You have a 80-watt, 120-volt light bulb. The coiled tungsten filament of the bul

Question

You have a 80-watt, 120-volt light bulb. The coiled tungsten filament of the bulb is 2.0 m long, and it heats to 3000° C during normal operation.
The extremely thin filament is 2.0 meters long and coiled so that it fits inside the bulb. The resistivity of tungsten is 5.28 ×10-8 O·m at room temperature (20° C), and its temperature coefficient of resistivity is 0.0045 C°-1. Assume the bulb is in a circuit that supplies a 120 V potential difference across the light bulb. Treat the light bulb as though it were a direct current (DC) device.

(a) Find the cross sectional area of the filament.

(b) Assuming that the filament has a circular cross section, find the diameter of the filament.

(c) Find the resistance of the filament at the operating temperature of 30000C and the

(d) current flowing through it at that temperature.

(e) Would the current be different at different temperatures, as the filament heats up to its operating temperature? Explain why.

Explanation / Answer

let the thermal resistivity be

temperature coefficient of resistivity be

we know that =room temp (1 + (temp))..........................................................1

  3000 = room temp(1 + temp) =5.28*10^(-8) (1+ .0045(3000-20) )=76.08 * 10^(-8)

resiistance =l/Area ....................................................................................2

power =V^2/R gives R=120*120/80 =180

a) from 2 we get 180 = 76.08 * 10^(-8) 2/A

hence Area=1.183* 10^(-8) m^2

b) d^2/4 =area hence diameter d=1.22 * 10^(-4) m

c) resistance at 3000 degrees temp =180

d)current power =i^2 *R hence i=.666 A

e)from 1.we can say that resistivity depends on operating temperature hence resistance of the filament varies and hence current also varies

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