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We are asked to find the rate at which the current \( I \) is changing, given that the voltage \( V \) and resistance \( R \) are both changing. The governing equation is Ohm's Law:

\[
V = I R
\]

To find how the current \( I \) is changing, we need to differentiate both sides of Ohm's Law with respect to time \( t \):

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

We are given the following values:
- \( R = 368 \, \Omega \)
- \( I = 0.01 \, A \)
- \( \frac{dV}{dt} = -0.04 \, V/s \)
- \( \frac{dR}{dt} = 0.05 \, \Omega/s \)

Now, substitute these values into the differentiated equation:

\[
-0.04 = (0.01)(0.05) + (368) \frac{dI}{dt}
\]

Simplify the equation:

\[
-0.04 = 0.0005 + 368 \frac{dI}{dt}
\]

Subtract 0.0005 from both sides:

\[
-0.0405 = 368 \frac{dI}{dt}
\]

Now solve for \( \frac{dI}{dt} \):

\[
\frac{dI}{dt} = \frac{-0.0405}{368} = -0.000110 \, A/s
\]

Thus, the current is decreasing at a rate of \( -0.000110 \, A/s \), to six decimal places.

Would you like any further details or explanations?

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Here are 5 related questions:
1. How does Ohm's law apply when both resistance and voltage are changing over time?
2. What would happen if the voltage were increasing instead of decreasing?
3. How would the answer change if \( \frac{dR}{dt} \) were negative?
4. What units are typically used for the rate of change of current?
5. How does temperature affect the resistance in real circuits?

**Tip**: Remember to carefully track units in problems involving rates of change!

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 (in A/s) at the moment when R = 368 Ω, I = 0.01 A, dV/dt = -0.04 V/s, and dR/dt = 0.05 Ω/s. (Round your answer to six decimal places.)

This problem demonstrates how to apply Ohm's Law to find the rate of change of current in a circuit where both voltage and resistance are changing over time. With given values for R = 368 Ω, I = 0.01 A, dV/dt = -0.04 V/s, and dR/dt = 0.05 Ω/s, we calculate dI/dt step by step.

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