Question

 A cylindrical tungsten filament 10.0 cm long with a diameter of 2.00 mm is to be used in a machine for which the temperature will range from the room temperature (20∘C) up to 120∘C. It will carry a current of 9.00 A at all temperatures. Temperature coefficient of resistivity of silver is 3.8 · 10−3 K −1 . (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?​

Answer 1

Given that,

Diameter = 2.00 mm

Length = 10.0 cm

Initial temperature = 20°C

Final temperature = 120°C

Current = 9.00 A

Temperature coefficient $\rho=3.8\times10^{-3}\ \K$

We need to calculate the resistivity

Using formula of resistivity

$\rho=\rho_{0}(1+\alpha(T_{f}-T_{i}))$

Put the value into the formula

$\rho=5.25\times10^{-8}+5.25\times10^{-8}\times3.8\times10^{-3}(120-20)$

$\rho=7.25\times10^{-8}\ \Omrga-m$

We need to calculate the area of the cylinder

Using formula of area

$A=\dfrac{\pi}{4}d^2$

Put the value into the formula

$A=\dfrac{\pi}{4}\times(2.00\times10^{-3})^2$

$A=3.141\times10^{-6}\ m^2$

(A). We need to calculate the maximum electric field in this filament

Using formula of electric filed

$E=\dfrac{I\rho}{A}$

Put the value into the formula

$E=\dfrac{9\times7.25\times10^{-8}}{3.141\times10^{-6}}$

$E=0.208\ N/C$

The maximum electric field in this filament is 0.208 N/C.

(B). We need to calculate the resistance with that field

Using formula of resistance

$R=\dfrac{\rho L}{A}$

Put the value into the formula

$R=\dfrac{10\times10^{-2}\times7.25\times10^{-8}}{3.141\times10^{-6}}$

$E=0.0023\ \Omega$

The resistance with that field is 0.0023 Ω.

(C). We need to calculate the maximum potential drop over the full length of the filament

Using formula of potential drop

$V=IR$

Put the value into the formula

$V=9\times0.0023$

$V=0.0207\ V$

The maximum potential drop over the full length of the filament is 0.0207 V.

Hence, (A). The maximum electric field in this filament is 0.208 N/C.

(B). The resistance with that field is 0.0023 Ω.

(C). The maximum potential drop over the full length of the filament is 0.0207 V.

Answer 2

Maximum electric field = 0.208 N/C

Resistance = 0.0023 Ω

Max potential drop over the length of the filament = 0.0207 V

Explanation:

Length L = 10cm = 10 * 10^-2 m

Diameter = 2mm = 2 * 10^-3 m

Cross-sectional Area = 3.14 * (2/2 * 10^-3)² = 3.14 * 10^-6

Temperature ranges from 20°C to 120°C.

Current = 9 A

Temperature co-efficient of resistivity of silver = 3.8 * 10^-3

Resistivity p = 5.25 * 10^-8 + 5.25 * 10^-8 * 3.8 * 10^-3 (120-20)

      = 7.25 * 10^-8

A) Maximum electric field = Ip/A = (9 * 7.25 * 10^-8) / 3.14 * 10^-6

    = 0.208 N/C

B) Resistance = pL/A = (10 * 10^-2 * 7.25 * 10^-8) / 3.14 * 10^-6

    = 0.0023 Ω

C) Max potential drop over the length of the filament V = IR

   = 9 * 0.0023

  = 0.0207 V