Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 21: Problem 6

A 4.0 -k \(W\) heater is designed to be connected to a \(120-V\) ms source. What is the power dissipated by the heater if it is instead connected to a \(120-\mathrm{V}\) de source?

Short Answer

Expert verified
Answer: The power dissipated by the heater when connected to a 120 V DC source is 4000 W, or 4 kW.

Step by step solution

01

Calculate the resistance of the heater

To find the resistance of the heater, we can use the formula for power in terms of voltage and resistance: \(P = \frac{V^2}{R}\). Rearrange this formula for resistance, \(R = \frac{V^2}{P}\). We know the power provided by the heater is 4 kW, or 4000 W, and the RMS voltage is 120 V. So we have: \(R = \frac{(120 V)^2}{4000 W}\) Calculating the resistance: \(R = \frac{14400 V^2}{4000 W} = 3.6 \,\Omega\)
02

Calculate the power dissipated by the heater using DC source

Now we have the resistance of the heater, and we are given a 120 V DC source. We can calculate the power dissipated by the heater using the same power formula, \(P = \frac{V^2}{R}\). Plug in the DC voltage of 120 V and the resistance of 3.6 Ohms: \(P = \frac{(120 V)^2}{3.6 \,\Omega}\) Calculating the power: \(P = \frac{14400 V^2}{3.6 \,\Omega} = 4000 W\) Thus, the power dissipated by the heater when connected to a 120 V DC source is 4000 W, or 4 kW.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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 capacitor is rated at \(0.025 \mu \mathrm{F}\). How much rms current flows when the capacitor is connected to a \(110-\mathrm{V}\) rms, \(60.0-\mathrm{Hz}\) line?
An RLC series circuit is connected to a \(240-\mathrm{V}\) ms power supply at a frequency of \(2.50 \mathrm{kHz}\). The elements in the circuit have the following values: \(R=12.0 \Omega, C=\) \(0.26 \mu \mathrm{F},\) and $L=15.2 \mathrm{mH},$ (a) What is the impedance of the circuit? (b) What is the rms current? (c) What is the phase angle? (d) Does the current lead or lag the voltage? (e) What are the rms voltages across each circuit element?
An RLC series circuit has \(L=0.300 \mathrm{H}\) and \(C=6.00 \mu \mathrm{F}\) The source has a peak voltage of \(440 \mathrm{V}\). (a) What is the angular resonant frequency? (b) When the source is set at the resonant frequency, the peak current in the circuit is \(0.560 \mathrm{A}\). What is the resistance in the circuit? (c) What are the peak voltages across the resistor, the inductor, and the capacitor at the resonant frequency?
A parallel plate capacitor has two plates, each of area $3.0 \times 10^{-4} \mathrm{m}^{2},\( separated by \)3.5 \times 10^{-4} \mathrm{m} .$ The space between the plates is filled with a dielectric. When the capacitor is connected to a source of \(120 \mathrm{V}\) rms at \(8.0 \mathrm{kHz}\) an mms current of \(1.5 \times 10^{-4} \mathrm{A}\) is measured. (a) What is the capacitive reactance? (b) What is the dielectric constant of the material between the plates of the capacitor?
An ac circuit contains a \(12.5-\Omega\) resistor, a \(5.00-\mu \mathrm{F}\) capacitor, and a \(3.60-\mathrm{mH}\) inductor connected in series to an ac generator with an output voltage of \(50.0 \mathrm{V}\) (peak) and frequency of \(1.59 \mathrm{kHz}\). Find the impedance,
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Electricity

Read Explanation

Electric Charge Field and Potential

Read Explanation

Fields in Physics

Read Explanation

Wave Optics

Read Explanation

Electrostatics

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.

Sign-up for free