Two lightbulbs are connected in series with the battery shown.

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

- 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.
- The additional lightbulb is added as shown below:
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 (

*i*_{2}/*i*_{1})^{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.