Chapter 2
Exercise 27 (orig.33):
Hint:
I believe that the load should be characterized as 10 ohm, rather than 10 W. Although loads are often characterized by the power they dissipate when connected to a 115 ac source, it is unusual to see it characterized this way for an arbitrary voltage. Also, notice that part b) asks what happens when the load is changed to 15 ohm.

Ans:
a. V.s = 50.8V, r.s = 0.37 ohm
b. With 15 ohm load, terminal voltage would be 49.6 V to three significant digits
C. short circuit current would be 137 A.

Exercise 46:
Hint:
So how much resistance can be tolerated if we wish to drop a voltage of only 5 V at a current of 7.1 A?
I do not find an adequate AWG guide in our textbook. The first one I found using a web search is here.

Ans:
280 ft of AWG 14 has a resistance of about 0.7 ohm.

Exercise 58:
Hint:
Use the terminal voltage across 10 ohm to find the current flowing through the battery. Then use the difference between the emf and the terminal voltage to find the internal resistance responsible for the voltage drop at this current.

Ans:
a) The initial internal resistance is about 0.044 ohm.
b) One year later the internal resistance has risen to 61 ohm.