-0.0021=389\frac{dI}{dt}$

$-0.0621=389\frac{dI}{dt}$

$\frac{dI}{dt}=\frac{-0.0621}{389}=-0.000160 A/s$

Hence, the current I is changing at the rate=-0.00016A/s