Problem 40 - Calculus Early Transcendentals

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, \(V=I R,\) to find how the current \(I\) is changing at the moment when \(R=400 \Omega\) \(I=0.08 \mathrm{A}, d V / d t=-0.01 \mathrm{V} / \mathrm{s},\) and \(d R / d t=0.03 \Omega / \mathrm{s}\)

Step-by-Step Solution

Answer

\(\frac{dI}{dt} = -3.1 \times 10^{-5} \; A/s\)

Step 1: Differentiate Ohm's Law

We start with Ohm's Law: \( V = I \cdot R \). Taking the derivative with respect to time \( t \), we have:\[ \frac{dV}{dt} = \frac{dI}{dt} \cdot R + I \cdot \frac{dR}{dt} \]This equation relates the rate of change of voltage to the rates of change of current and resistance.

Step 2: Substitute Known Values

Substitute the known values into the differentiated equation:\[ \frac{dV}{dt} = -0.01 \; \text{V/s}, \quad \frac{dR}{dt} = 0.03 \; \Omega/s, \quad R = 400\; \Omega, \quad I = 0.08\; A \]This gives us:\[ -0.01 = \frac{dI}{dt} \cdot 400 + 0.08 \cdot 0.03 \]

Step 3: Simplify the Equation

Calculate the expression for the term involving the rate of change of resistance:\[ 0.08 \cdot 0.03 = 0.0024 \]Substitute this back into the equation:\[ -0.01 = 400 \cdot \frac{dI}{dt} + 0.0024 \]

Step 4: Solve for \(\frac{dI}{dt}\)

Rearrange the equation to solve for \(\frac{dI}{dt}\):\[ -0.01 - 0.0024 = 400 \cdot \frac{dI}{dt} \]\[ -0.0124 = 400 \cdot \frac{dI}{dt} \]Divide both sides by 400:\[ \frac{dI}{dt} = \frac{-0.0124}{400} \]

Step 5: Calculate \(\frac{dI}{dt}\)

Calculate the value of \(\frac{dI}{dt}\):\[ \frac{dI}{dt} = -3.1 \times 10^{-5} \; A/s \]

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in electronics, expressing the relationship between voltage (\( V \)), current (\( I \)), and resistance (\( R \)) in an electric circuit as \( V = I \cdot R \). This law can be incredibly useful for understanding how electricity behaves across electrical components. By using Ohm's Law, you can determine any one of the three parameters—provided the other two are known.

• Voltage (\( V \)): the potential difference across the circuit component.
• Current (\( I \)): the flow of electric charge through the circuit.
• Resistance (\( R \)): the opposition to the flow of current, measured in ohms (\(\Omega\)).

In practical situations like the one in our problem, Ohm's Law allows us to model how changes in voltage and resistance affect the current flow in a circuit. Understanding this relationship is key when components in a circuit start to change their properties over time, such as a resistor heating up or a battery running low.

Time Derivatives

Time derivatives are used in calculus to describe how a quantity changes over time. In our exercise, the time derivatives of voltage, current, and resistance are crucial for solving the problem.

• \(\frac{dV}{dt}\): Represents the rate of change of voltage with respect to time, indicating how quickly the voltage is decreasing or increasing.
• \(\frac{dI}{dt}\): Represents the rate of change of current with respect to time, which our problem asks us to find.
• \(\frac{dR}{dt}\): Represents the rate of change of resistance, indicating how the resistance is changing as the circuit operates.

Understanding these derivatives helps predict how the parameters of the circuit will change over time. It's like a snapshot of what is happening in the circuit at any given moment, allowing us to adjust or design our circuits accordingly.

Rate of Change

The rate of change is a concept that describes how one quantity changes in relation to another. The exercise involves solving for the rate of change of the current (\( \frac{dI}{dt} \)) using known rates of change for voltage and resistance.
To find \( \frac{dI}{dt} \), we differentiate Ohm's Law with respect to time, resulting in an equation that includes all relevant rates of change. This equation was:\[ \frac{dV}{dt} = \frac{dI}{dt} \cdot R + I \cdot \frac{dR}{dt} \]This formula shows that:

• The rate at which voltage changes is influenced by how quickly the current and resistance are changing.
• By finding \( \frac{dI}{dt} \), we can determine how the current is adjusting because of the changes in voltage and resistance.

This concept is frequently used in physics and engineering to understand dynamic systems where multiple properties change simultaneously.

Electrical Circuits

Electrical circuits are pathways through which electricity flows, consisting of components like resistors, batteries, and wires. Each component plays a role in maintaining and controlling the flow of electricity.
In the problem, we considered a circuit where both the voltage supplied by a battery and the resistance of a resistor were changing over time. These changes influenced the electric current flowing through the circuit.

• A battery provides the voltage, a necessary push that moves electrons through the circuit.
• A resistor opposes this flow, creating a restriction that affects the overall current.

Electrical characteristics can change due to various factors, such as temperature, component aging, or intentional adjustment of component values. Understanding these changes is essential for managing circuit performance and stability, making the study of circuits a core aspect of electrical engineering.