/*! 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 4 (a) Express the power dissipated... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Express the power dissipated by a resistor in terms of \(R\) and \(\Delta V\) only, eliminating \(I\). (answer check available at lightandmatter.com) (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 W. How much power does it actually draw now? (answer check available at lightandmatter.com)

Short Answer

Expert verified
The power dissipated by the resistor, expressed in terms of \( R \) and \( \Delta V \), is \( P = \frac{(\Delta V)^2}{R} \). When connected to a 110 V outlet, the drier draws 50 W.

Step by step solution

01

Understanding Power Dissipation Formula

The power dissipated by a resistor can be calculated using the formula \( P = I^2 R \), where \( P \) is the power, \( I \) is the current, and \( R \) is the resistance.
02

Expressing Power in Terms of \( R \) and \( \Delta V \)

First, use Ohm's law, \( \Delta V = IR \), to express \( I \) as \( I = \frac{\Delta V}{R} \). Next, substitute \( I \) in the power formula: \( P = \left(\frac{\Delta V}{R}\right)^2 R \). This simplifies to \( P = \frac{(\Delta V)^2}{R} \).
03

Analyzing the Drier's Power Requirement

The drier is normally plugged into a \( 220 \text{ V} \) circuit and draws \( 200 \text{ W} \). Therefore, applying \( P = \frac{(\Delta V)^2}{R} \), we find \( R = \frac{(220)^2}{200} \).
04

Recalculating Power at \( 110 \text{ V} \)

Using the calculated resistance from the drier's normal operation, compute the power when it's plugged into a \( 110 \text{ V} \) outlet using \( P = \frac{(110)^2}{R} \).
05

Solution Verification

Now verify the calculations. At \( 110 \text{ V} \), the power drawn by the drier is much lower due to the halved voltage compared to its design voltage.

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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 used to describe the relationship between voltage, current, and resistance in an electric circuit. It is a simple formula: \( V = IR \), where \( V \) represents voltage in volts, \( I \) stands for the current in amperes, and \( R \) denotes resistance in ohms. This principle helps in determining one value as long as the other two are known. For example, if you have a resistance of 5 ohms and a voltage of 10 volts, the current flowing through the circuit can be calculated as \( I = \frac{10}{5} = 2 \) amperes. This law is crucial in solving many problems related to electric circuits and is often the starting point for more complex calculations.
Power dissipation
Power dissipation in a resistor is the process by which electrical power is converted to heat energy and lost from the system. This can be calculated using the formula \( P = I^2 R \), where \( P \) is power in watts, \( I \) is current, and \( R \) is resistance. However, with Ohm's Law, we can express the power dissipation in terms of voltage and resistance:
  • Substitute \( I \) from Ohm's Law: \( I = \frac{V}{R} \) into \( P = I^2 R \)
  • This results in \( P = \left(\frac{V}{R}\right)^2 R \) which simplifies to \( P = \frac{V^2}{R} \).
This means for a given voltage across a resistor, the power dissipated can be determined by the square of the voltage divided by the resistance. This is particularly useful for designing circuits and ensuring components can safely handle the power levels involved.
Resistance
Resistance is a property of an electrical component that resists the flow of electrical current. It is measured in ohms (Ω) and can vary depending on the material, temperature, and physical dimensions of the component. In the context of our problem, we determine resistance using the power dissipation formula derived earlier:
  • If a drier normally operates at 220 V and dissipates 200 W, we can find its resistance using \( R = \frac{V^2}{P} \)
  • This results in \( R = \frac{(220)^2}{200} \), which simplifies to \( R = 242 \) ohms.
Knowing the resistance helps to redesign or troubleshoot circuits to ensure they function correctly under different conditions.
Voltage
Voltage, often called electric potential difference, is the force that pushes electric current through a circuit. It is measured in volts. Voltage can be thought of as the pressure from an electrical circuit's power source that pushes charged electrons (current) through a conducting loop, enabling them to do work such as lighting a lamp. In our scenario, household receptacles primarily provide 110 V, while certain appliances require 220 V. When an appliance designed for 220 V is instead plugged into a 110 V outlet, it will operate with different electrical characteristics. For our drier, this changes the power it draws, demonstrating the importance of ensuring devices are connected to their intended voltage levels. Voltage is critical for calculating power and current in circuits, impacting overall performance.
Power calculation
Calculating power in a circuit involves using the formula \( P = \frac{V^2}{R} \) or using \( P = VI \), depending on known values. It is measured in watts and reflects the rate at which energy is used. In our exercise, by plugging a drier normally using 220 V into a 110 V supply, its power draw changes.
  • From previously calculated resistance, \( R = 242 \) ohms, the power at 110 V is \( P = \frac{(110)^2}{242} \).
  • This calculates to approximately 50 watts, a quarter of the designed 200 watts.
This demonstrates the dependency of power on both voltage and resistance, emphasizing the need for correct voltage levels for optimal appliance function.

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Most popular questions from this chapter

(a) Recall that the gravitational energy of two gravitationally interacting spheres is given by \(P E=-G m_{1} m_{2} / r\), where \(r\) is the center-to- center distance. Sketch a graph of \(P E\) as a function of \(r\), making sure that your graph behaves properly at small values of \(r\), where you're dividing by a small number, and at large ones, where you're dividing by a large one. Check that your graph behaves properly when a rock is dropped from a larger \(r\) to a smaller one; the rock should lose potential energy as it gains kinetic energy. (b) Electrical forces are closely analogous to gravitational ones, since both depend on \(1 / r^{2}\). Since the forces are analogous, the potential energies should also behave analogously. Using this analogy, write down the expression for the electrical potential energy of two interacting charged particles. The main uncertainty here is the sign out in front. Like masses attract, but like charges repel. To figure out whether you have the right sign in your equation, sketch graphs in the case where both charges are positive, and also in the case where one is positive and one negative; make sure that in both cases, when the charges are released near one another, their motion causes them to lose PE while gaining KE.(answer check available at lightandmatter.com)

Wire is sold in a series of standard diameters, called "gauges." The difference in diameter between one gauge and the next in the series is about \(20 \%\). How would the resistance of a given length of wire compare with the resistance of the same length of wire in the next gauge in the series? (answer check available at lightandmatter.com)

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 resistor has a voltage difference \(\Delta V\) across it, causing a current \(I\) to flow. (a) Find an equation for the power it dissipates as heat in terms of the variables \(I\) and \(R\) only, eliminating \(\Delta V\). (answer check available at lightandmatter.com) (b) If an electrical line coming to your house is to carry a given amount of current, interpret your equation from part a to explain whether the wire's resistance should be small, or large.

How many different resistance values can be created by combining three unequal resistors? (Don't count possibilities in which not all the resistors are used, i.e., ones in which there is zero current in one or more of them.)

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