### Problem StatementA 1.1 cm length of tungsten filament in a small lightbulb has a resistance of 0.029 Ω. Find its diameter. (The resistivity is \( 5.6 \times 10^{-8} \, \Omega \cdot \text{m} \)). Answer in units of mm.### Detailed Explanation and CalculationsTo find the diameter of the tungsten filament, we can use the formula for electrical resistance in a cylindrical conductor:\[ R = \rho \frac{L}{A} \]Where:- \( R \) is the resistance (0.029 Ω)- \( \rho \) is the resistivity (\( 5.6 \times 10^{-8} \, \Omega \cdot \text{m} \))- \( L \) is the length (1.1 cm = 0.011 m)- \( A \) is the cross-sectional area of the filamentGiven that the filament is cylindrical, the cross-sectional area \( A \) can be expressed in terms of the diameter \( d \):\[ A = \pi \left(\frac{d}{2}\right)^2 \]Substituting \( A \) in the resistance formula and solving for the diameter \( d \):\[ R = \rho \frac{L}{\pi \left(\frac{d}{2}\right)^2} \]Rearranging the equation to solve for \( d \):\[ d = 2 \sqrt{\left(\frac{\rho L}{R \pi}\right)} \]Now substitute the given values:\[ d = 2 \sqrt{\left(\frac{(5.6 \times 10^{-8}) \times (0.011)}{0.029 \times \pi}\right)} \]Simplify the calculation inside the square root:\[ d = 2 \sqrt{\left(\frac{6.16 \times 10^{-10}}{0.029 \times 3.1416}\right)} \]\[ d = 2 \sqrt{\left(\frac{6.16 \times 10^{-10}}{0.0911064}\right)} \]\[ d = 2 \sqrt{6.76 \times 10^{-9}} \]\[ d \approx 2 \times 2.6 \times 10^{-5} \]\[

Length of filament, l=1.1 cm=0.011 m Resistance, R=0.029 ohm

Answered: A 1.1 cm length of tungsten filament in…

A 1.1 cm length of tungsten filament in a small lightbulb has a resistance of 0.029 N. Find its diameter. (The resistivity is 5.6 × 10-8 · m). Answer in units of mm.

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter20: Electric Current, Resistance, And Ohm's Law

Section: Chapter Questions

Problem 7PE: (a) A defibrillator sends a 6.00-A current through the chest of a patient by applying a 10,000-V...

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