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.)

Following are the solution to the given question:

Given:

To find:

Solution:

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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 = ?

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: option c. i.e. electric current flowing “downhill” from the negative electrode to the positive electrode in a battery

explanation:

the water analogy to the circuits define the pipes (water flows) to the electric current. now, we know that the electric current flows from the positive terminal to the circuit (as shown in the diagram attached).

hence, option c is the correct answer.