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 = 369Ω, IS = 0.08A, dV/dt = -0.09V/s, and dR/dt = 0.01Ω/s. (Round your answer to six decimal places.)

Answer 1

Given Information:

Current = I = 0.08 A

Resistance = R = 369 Ω

Rate of change of voltage = dV/dt = -0.09 V/s

Rate of change of resistance = dR/dt = 0.01 Ω/s

Required Information:

Rate of change of current = dI/dt = ?

Answer:

Rate of change of current = dI/dt = 0.000241 A/s

Explanation:

As we know Ohm's law is given by

V = IR

or

I = V/R

Where V is the voltage, I is the current and R is the resistance.

Taking derivative with respect to time yields,

dI/dt = -V/R²(dR/dt) + 1/R(dV/dt)

We have V = IR = 0.09*369 = 33.21 V

So the rate of change of current becomes,

dI/dt = -(33.21)/(369)²(0.01) + 1/369(-0.09)

dI/dt = -0.000241 A/s

Therefore, the current in the given electrical circuit is decreasing at the rate of 0.000241 A/s

Answer 2

Answer:

dI/dt = -2.4*10^{-4}A/s

Explanation:

In this case both I and R depends of time. We have information about derivatives and constants.  by deriving V (Ohm's law) we can obtain the change of the current dI/dt:

$V(t)=I(t)R(t)\\\\\frac{dV}{dt}=I\frac{dR}{dt}+R\frac{dI}{dt}\\\\\frac{dI}{dt}=\frac{1}{R}[\frac{dV}{dt}-I\frac{dR}{dt}]$

Finally, by replacing the values of R, Is, dV/dt, and dR/dt we obtain:

$\frac{dI}{dt}=\frac{1}{369\Omega}[-0.09V/s-(0.08A)(0.01\Omega)]=-2.4*10^{-4}A/s$

Hence, the current is decreasing with a value of -2.4*10^{-4}A/s