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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

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