## PHYS208 Fundamentals of Physics II

### Answer for Ch. 28, 43P

3.33 V

### Answer for Ch. 28, 44P

a)

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).

b)

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 i_{1} (through emf_{1} and R_{3}),
i_{2} (through emf_{2} and R_{2}), and i_{3} (through R_{1}).
*Why did **HRW* not switch the labels of R_{1} and R_{3} -- it would have made a lot more sense!

**Currents:** |

i_{1}: |
421 mA |
to right |

i_{2}: |
158 mA |
to right |

i_{3}: |
263 mA |
down |

**Powers:** |

of R_{1: } |
346 mW |

of R_{2}: |
50 mW |

of R_{3}: |
709 mW |

total dissipated: |
1.10 W |

total provided by emf_{1}: |
1.26 W |

total absorbed by emf_{2}: |
0.16 W |

Note that since the current through emf_{2} is going in the opposite direction
to its polarity, it is being re-charged (receiving power from emf_{1}).

### Answer for Ch. 28, 57P

a) 12.5 V
b) 50 A

"http://www.physics.udel.edu/~watson/phys208/exercises/answers/28-43P.html"

Last updated Feb. 21, 1998.

Copyright George Watson, Univ. of Delaware, 1997.