cylindrical tungsten filament 19.0 cm long with a diameter of 1.35 mm is to be u
ID: 2268981 • Letter: C
Question
cylindrical tungsten filament 19.0 cm long with a diameter of 1.35 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 15.0 A at all temperatures (consult tables [a] and [b] below).
[a] Resistivities at Room Temperature (20 °C) Substance (Q.rm ) Substance (Q.rm ) Conductors Metals Semiconductors 1.47 × 10-8 1.72 × 108 2.44 × 10-8 2.75 × 10-8 5.25 × 10-8 20 × 10-8 22 × 10-8 95 x 108 44 × 10-8 49 × 10-8 100× 10-8 3.5×105 0.60 2300 Silver Copper Gold Aluminum 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 Amber Glass Lucite Mica Quartz (fused) Sulfur Teflon Wood 101-1015 75 1016 1015 1013 108-1011 Alloys [b] Temperature Coefficients of Resistivity (Approximate Values Near Room Temperature) Material Aluminunm Material 0.0039 0.0020 Lead Manganin Mercury Nichrome Silver Tungsten 0.0043 0.00000 0.00088 0.0004 0.0038 0.0045 Brass Carbon (graphite) 0.0005 Constantan 0.00001 0.00393 0.0050 Copper IronExplanation / Answer
at 20 deg C,
R0 = rho L / A = (5.25 x 10^-8)(0.19)/(pi(1.35 x 10^-3 / 2)^2)
R0 = 6.97 x 10^-3 ohm
R = R0 [1 + alpha delatT ]
R = (6.97 x 10^-3)[1 + (0.0045)(120 - 20)]
R = 0.01 ohm
(A) V_max= I R = 15 x 0.01 = 0.152 Volt
E_max = V_max / L = 0.80 V/m
(B) R = 0.01 ohm
(C) V = 0.152 Volt
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