Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Question: The heating element of an electric dryer is rated at 4.1 kW when connected to a 240-V line. (a) What is the current in the heating element? Is 12-gauge wire large enough to supply this current? (b) What is the resistance of the dryer’s heating element at its operating temperature? (c) At 11 cents per kWh, how much does it cost per hour to operate the dryer?

Short Answer

Expert verified

Answer:

(a) The current in the heating element is and safe for 12-gauge wire.

(b) The resistance of the dryer’s heating element at its operating temperature is

(c) Cost per hour to operate the dryer is 45.1 Cents.

Step by step solution

01

Definition of the heating element.

The heating element is part of an electric heating appliance in which electric energy is transformed into heat.

Heating element power P=4.1kW103W1kW=4.1×103W

Connected line voltage = 240 V

02

Calculation of current

(a)

The current is given as,

I=PV……………….(1)

Substitute the values of P and V in equation (1) to get the current,

=4.1×103W240V=17.08A

Thus, the current in the heating element is which is safe for 12-gauge wire.

03

Step 3:

(b)

The power dissipated is given as,

P=V2R (2)

Substitute all the values in equation (2) which gives resistance,

R=240V24.1×103W=14.1Ω

Thus, the resistance of the dryer’s heating element at its operating temperature is 14.1Ω

04

Step 4:

(c)

The dryer costs 11 cents for the energy of 1 kWh.energy consumed by the dryer is,

E=P×tE=4.1kW×1hE=4.1kWhE=pt.=(4.1kW) (1h) =4.1kWh

The cost of operating the dryer for 1 h can be calculated as,

Operatingcost=41KWh×11Cent/KWh=451Cent

Thus, the cost per hour to operate the dryer is 45.1 Cents

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 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 resistor with resistance Ris connected to a battery that has emf 12.0 V and internal resistance r=0.40Ω. For what two values of R will the power dissipated in the resistor be 80.0 W ?

A fuse is a device designed to break a circuit, usually by melting when the current exceeds a certain value. What characteristics should the material of the fuse have?

A typical small flashlight contains two batteries, each having an emf of1.5V, connected in series with a bulb having resistance17Ω. (a) If the internal resistance of the batteries is negligible, what power is delivered to the bulb? (b) If the batteries last for1.5hwhat is the total energy delivered to the bulb? (c) The resistance of real batteries increases as they run down. If the initial internal resistance is negligible, what is the combined internal resistance of both batteries when the power to the bulb has decreased to half its initial value? (Assume that the resistance of the bulb is constant. Actually, it will change somewhat when the current through the filament changes, because this changes the temperature of the filament and hence the resistivity of the filament wire.)

The energy that can be extracted from a storage battery is always less than the energy that goes into it while it is being charged. Why?

CALC The region between two concentric conducting spheres with radii and is filled with a conducting material with resistivity ρ. (a) Show that the resistance between the spheres is given by

R=ρ4π(1a-1b)

(b) Derive an expression for the current density as a function of radius, in terms of the potential differenceVab between the spheres. (c) Show that the result in part (a) reduces to Eq. (25.10) when the separation L=b-abetween the spheres is small.
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.

Sign-up for free