chp27_explained

chp27_explained - Week 4 Chapter 27 Problem 10 The filament...

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Week 4 Chapter 27 Problem 10 The filament of the light bulb in example 27-6 has a diameter of 0.050 mm. What is the current density in the filament? Compare with the current density in a 12-gauge wire (diameter 0.21 cm) supplying current to the light bulb. First you need to know that the current in example 27-6 (pg. 702) is 0.833 Amps. Also observe that the radii are half the diameters, so they are 0.025 mm and 0.105 cm respectively. From there you only need the formula for current density: J=I/A, where A is the cross sectional area. ( ) ( ) 2 5 2 2 2 8 2 3 / 10 41 . 2 10 105 . 0 833 . 0 / 10 24 . 4 10 025 . 0 833 . 0 m A J A J m A J m A J A I J wire wire filament filament ! = ! = ! = ! = = " " # So the current density in the wire is a lot smaller. The larger current density in the filament causes it to heat up and when it gets hot enough, it glows, providing light with which to see. Problem 36 You have a cylindrical piece of material 2.4 cm long and 2.0 mm in diameter. When you attach a 9-V battery to the ends of the piece, a current of 2.6 mA results. Which material from Table 27-1 do you have? Table 27-1 is on page 690. To solve this equation, you’ll need to find the resistance, and
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chp27_explained - Week 4 Chapter 27 Problem 10 The filament...

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