PHYS208 Fundamentals of Physics II

Quiz 2 - dc Circuits

simple circuit

Two lightbulbs are connected in series with the battery shown.

  1. What is the voltage difference between a and b? (That is, what is the terminal voltage of the battery?)

  2. A third identical lightbulb is now connected in parallel to one of the other lightbulbs. By what percentage does the power dissipated by the internal resistance change?


Solution:

  1. The equivalent resistance of the circuit above is (2.0 + 7.0 + 7.0) ohm = 16.0 ohm. Accordingly the current through the battery is (9 V) / (16 ohm) = 0.56 A. The voltage drop across the internal resistance is (0.56 A)(2.0 ohm) = 1.1 V; the terminal voltage of the battery is thus 9.0 V - 1.1 V = 7.9 V.

  2. The additional lightbulb is added as shown below:

    Third resistor added.

    The two 7.0 ohm resistors in parallel are equivalent to a 3.5 ohm resistance; the total equivalent resistance "seen" by the emf is thus (2.0 + 3.5 + 7.0) ohm = 12.5 ohm. Accordingly the current through the battery is now (9 V) / (12.5 ohm) = 0.72 A.

    Since the power dissipated by a resistor is proportional to the square of the current, the power ratio of the internal resistance will be (i2/i1)2 = (0.72/0.56)2 = 1.65. The power dissipated by the internal resistance increases by 65% when the third lightbulb is added.


"http://www.physics.udel.edu/~watson/phys208/quiz2soln.html"
Last updated Feb. 27, 1998.
Copyright George Watson, Univ. of Delaware, 1998.