Question

A cylindrical tungsten filament 16.0 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 (20⁰ C) up to 120⁰ C. It will carry a current of 12.5 A at all temperatures ( Resistivity of tungsten at 20⁰ C is 5.25×10−8Ωm, the temperature coefficient of resistivity at 20⁰ C is 0.0045 ⁰ C-1)
a) What will be the maximum electric field in this filament? Express your answer using two significant figures.
b) What will be its resistance with that field? Express your answer using two significant figures.
c) What will be the maximum potential drop over the full length of the filament? Express your answer using two significant figures.

Answer 1

Answer:

Explanation:

Resistance of the tungsten wire

R = resistivity x length / cross sectional area

= $\frac{5.25\times10^{-8}\times16\times10^{-2}}{3.14\times(.5\times10^{-3})^2}$

= 107 x 10⁻⁴ ohm

Resistance at 120 degree can be obtained from the following formula

$R_t = R_0( 1 + \alpha\times t )$

$R_t = 107\times10^{-4}( 1 + .0045\times 100)$

= 155.15 x 10⁻⁴ ohm

= 160 x 10⁻⁴ ohm ( rounding off to two syg fig )

current = 12.5

potential diff = 12.5 x 155.15 x 10⁻⁴ V

=  0 .1939 V

= .19 V

required electric field = potential diff / length of wire

= .1939 / 16 x 10⁻²

= 1.2 N / C

Answer 2

Answer:

a) maximum electric field in this filament is 1.26 V/m

b) resistance with that field is 0.016 ohms (using two significant figures).

c) 0.2016 V is the maximum potential drop over the full length of the filament.

Explanation:

a)

E max is at higher temperature

p(120) = p(20)(1+α(T-T₀))

p(120) = 7.61x10⁻⁸

E = pJ = p(i/A)

  ⇒7.61×10⁻⁸ (13 / 3.14×0.0005²) = 1.26 V/m

b)

R = pL/A ⇒ pL/πr²

R = [7.61×10⁻⁸×0.16] ÷ [3.14×0.5²×10⁻⁶] = 0.0155 ohm

c)

V = Ed

⇒ 1.26 × 0.16 = 0.2016 V