PHYS208 Fundamentals of Physics II

Answer for Ch. 28, 43P

3.33 V

Answer for Ch. 28, 44P


parallel configuration:  i = emf / (R + r/2)
OR:  i(x) = 1 / (1 + x/2)
series configuration:  i = emf / (r + R/2)
OR:  i(x) = 1 / (x + 1/2)

where the dimensionless involves a dimensionless internal resistance x = r / R and a dimensionless current which is the actual current in ratio to the current that would be present if r = 0. This formulation facilitates the computer generated graph below (or whatever algebraic operations which you might wish to carry out in finding the optimal configuration).


When the load resistance is larger than internal resistance (R > r, lighter load, x < 1), use the series configuration.

When load resistance is smaller than internal resistance (R < r, heavier load, x > 1), use the parallel configuration.


Graph created with Origin for Windows

Answer for Ch. 28, 48P

I labeled the currents as i1 (through emf1 and R3), i2 (through emf2 and R2), and i3 (through R1). Why did HRW not switch the labels of R1 and R3 -- it would have made a lot more sense!

i1 421 mA to right
i2 158 mA to right
i3 263 mA down
of R1 346 mW
of R2 50 mW
of R3 709 mW
total dissipated:  1.10 W
total provided by emf1 1.26 W
total absorbed by emf2 0.16 W

Note that since the current through emf2 is going in the opposite direction to its polarity, it is being re-charged (receiving power from emf1).

Answer for Ch. 28, 57P

a) 12.5 V

b) 50 A

Last updated Feb. 21, 1998.
Copyright George Watson, Univ. of Delaware, 1997.