/*! 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 14 (a) Express the power dissipated... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 21: Problem 14

(a) Express the power dissipated by a resistor in terms of \(R\) and \(\Delta V\) only, eliminating \(I\). (b) Electrical receptacles in your home are mostly \(110 \mathrm{~V}\), but circuits for electric stoves, air conditioners, and washers and driers are usually \(220 \mathrm{~V}\). The two types of circuits have differently shaped receptacles. Suppose you rewire the plug of a drier so that it can be plugged in to a \(110 \mathrm{~V}\) receptacle. The resistor that forms the heating element of the drier would normally draw \(200 \mathrm{~W}\). How much power does it actually draw now?

Short Answer

Expert verified
The power dissipated is now approximately 50 W.

Step by step solution

01

Power Formula using Current and Voltage

The power dissipated by a resistor can initially be expressed using the formula: \[ P = I \times \Delta V \]where \(P\) is the power, \(I\) is the current, and \(\Delta V\) is the voltage across the resistor.
02

Substitute Ohm's Law

Ohm's Law provides the relationship between voltage, current, and resistance: \[\Delta V = I \times R\]Solving for \(I\), we get:\[ I = \frac{\Delta V}{R} \]Substitute this into the power formula: \[ P = \left(\frac{\Delta V}{R}\right) \times \Delta V \]
03

Simplify Power Formula

Simplify the expression for power:\[ P = \frac{(\Delta V)^2}{R} \]This formula expresses the power \(P\) in terms of \(R\) and \(\Delta V\) only, eliminating \(I\).
04

Calculate Original Power Dissipation

Originally, the drier operates at \(220\, \mathrm{V}\) and draws \(200\, \mathrm{W}\), so we use the formula \(P = \frac{(\Delta V)^2}{R}\) to find the resistance \(R\):\[ 200 = \frac{(220)^2}{R} \]Solving for \(R\), we get:\[ R = \frac{(220)^2}{200} = 242 \space \Omega \]
05

Calculate Power at New Voltage

Now, plug in the new voltage \(\Delta V = 110\, \mathrm{V}\) into the power formula \(P = \frac{(\Delta V)^2}{R}\):\[ P = \frac{(110)^2}{242} \]Calculate:\[ P = \frac{12100}{242} \approx 50\, \mathrm{W} \]
06

Conclusion

Rewriting the plug for a \(110 \mathrm{~V}\) receptacle causes the drier to draw only \(50 \mathrm{~W}\). This significantly reduces the power it was supposed to consume, affecting its performance.

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 91影视!

Key Concepts

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

Ohm's Law
Ohm's Law is a fundamental principle in electrical circuits that describes the relationship between voltage, current, and resistance. It's often expressed with the formula \( \Delta V = I \times R \), where \( \Delta V \) is the voltage across a component, \( I \) is the current flowing through it, and \( R \) is the resistance.
This law essentially tells us how much voltage is needed to push a certain amount of current through a resistance.
  • If you know any two of these quantities, you can calculate the third.
  • This relationship helps in designing circuits and understanding how they will behave under different conditions.
It's quite handy because it allows us to express power in terms of just two variables instead of three, giving clarity and simplicity in problem-solving.
Voltage
Voltage is the electric potential difference between two points in a circuit. It's essentially the "pressure" that pushes electric charges through a conductor.
Imagine it like water pressure in a hose, which moves water through the length of the hose.
  • Measured in volts (V), voltage is crucial for the functioning of any electrical circuit.
  • The higher the voltage, the more potential energy per charge, which can drive larger currents through the circuit resistances.
  • In the context of home circuits, common voltage ratings are \(110 \text{ V}\) and \(220 \text{ V}\), which are key to understanding how different appliances are powered.
Voltage determines how much power an appliance can draw from a circuit, affecting its performance and efficiency.
Resistance
Resistance is a measure of how much a material opposes the flow of electric current. It's akin to trying to walk through a narrow hallway with lots of obstacles.
Materials have different levels of resistance based on their properties.
  • Measured in ohms (\( \Omega \)), resistance is integral in limiting current flow and changing energy into other forms like heat, as seen in heating elements or resistors.
  • When applied to circuits, it determines how much current can flow for a given voltage (as per Ohm's Law).
  • In practical applications, manipulating resistance can control the power usage of electrical devices.
Understanding resistance helps in matching appliances to suitable power sources, ensuring efficiency and safety.
Electrical Circuits
An electrical circuit is a closed loop path that allows current to flow, comprising a power source, conductors, and loads (like lights or appliances).
Everyone has interacted with electrical circuits in everyday life, even if they don't realize it.
  • Circuit components are strategically connected to achieve specific functions, such as lighting or heating.
  • Power dissipation within a circuit is when electrical energy is converted into thermal energy, often in resistive components.
  • Components need to be matched for voltage and resistance to ensure proper function and prevent damage.
Understanding the basics of electrical circuits is essential for diagnosing problems and properly setting up modern devices in a safe and energy-efficient way.

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

A person in a rural area who has no electricity runs an extremely long extension cord to a friend's house down the road so she can run an electric light. The cord is so long that its resistance, \(x\), is not negligible. Show that the lamp's brightness is greatest if its resistance, \(y\), is equal to \(x\). Explain physically why the lamp is dim for values of \(y\) that are too small or too large.

A silk thread is uniformly charged by rubbing it with llama fur. The thread is then dangled vertically above a metal plate and released. As each part of the thread makes contact with the conducting plate, its charge is deposited onto the plate. Since the thread is accelerating due to gravity, the rate of charge deposition increases with time, and by time \(t\) the cumulative amount of charge is \(q=c t^{2}\) where \(c\) is a constant. (a) Find the current flowing onto the plate. (b) Suppose that the charge is immediately carried away through a resistance \(R\). Find the power dissipated as heat.

(a) You take an LP record out of its sleeve, and it acquires a static charge of \(1 \mathrm{nC}\). You play it at the normal speed of \(33 \frac{1}{3}\) r.p.m. and the charge moving in a circle creates an electric current. What is the current, in amperes? (b) Although the planetary model of the atom can be made to work with any value for the radius of the electrons' orbits, more advanced models that we will study later in this course predict definite radii. If the electron is imagined as circling around the proton at a speed of \(2.2 \times 10^{6} \mathrm{~m} / \mathrm{s}\), in an orbit with a radius of \(0.05 \mathrm{~nm}\), what electric current is created? The charge of an electron is \(-e=-1.60 \times\) \(10^{-19} \mathrm{C}\)

If a typical light bulb draws about \(900 \mathrm{~mA}\) from a \(110 \mathrm{~V}\) household circuit, what is its resistance?

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?
See all solutions

Recommended explanations on Physics Textbooks

Dynamics

Read Explanation

Geometrical and Physical Optics

Read Explanation

Waves Physics

Read Explanation

Electricity

Read Explanation

Physical Quantities And Units

Read Explanation

Conservation of Energy and Momentum

Read Explanation
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.