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

A cylindrical tungsten filament 15.0 cm long with a diameter of 1.00 mm is to be used in a machine for which the temperature will range from room temperature 120°C2 up to 120°C. It will carry a current of 12.5 A at all temperatures (consult Tables 25.1 and 25.2). (a) What will be the maximum electric field in this filament, and (b) what will be its resistance with that field? (c) What will be the maximum potential drop over the full length of the filament?

Short Answer

Expert verified

(a) Maximum electric field in the filament isE=1.21V/m

(b) Resistance isR=0.0145Ω

(c) Maximum potential drop isV=0.181V

Step by step solution

01

Step 1:

Given data:

Length of the cylinder of tungstenL=15.0cm=0.15m

And the radius isr=1.0mm/2=0.50mm=0.00050m

Temperature rangeT0=20.0C°uptoT=120.0C°

Currentl=12.5A

02

Step 2:

(a) To determine the electric field passing through the filament Where the electric field is connected to the resistivity, the greater the resistivity, the greater the electric field required to induce a particular current.

E=ρJ=ρlA=ρlπr2

Here,is the resistivity of the tungsten at a temperatureT=120.0C°

As the increase in temperature, the resistivity of metal increases at a temperature T0=20.0C°and resistivity ρo=5.25×10-8Ω.m

So ρis;

ρ=ρo1+αT-To=5.25×10-8Ω.m1+0.0045C°-1120C°-20C°=7.6125×10-8Ω.m

Here αis the temperature coefficient andα=0.0045C°-1

Putting the values ofρ,landr :

E=ρlπr2=7.6125×10-8Ω.m12.5Aπ0.0005m2=1.21V/m

Hence, the maximum electric field in the filament isE=1.21V/m

03

Step 3:

(b) For determining the resistance R of the filament, where the resistance of the cylinder is determined by its area, resistivity, and length,

R=ρLA

Putting the values:

R=ρLπr2=7.6125×10-8Ω.m0.15mπ0.0005m2=0.0145Ω

Hence, the resistance isR=0.0145Ω

04

Step 4:

(c) For the voltage of the rod, putting the value of R in equation V=IR;

V=IR=12.5A0.0145Ω=0.181V

Therefore, the maximum potential drop isV=0.181V

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

In an L-R-C series circuit, what criteria could be used to decide whether the system is over damped or underdamped? For example, could we compare the maximum energy stored during one cycle to the energy dissipated during one cycle? Explain.

(a) What is the potential difference Vadin the circuit of Fig. P25.62? (b) What is the terminal voltage of the 4.00-Vbattery? (c) A battery with emf and internal resistance 0.50Ωis inserted in the circuit at d, with its negative terminal connected to the negative terminal of the 8.00-Vbattery. What is the difference of potential Vbcbetween the terminals of the 4.00-Vbattery now?

A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction pependicurlar to its original direction (Fig. E27.24). The beam travels a distance of 1.10 cm while in the field. What is the magnitude of the magnetic field?

Two copper wires with different diameter are joined end to end. If a current flow in the wire combination, what happens to electrons when they move from the large diameter wire into the smaller diameter wire? Does their drift speed increase, decrease, or stay the same? If the drift speed change, what is the role the force that causes the change? Explain your reasoning.

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