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

Following are the solution to the given question:

Given:

$R = 369\ \Omega \\\\ I = 0.08\ A \\\\\frac{dV}{dt} = -0.09 \ \frac{V}{s}\\\\ \frac{dR}{dt} = 0.01 \ \frac{\Omega}{s}$

To find:

$\frac{dI}{dT}=?$

Solution:

$\to V=IR\\\\\to \frac{dV}{dt}=I \frac{dR}{dt}+R\frac{dI}{dT}\\\\\to -0.09=0.08 \times 0.01+369\times \frac{dI}{dT}\\\\ \to -0.09=0.0008 +369\times \frac{dI}{dT}\\\\\to -0.09-0.0008 =369\times \frac{dI}{dT}\\\\\to \frac{dI}{dT} = \frac{ -0.0908}{369}\\\\\to \frac{dI}{dT} = -0.0002460 \ \frac{A}{s}$

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