/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 It's fairly common in electrical... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

It's fairly common in electrical circuits for additional, undesirable resistances to occur because of factors such as dirty, corroded, or loose connections. Suppose that a device with resistance \(R\) normally dissipates power \(P\), but due to an additional series resistance \(r\) the total power is reduced to \(P^{\prime}\). We might, for example, detect this change because the battery powering our device ran down more slowly than normal. (a) Find the unknown resistance \(r\). (b) Check that the units of your result make sense. (c) Check that your result makes sense in the special cases \(P^{\prime}=P\) and \(P^{\prime}=0\). (d) Suppose we redefine \(P^{\prime}\) as the useful power dissipated in \(R\). For example, this would be the change we would notice because a flashlight was dimmer. Find \(r\).

Short Answer

Expert verified
The unknown resistance \( r \) due to power change is \( R\left(\frac{P}{P'} - 1\right) \), and when using useful power, it's \( R\left(\left(\frac{P}{P'}\right)^{1/2} - 1\right) \).

Step by step solution

01

Write the Formula for Power Dissipation

The power dissipated by a resistance \( R \) when a voltage \( V \) is applied is given by \( P = \frac{V^2}{R} \). When an additional resistance \( r \) is added in series, the total resistance becomes \( R + r \). Therefore, the new power dissipated is \( P^{\prime} = \frac{V^2}{R + r} \).
02

Rearrange for Unknown Resistance

To find the unknown resistance \( r \), equate the two expressions and solve for \( r \). Start with \( \frac{V^2}{R + r} = P^{\prime} \). Rearrange to get \( R + r = \frac{V^2}{P^{\prime}} \). Subtracting \( R \) from both sides gives \( r = \frac{V^2}{P^{\prime}} - R \).
03

Relate to Power and Solve for r

Using the original power expression \( P = \frac{V^2}{R} \), we solve for \( V^2 \): \( V^2 = PR \). Substitute this into the equation for \( r \): \( r = \frac{PR}{P^{\prime}} - R \). Simplify to find \( r = R\left(\frac{P}{P^{\prime}} - 1\right) \).
04

Unit Analysis

Given that both \( R \) and \( r \) have units of ohms (\( \Omega \)), and \( P \) and \( P^{\prime} \) have units of watts (\( W \)), notice \( \frac{P}{P^{\prime}} \) is dimensionless. Thus, \( R\left( \frac{P}{P^{\prime}} - 1 \right) \) maintains units of ohms, ensuring the formula for \( r \) has consistent units.
05

Check Special Case 1: \( P^{\prime} = P \)

In the case where \( P^{\prime} = P \), \( \frac{P}{P^{\prime}} = 1 \), therefore \( r = R (1 - 1) = 0 \). This indicates no additional resistance, which matches the scenario.
06

Check Special Case 2: \( P^{\prime} = 0 \)

For \( P^{\prime} = 0 \), the expression \( \frac{P}{P^{\prime}} \) becomes undefined, theoretically suggesting infinite resistance is required to reduce power dissipation to zero, aligning logically.
07

Redefine \( P' \) and Derive \( r \) for Useful Power

If \( P' \) is the useful power in \( R \), formulate the power used in \( R \) as \( P' = \frac{V^2 R}{(R + r)^2} \). Set up \( (R + r)^2 = \frac{V^2 R}{P'} \), leading to \( R + r = \frac{\sqrt{V^2 R}}{\sqrt{P'}} = \frac{V\sqrt{R}}{\sqrt{P'}} \). Solve for \( r \): \( r = \frac{V\sqrt{R}}{\sqrt{P'}} - R \). Finally, using \( V^2 = PR \), get \( r = R\left(\frac{P}{P'}\right)^{1/2} - R \).
08

Simplify and Interpret Result

The result \( r = R\left(\left(\frac{P}{P'}\right)^{1/2} - 1\right) \) indicates how \( R \) can adjust based on desired and useful power measures. This formula captures the essence of practical resistance occurring when actual output diverges from designed efficiency.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Power Dissipation
Power dissipation in an electrical circuit is a critical concept often described by how much energy is expended as heat by a resistor. When a resistor with resistance \( R \) is connected to a voltage \( V \), the power dissipated can be calculated using the equation \( P = \frac{V^2}{R} \). This equation essentially shows how efficiently our device uses the provided electrical energy, converting it into heat.
When an undesired series resistance \( r \) is introduced, the total resistance becomes \( R + r \). Therefore, the new power dissipated in the system changes. This results in a lower power dissipation, \( P' \), occurring because the additional resistance impacts the current flowing through the circuit.
This principle helps troubleshoot and understand when appliances, like a flashlight, experience lower efficiency (dimmer light). Recognizing power dissipation in this way is crucial for designing systems that efficiently harness electrical energy.
Series Resistance
Series resistance occurs when resistors are connected in such a way that the current flows through one resistor and then another in a single path or sequence. This setup means that the total resistance of the circuit is the sum of the individual resistances.
In our problem, an extra resistance \( r \) is added to the existing resistance \( R \). The combined resistance becomes \( R + r \), which affects the total power dissipated as discussed previously. This models real-life issues like dirty or loose connections in a circuit, introducing unwanted series resistance.
The greater the series resistance, the more it can impact the circuit's performance. The concept helps engineers design circuits that avoid unnecessary power loss, ensuring efficient use of electricity. It's also a fundamental factor in determining the overall behavior of circuit components.
Ohm's Law
Ohm's Law is one of the fundamental principles in electrical engineering. It states that the current \( I \) through a conductor between two points is directly proportional to the voltage \( V \) across the two points and inversely proportional to the resistance \( R \). This relationship is expressed as \( V = IR \).
The law is crucial for understanding how variations in resistance (like the unwanted series resistance \( r \)) can affect current flow and power dissipation. For example, adding resistance causes the current to drop if the voltage is constant, directly influencing the power dissipation according to the power formulas mentioned above.
Understanding Ohm's Law enables us to predict how changes in a circuit's components will affect its performance. By applying it, engineers can troubleshoot and mitigate problems arising from components like additional resistances that we don't want.
Unit Analysis
Unit analysis is vital for validating the dimensions in equations, ensuring calculations are physically meaningful. Consider resistances measured in ohms \( \Omega \) and power in watts \( W \).
In the solution, we ensure the formula for \( r \) retains the correct units. Given \( R \) and \( r \) both have units of ohms, we express \( r \) as \( R \left(\frac{P}{P'} - 1\right) \). Notice \( \frac{P}{P'} \) is dimensionless because both \( P \) and \( P' \) are in watts, hence subtracting 1 maintains no unit changes.
This coherent unit analysis confirms that our equations are logically consistent and ensures that no errors in dimensions are made. Overall, unit analysis acts as a necessary tool for engineers to consistently verify their work, ensuring that mathematical expressions correspond correctly to real-world physical phenomena.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

How many different resistance values can be created by combining three unequal resistors? (Don't count possibilities where not all the resistors are used.)

What resistance values can be created by combining a \(1 \mathrm{k} \Omega\) resistor and a \(10 \mathrm{k} \Omega\) resistor?

Hybrid and electric cars have been gradually gaining market share, but during the same period of time, manufacturers such as Porsche have also begun designing and selling cars with "mild hybrid" systems, in which power-hungry parts like water pumps are powered by a higher-voltage battery rather than running directly on shafts from the motor. Traditionally, car batteries have been 12 volts. Car companies have dithered over what voltage to use as the standard for mild hybrids, building systems based on \(36 \mathrm{~V}, 42 \mathrm{~V}\), and \(48 \mathrm{~V}\). For the purposes of this problem, we consider \(36 \mathrm{~V}\). (a) Suppose the battery in a new car is used to run a device that requires the same amount of power as the corresponding device in the old car. Based on the sample figures above, how would the currents handled by the wires in one of the new cars compare with the currents in the old ones? (b) The real purpose of the greater voltage is to handle devices that need more power. Can you guess why they decided to change to higher-voltage batteries rather than increasing the power without increasing the voltage?

You are given a battery, a flashlight bulb, and a single piece of wire. Draw at least two configurations of these items that would result in lighting up the bulb, and at least two that would not light it. (Don't draw schematics.) Note that the bulb has two electrical contacts: one is the threaded metal jacket, and the other is the tip (at the bottom in the figure). If you're not sure what's going on, there are a couple of ways to check. The best is to try it in real life by either borrowing the materials from your instructor or scrounging the materials from around the house. (If you have a flashlight with this type of bulb, you can remove the bulb.) Another method is to use the simulation at phet.

The subatomic particles called muons behave exactly like electrons, except that a muon's mass is greater by a factor of \(206.77\). Muons are continually bombarding the Earth as part of the stream of particles from space known as cosmic rays. When a muon strikes an atom, it can displace one of its electrons. If the atom happens to be a hydrogen atom, then the muon takes up an orbit that is on the average \(206.77\) times closer to the proton than the orbit of the ejected electron. How many times greater is the electric force experienced by the muon than that previously felt by the electron?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.