The conductivity of tungsten at room temperature, = 1.8 107 (A/m2)/(V/m), is sig

Question

The conductivity of tungsten at room temperature, = 1.8 107 (A/m2)/(V/m), is significantly smaller than that of copper. At the very high temperature of a glowing light-bulb filament (nearly 3000 degrees Kelvin!), the conductivity of tungsten is 18 times smaller than it is at room temperature. The tungsten filament of one of your round bulbs has a radius of about 0.023 mm. Calculate the magnitude of the electric field required to drive 0.39 amperes of current through the glowing bulb.

The answer is 235 V/m. Could you explain (solve) this?

Explanation / Answer

conductivity of tungsten at room temperature, = 1.8 10^7 (A/m2)/(V/m)

the conductivity of tungsten at 3000 K = 1.8*10^7/18 = 0.1*10^7 = 10^6 (A/m2)/(V/m)

area of tungusten filament = pi*r*r = 3.14 * 0.023 * 0.023 *10^-6 = 0.00166106 *10^-6 m^2

conductance of tungstem filament = area * conductivity = 0.00166106 *10^-6 *10^6 = 0.00166106 mho-m

resistance/metre = 1/conductance = 1/0.00166106 ohm/m

voltage/ metre = electric field = resistance/m * current = 0.39/0.00167 = 234.78 ~ 235 V/m

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