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Chapter 14: Problem 39

Changing voltage in a circuit The voltage \(V\) in a circuit that satisfies the law \(V=I R\) is slowly dropping as the battery wears out. At the same time, the resistance \(R\) is increasing as the resistor heats up. Use the equation $$ \frac{d V}{d t}=\frac{\partial V}{\partial I} \frac{d I}{d t}+\frac{\partial V}{\partial R} \frac{d R}{d t} $$ to find how the current is changing at the instant when \(R=\) 600 ohms, \(I=0.04\) amp, \(d R / d t=0.5 \mathrm{ohm} / \mathrm{sec},\) and \(d V / d t=\) \(-0.01\) volt/sec.

Short Answer

Expert verified
The current \( I \) is decreasing at a rate of \(-0.00005\) A/sec.

Step by step solution

01

Differentiate V with Respect to I

Using the relationship \( V = IR \), we first find the partial derivative of \( V \) with respect to \( I \). This gives us \( \frac{\partial V}{\partial I} = R \) because \( V = IR \) and \( R \) is treated as a constant when differentiating with respect to \( I \).
02

Differentiate V with Respect to R

Next, we find the partial derivative of \( V \) with respect to \( R \). Similarly, treating \( I \) as a constant, \( \frac{\partial V}{\partial R} = I \) because \( V = IR \).
03

Substitute Known Values into the Equation

Now substitute the known values into the equation \( \frac{d V}{d t} = \frac{\partial V}{\partial I}\frac{d I}{d t} + \frac{\partial V}{\partial R}\frac{d R}{d t} \). We have \( R = 600 \) ohms, \( I = 0.04 \) amp, \( \frac{d R}{d t} = 0.5 \) ohms/sec, and \( \frac{d V}{d t} = -0.01 \) volt/sec.
04

Substitute Partial Derivatives into the Formula

Substitute the partial derivatives from Step 1 and Step 2: \( \frac{\partial V}{\partial I} = 600 \) and \( \frac{\partial V}{\partial R} = 0.04 \).
05

Set Up the Equation

Now, place these into the equation: \( -0.01 = 600 \times \frac{d I}{d t} + 0.04 \times 0.5 \).
06

Simplify the Equation

First, calculate \( 0.04 \times 0.5 = 0.02 \). This gives us the equation: \( -0.01 = 600 \times \frac{d I}{d t} + 0.02 \).
07

Solve for \( \frac{d I}{d t}\)

Rearrange the equation to solve for \( \frac{d I}{d t} \): \( 600 \frac{d I}{d t} = -0.01 - 0.02 = -0.03 \). Thus, \( \frac{d I}{d t} = \frac{-0.03}{600} = -0.00005 \) A/sec.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a fundamental concept in calculus that involves finding the rate at which a function is changing at any given point. In the context of related rates, differentiation helps us understand how one variable changes with respect to another over time. For electrical circuits, this is crucial as it allows for the calculation of changing parameters like current or voltage.To differentiate a function like voltage in a circuit, which is expressed as \( V = IR \), we apply partial differentiation to find its rate of change with respect to other variables like current \( I \) and resistance \( R \).
  • When differentiating with respect to \( I \), we treat \( R \) as a constant, resulting in \( \frac{\partial V}{\partial I} = R \).
  • Similarly, differentiating with respect to \( R \) while treating \( I \) as a constant gives \( \frac{\partial V}{\partial R} = I \).
Understanding these partial derivatives is vital in applying related rates in real-life scenarios like changing circuit conditions over time.
Circuit Analysis
Circuit analysis is the process of quantifying how electrical currents and voltages are distributed across electrical circuit components. A typical circuit analysis problem involves determining voltages and currents at various points in the circuit using fundamental laws such as Ohm's Law and Kirchhoff's Laws. In the problem at hand, we use circuit analysis to monitor how changes in voltage and resistance impact the current within a circuit. Since the voltage is dropping and the resistance is increasing, we apply the related rates formula. One advantage of using circuit analysis with related rates is:
  • It reveals exact relationships between different changing aspects, such as voltage and resistance affecting current over time.
This is critical for designing circuits that are resilient to changes and maintaining optimal performance of electrical systems.
Voltage and Current
Voltage and current are essential parameters in understanding and analyzing electrical circuits. Voltage, measured in volts, is the potential difference that drives current through a circuit. Current, measured in amperes, represents the flow of electric charge.In our exercise, the relationship between voltage \( V \), current \( I \), and resistance \( R \) is defined by Ohm's Law: \( V = IR \). This equation shows that for a given resistance, changes in voltage directly impact the current and vice versa.
  • As voltage decreases due to a wearing battery, maintaining a constant current becomes challenging without adjusting resistance.
  • Optimal circuit operation often requires balancing these two to ensure energy is efficiently managed and delivered.
Understanding how voltage and current behave under different conditions helps engineers and technicians create systems that are both efficient and safe.

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Most popular questions from this chapter

a. Maximum on a sphere Show that the maximum value of \(a^{2} b^{2} c^{2}\) on a sphere of radius \(r\) centered at the origin of a Cartesian \(a b c\) -coordinate system is \(\left(r^{2} / 3\right)^{3}\) b. Geometric and arithmetic means Using part (a), show that for nonnegative numbers \(a, b,\) and \(c,\) $$(a b c)^{1 / 3} \leq \frac{a+b+c}{3}$$ that is, the geometric mean of three nonnegative numbers is less than or equal to their arithmetic mean.

In Exercises \(65-70,\) you will explore functions to identify their local extrema. Use a CAS to perform the following steps: a. Plot the function over the given rectangle. b. Plot some level curves in the rectangle. c. Calculate the function's first partial derivatives and use the CAS equation solver to find the critical points. How do the critical points relate to the level critical plotted in part (b)? Which critical points, if any, appear to give a saddle point? Give reasons for your answer. $$ \begin{array}{l}{f(x, y)=x^{4}+y^{2}-8 x^{2}-6 y+16, \quad-3 \leq x \leq 3} \\\ {-6 \leq y \leq 6}\end{array} $$

Use a CAS to perform the following steps implementing the method of Lagrange multipliers for finding constrained extrema: a. Form the function \(h=f-\lambda_{1} g_{1}-\lambda_{2} g_{2},\) where \(f\) is the function to optimize subject to the constraints \(g_{1}=0\) and \(g_{2}=0 .\) b. Determine all the first partial derivatives of \(h\) , including the partials with respect to \(\lambda_{1}\) and \(\lambda_{2},\) and set them equal to \(0 .\) c. Solve the system of equations found in part (b) for all the unknowns, including \(\lambda_{1}\) and \(\lambda_{2} .\) d. Evaluate \(f\) at each of the solution points found in part (c) and select the extreme value subject to the constraints asked for in the exercise. Minimize \(f(x, y, z)=x^{2}+y^{2}+z^{2}\) subject to the constraints \(x^{2}-x y+y^{2}-z^{2}-1=0\) and \(x^{2}+y^{2}-1=0\)

The Sandwich Theorem for functions of two variables states that if \(g(x, y) \leq f(x, y) \leq h(x, y)\) for all \((x, y) \neq\left(x_{0}, y_{0}\right)\) in a disk centered at \(\left(x_{0}, y_{0}\right)\) and if \(g\) and \(h\) have the same finite limit \(L\) as \((x, y) \rightarrow\left(x_{0}, y_{0}\right),\) then $$ \lim _{(x, y) \rightarrow\left(x_{0}, y_{0}\right)} f(x, y)=L $$ Use this result to support your answers to the questions in Exercises \(45-48 .\) Does knowing that $$ 1-\frac{x^{2} y^{2}}{3}<\frac{\tan ^{-1} x y}{x y}<1 $$ tell you anything about $$\lim _{(x, y) \rightarrow(0,0)} \frac{\tan ^{-1} x y}{x y} ?$$ Give reasons for your answer.

In Exercises \(31-38\) , find the absolute maxima and minima of the functions on the given domains. \(f(x, y)=\left(4 x-x^{2}\right) \cos y\) on the rectangular plate \(1 \leq x \leq 3\) \(-\pi / 4 \leq y \leq \pi / 4(\text { see accompanying figure). }\)
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