Question

In: Physics

An incandescent light bulb uses a coiled filament of tungsten that is 580 mm long with...

An incandescent light bulb uses a coiled filament of tungsten that is 580 mm long with a diameter of 46.0 μm. At 20.0∘C tungsten has a resistivity of 5.25×10−8Ω⋅m. Its temperature coefficient of resistivity is 0.0045 (C∘)−1, and this remains accurate even at high temperatures. The temperature of the filament increases linearly with current, from 20∘C when no current flows to 2520∘C at 1.00 A of current. What is the resistance of the light bulb at 20∘C? What is the current through the light bulb when the potential difference across its terminals is 120 V? (Hint: First determine the temperature as a function of the current; then use this to determine the resistance as a function of the current. Substitute this result into the equation V=IR and solve for the current I.) Express your answer to two significant figures and include the appropriate units. What is the resistance when the potential is 120 V? Express your answer to two significant figures and include the appropriate units. How much energy does the light bulb dissipate in 1 min when 120 V is supplied across its terminals? Express your answer in kilojoules to two significant figures.

Solutions

Expert Solution

Temperature as a Function of Current

It is given that temperature increases linearly with current.

Given at zero current, the temperature is 20oC

And 1.00A current, the temperature is 2520oC.

The temperature as a function of current is

Resistance as a Function of Current

The diameter of the wire is

The radius of the wire

The cross sectional area of the wire is

The resistivity as a function of temperature is

where is the resistivity of the wire at temperature To.

The resistance of a wire of resistivity , length L, and cross-sectional area A is


Using (1)

Using To=20oC

Using Ohm's law, the current V=IR

Substituting , , , , we get

This is quadratic equation in current. The roots of quadratic equation are

The current in the wire is

The other root is negative (-0.81), so we neglected it.

Using Ohm's law, the resistance is

The power dissipated is

The energy dissipated in time t is


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