Answer:
Explanation:
Resistance of the tungsten wire
R = resistivity x length / cross sectional area
= [tex]\frac{5.25\times10^{-8}\times16\times10^{-2}}{3.14\times(.5\times10^{-3})^2}[/tex]
= 107 x 10⁻⁴ ohm
Resistance at 120 degree can be obtained from the following formula
[tex]R_t = R_0( 1 + \alpha\times t )[/tex]
[tex]R_t = 107\times10^{-4}( 1 + .0045\times 100)[/tex]
= 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:
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
