A cylindrical tungsten filament 195 cm long with a diameter

A cylindrical tungsten filament 19.5 cm long with a diameter of 1.15 mm is to be used in a machine for which the temperature will range from room temperature (20°C) up to 120°C. It will carry a current of 14.0 A at all temperatures (consult tables [a] and [b] below)

V/m

(b) What will be its resistance with that field?
?
(c) What will be the maximum potential drop over the full length of the filament?

[a Resistivities at Room Temperature (20 °C) Substance (·m) Substance 012 . m ) Conductors Metals Semiconductors 1.47 × 10-8 1.72 × 10-8 2.44 × 10-8 2.75 × 10-8 5.25 × 10-8 20 × 10-8 22 × 10-8 95 × 10-8 44 × 10-8 49 × 10-8 100× 10-8 3.5×105 0.60 2300 Silver Copper Gold Aluminunm Tungsten Steel Lead Mercury Manganin (Cu 84%, Mn 12%, Ni 4%) Constantan (Cu 60%, Ni 40%) Nichrome Pure carbon (graphite) Pure germanium Pure silicon Insulators 5 × 1014 1010-1014 >1013 101-1015 75 1016 1015 1013 108-1011 Amber Glass Lucite ica Quartz (fused) Sulfur Teflon Wood Alloys [b] Temperature Coefficients of Resistivity (Approximate Values Near Room Temperature) Material Material 0.0039 0.0020 Aluminunm Brass Carbon (graphite) Constantan Copper Iron Lead Manganin Mercury Nichrome Silver Tungsten 0.0043 0.00000 0.00088 0.0004 0.0038 0.0045 0.0005 0.00001 0.00393 0.0050

Solution

A cylindrical tungsten filament 19.5 cm long with a diameter of 1.15 mm is to be used in a machine for which the temperature will range from room temperature (20°C) up to 120°C. It will carry a current of 14.0 A at all temperatures (consult tables [a] and [b] below)

resistivity of Tungsten = 5.25e-8 ohm-m

length of the filament l= 19.5cm

dia of the wire = 1.15mm

area of cross section A =(1.15e-3/2)² = 1.04e-6 sq.m

Resistance of the filament R = l/A = 5.25e-8 *0.195/1.04e-6

                                          = 9.84e-3 ohms at 20⁰C

current = 14.0A

temperature co-efficient = 0.0045/⁰C

resistivity and resistance increases with temperature, operating temperature is from 20 to 120⁰C and it will have maximum at 120⁰C

R(T) = R₀[1+(T-T₀)]

R(120) = 9.84e-3[1+0.0045(120-20)] = 14.3e-3 ohms

current is constant and the E will be maximum with maximum resistance

at 120C

potential drop across the filament length = IR = 14*14.3e-3 = 0.2 V

maximum E = V/l = 0.2/0.195 = 1.027 V/m

resistance = 14.3e-3 ohms

maximum potential drop across the filament = -0.2V