Hints 

  Ch. 27

 12 34 39 42 66

 
 
Summary equations for Chapter 27:

current = I = nqvdA where A = crossection area of wire, q = elemental charge, n = charge density and vd = drift speed.  Remember that electrons travel opposite the positive charges by convention!

J = I/A = current density = nqvd

This how resistivity changes with temperature.

This is the rate that charges in the battery gain potential energy. The voltage change is across the battery. 
This is the magnitude of the rate that charges lose potential energy in the resistor. This is also equal to the rate of heat generation in the resistor. The current  and voltage change are for the resistor.
12. See example 27.1. All of the methods that you need for this problem are there.
34.
 Use :
 

R0 = 1.00 ohm corresponding to a temperature of 20oC. Calculate the change in temperature and solve for R. 
39. The power supplied by the heater is equal to the rate that heat is added to the water:

On the right hand side of the equation, we divide the heat gained by the water by the time = 10 minutes. Remember to convert to seconds.
Find R.
42. Find R from the power P using the relationship:

(a):

(b): Recalculate R:

Then recalculate the power P using the new value of R.
66. (a):


The area A is equal to the difference between the area of a circle of radius rb and the area of a circle of radius ra.
(b): Evaluate using the numbers given.
(c) and (d): Repeat parts (a) and (b) of the problem assuming the current flows radially outward. In this case, you must use calculus. Consider a differential shell of inner radius r, thickness dr and length equal to the length of the cylinder. The area of the inner surface is:

Thus,

This is the differential resistance of the shell. Integrate this between the limits of the inner and outer radii and you will get the answer in the back of the book. Enjoy your studying!
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