PHYS208 Fundamentals of Physics II

Answer for Ch. 28, 62P

a) current:  70.9 mA
  voltage:  4.70 V
  resistance:  66.3 ohm
b) current:  55.2 mA
  voltage:  4.86 V
  resistance:  88.0 ohm

In case b) there was 71.4 mA through the battery, slightly different from case a), as expected; however, the current splits between the branch with the resistor and ammeter and the branch with the voltmeter. The voltage of 4.86 V across the 88 ohms of the resistor being measured in series with the ammeter yields a current of (4.86 V)/(88 ohm) = 55.2 mA.


"http://www.physics.udel.edu/~watson/phys208/exercises/answers/28-62P.html"
Last updated Sept. 16, 1997.
Copyright George Watson, Univ. of Delaware, 1997.