/*! 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 67 A \(500 \mathrm{~W}\) heating un... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A \(500 \mathrm{~W}\) heating unit is designed to operate with an applied potential difference of \(115 \mathrm{~V}\). (a) By what percentage will its heat output drop if the applied potential difference drops to \(110 \mathrm{~V}\) ? Assume no change in resistance. (b) If you took the variation of resistance with temperature into account, would the actual drop in heat output be larger or smaller than that calculated in (a)?

Short Answer

Expert verified
(a) The heat output decreases by approximately 8.48%. (b) The actual drop would be larger if resistance varies with temperature.

Step by step solution

01

Calculate Initial Current

Using the formula for power, \( P = IV \), where \( P = 500 \, \text{W} \) and \( V = 115 \, \text{V} \), we find the initial current \( I_1 = \frac{P}{V} = \frac{500}{115} \approx 4.35 \, \text{A} \).
02

Calculate Resistance

Using Ohm's law (\( V = IR \)), the resistance is \( R = \frac{V}{I} = \frac{115}{4.35} \approx 26.44 \, \Omega \).
03

Calculate New Current

With the new voltage \( V_2 = 110 \, \text{V} \) and the resistance \( R = 26.44 \, \Omega \) unchanged, the new current is \( I_2 = \frac{V_2}{R} = \frac{110}{26.44} \approx 4.16 \, \text{A} \).
04

Calculate New Power Output

Using \( P = IV \) again, the new power output is \( P_2 = I_2 V_2 = 4.16 \times 110 \approx 457.6 \, \text{W} \).
05

Calculate Percentage Decrease

The percentage decrease in power is \( \frac{P_1 - P_2}{P_1} \times 100\% = \frac{500 - 457.6}{500} \times 100\% \approx 8.48\% \).
06

Consider Resistance Variation

If resistance increases with temperature, the drop in heat output will be larger than calculated because increased resistance would result in lower current and thus lower power output, beyond the initial calculation.

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.

Electrical Power
Electrical power is the rate at which electrical energy is converted into another form of energy, such as heat. It is given by the formula \( P = IV \), where \( P \) is the power in watts (W), \( I \) is the current in amperes (A), and \( V \) is the potential difference in volts (V). This formula can also be modified using Ohm's law (\( V = IR \)) to give \( P = I^2R \) or \( P = \frac{V^2}{R} \). These variations are useful depending on the known quantities.
  • When a heating device operates at a specific power, it means this is the rate at which it converts electrical energy into heat.
  • The device in our exercise is a 500 W unit, meaning it converts 500 joules of energy per second into heat.
A change in voltage will affect the current and consequently the power. In our example, reducing the voltage from 115 V to 110 V decreased the power, showing how sensitive electrical devices can be to changes in voltage.
Resistance and Temperature
Resistance is a measure of how much a material opposes the flow of electric current. In conductors, resistance increases with temperature, generally because the atoms in the conductor vibrate more at higher temperatures, impeding the flow of electrons.
In the exercise, it was assumed that resistance stayed constant when calculating the percentage change in power. This simplification is useful for basic calculations, but in reality, resistance would increase as the device heats.
  • In many materials, \( R \) can be approximated as \( R = R_0(1 + \alpha \Delta T) \), where \( R_0 \) is the original resistance, \( \alpha \) is the temperature coefficient, and \( \Delta T \) is the change in temperature.
  • This means that if the temperature rises, the resistance could increase, causing a decrease in current even if the voltage is unchanged.
When considering temperature effects, the actual drop in power output could be more significant than initially calculated, because the increased resistance further lowers the current.
Potential Difference
Potential difference, also known as voltage, is the difference in electric potential between two points in a circuit. It is what drives the flow of electrons, or current, through a conductor. Measured in volts (V), it determines how much energy is given to the electrons as they move through the circuit.
In the case of the heating unit, the potential difference of 115 V is what initially powers the device to produce 500 W of heat. When the potential difference drops to 110 V, less energy is supplied to the device per coulomb of charge, reducing both the current and the output power.
  • The relationship between potential difference and power can be particularly observed through the formula \( P = IV \).
  • This highlights that any drop in voltage without compensating changes will lead directly to reduced power output.
Understanding how potential difference influences power and performance is crucial, especially in devices sensitive to changes in voltage like household appliances.

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 resistor with a potential difference of \(200 \mathrm{~V}\) across it transfers electrical energy to thermal energy at the rate of \(3000 \mathrm{~W}\). What is the resistance of the resistor?

A common flashlight bulb is rated at \(0.30 \mathrm{~A}\) and \(2.9 \mathrm{~V}\) (the values of the current and voltage under operating conditions). If the resistance of the tungsten bulb filament at room temperature \(\left(20^{\circ} \mathrm{C}\right)\) is \(1.1 \Omega\), what is the temperature of the filament when the bulb is on?

In a hypothetical fusion research lab, high temperature helium gas is completely ionized and each helium atom is separated into two free electrons and the remaining positively charged nucleus, which is called an alpha particle. An applied electric field causes the alpha particles to drift to the east at \(25.0 \mathrm{~m} / \mathrm{s}\) while the electrons drift to the west at \(88.0 \mathrm{~m} / \mathrm{s}\). The alpha particle density is \(2.80 \times 10^{15} \mathrm{~cm}^{-3}\). What are (a) the net current density and (b) the current direction?

An electric immersion heater normally takes 100 min to bring cold water in a well-insulated container to a certain temperature, after which a thermostat switches the heater off. One day the line voltage is reduced by \(6.00 \%\) because of a laboratory overload. How long does heating the water now take? Assume that the resistance of the heating element does not change.

What is the resistivity of a wire of \(1.0 \mathrm{~mm}\) diameter, \(2.0 \mathrm{~m}\) length, and \(50 \mathrm{~m} \Omega\) resistance?

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.